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How does RNA polymerase II CTD bind to the RNA modification proteins if the tail is flexible?

How does RNA polymerase II CTD bind to the RNA modification proteins if the tail is flexible?


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The tail of RNA polymerase II is flexible, not folded into a fixed structure , but does each repeat have more "rigid" structure (i.e. fold into a structure that has less rotation freedom inside a repeat)? If not, then how can the proteins bind some repeat with high specificity? (you can't make lock and key with soft material)


How does RNA polymerase II CTD bind to the RNA modification proteins if the tail is flexible? - Biology

Initiation is the first step of eukaryotic transcription and requires RNAP and several transcription factors to proceed.

Learning Objectives

Describe how transcription is initiated and proceeds along the DNA strand

Key Takeaways

Key Points

  • Eukaryotic transcription is carried out in the nucleus of the cell and proceeds in three sequential stages: initiation, elongation, and termination.
  • Eukaryotes require transcription factors to first bind to the promoter region and then help recruit the appropriate polymerase.
  • RNA Polymerase II is the polymerase responsible for transcribing mRNA.

Key Terms

  • repressor: any protein that binds to DNA and thus regulates the expression of genes by decreasing the rate of transcription
  • activator: any chemical or agent which regulates one or more genes by increasing the rate of transcription
  • polymerase: any of various enzymes that catalyze the formation of polymers of DNA or RNA using an existing strand of DNA or RNA as a template

Steps in Eukaryotic Transcription

Eukaryotic transcription is carried out in the nucleus of the cell by one of three RNA polymerases, depending on the RNA being transcribed, and proceeds in three sequential stages:

Initiation of Transcription in Eukaryotes

Unlike the prokaryotic RNA polymerase that can bind to a DNA template on its own, eukaryotes require several other proteins, called transcription factors, to first bind to the promoter region and then help recruit the appropriate polymerase. The completed assembly of transcription factors and RNA polymerase bind to the promoter, forming a transcription pre-initiation complex (PIC).

The most-extensively studied core promoter element in eukaryotes is a short DNA sequence known as a TATA box, found 25-30 base pairs upstream from the start site of transcription. Only about 10-15% of mammalian genes contain TATA boxes, while the rest contain other core promoter elements, but the mechanisms by which transcription is initiated at promoters with TATA boxes is well characterized.

The TATA box, as a core promoter element, is the binding site for a transcription factor known as TATA-binding protein (TBP), which is itself a subunit of another transcription factor: Transcription Factor II D (TFIID). After TFIID binds to the TATA box via the TBP, five more transcription factors and RNA polymerase combine around the TATA box in a series of stages to form a pre-initiation complex. One transcription factor, Transcription Factor II H (TFIIH), is involved in separating opposing strands of double-stranded DNA to provide the RNA Polymerase access to a single-stranded DNA template. However, only a low, or basal, rate of transcription is driven by the pre-initiation complex alone. Other proteins known as activators and repressors, along with any associated coactivators or corepressors, are responsible for modulating transcription rate. Activator proteins increase the transcription rate, and repressor proteins decrease the transcription rate.

Eukaryotic Transcription Initiation: A generalized promoter of a gene transcribed by RNA polymerase II is shown. Transcription factors recognize the promoter, RNA polymerase II then binds and forms the transcription initiation complex.

The Three Eukaryotic RNA Polymerases (RNAPs)

The features of eukaryotic mRNA synthesis are markedly more complex those of prokaryotes. Instead of a single polymerase comprising five subunits, the eukaryotes have three polymerases that are each made up of 10 subunits or more. Each eukaryotic polymerase also requires a distinct set of transcription factors to bring it to the DNA template.

RNA polymerase I is located in the nucleolus, a specialized nuclear substructure in which ribosomal RNA (rRNA) is transcribed, processed, and assembled into ribosomes. The rRNA molecules are considered structural RNAs because they have a cellular role but are not translated into protein. The rRNAs are components of the ribosome and are essential to the process of translation. RNA polymerase I synthesizes all of the rRNAs except for the 5S rRNA molecule.

RNA polymerase II is located in the nucleus and synthesizes all protein-coding nuclear pre-mRNAs. Eukaryotic pre-mRNAs undergo extensive processing after transcription, but before translation. RNA polymerase II is responsible for transcribing the overwhelming majority of eukaryotic genes, including all of the protein-encoding genes which ultimately are translated into proteins and genes for several types of regulatory RNAs, including microRNAs (miRNAs) and long-coding RNAs (lncRNAs).

RNA polymerase III is also located in the nucleus. This polymerase transcribes a variety of structural RNAs that includes the 5S pre-rRNA, transfer pre-RNAs (pre-tRNAs), and small nuclear pre-RNAs. The tRNAs have a critical role in translation: they serve as the adaptor molecules between the mRNA template and the growing polypeptide chain. Small nuclear RNAs have a variety of functions, including “splicing” pre-mRNAs and regulating transcription factors. Not all miRNAs are transcribed by RNA Polymerase II, RNA Polymerase III transcribes some of them.

Modeling transcription: This interactive models the process of DNA transcription in a eukaryotic cell.


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Results

Efficient and rapid depletion of GTFs in vivo

To systematically study the role of GTFs and Pol II on transcription, we constructed anchor-away strains for subunits of TBP (Spt15), TFIIA (Toa1 and Toa2), TFIIB (Sua7), TFIIE (Tfa1 and Tfa2), TFIIF (Tfg1), TFIIH (Ssl1 and Ssl2), and Pol II (Rpb1). In the absence of rapamycin, growth of these strains is comparable to that of an untagged parental strain (Figure 1A), indicating that the fusion of the FRB domain to the targeted factors does not significantly affect their function. In contrast, when these proteins are removed from the nucleus by treatment with rapamycin, these strains fail to grow (Figure 1A), as expected from the essential roles of GTFs. Binding of the targeted GTFs at active promoters (PMA1 and CCW12) is reduced to background or near-background levels after rapamycin treatment for 1 hr (Figure 1B), indicating that depletion of GTFs is highly effective.

Conditional depletion of GTFs causes severe growth and transcriptional defects.

(A) Growth of the indicated anchor-away cells (5-fold serial dilutions) in the presence or absence of rapamycin. (B) Occupancy of the indicated FRB-tagged GTFs at the PMA1 and CCW12 promoters in the corresponding strains and an untagged control strain in the presence of absence of rapamycin. (C) Pol II occupancy at the CCW12 and RPS14B coding regions in the indicated strains grown in the presence or absence of rapamycin. Error bars represent the standard error of at least three independent experiments.

All GTFs are required for pol II transcription in vivo

To examine the effect of depleting individual GTFs on Pol II transcription, we first measured Pol II occupancy at the coding regions of several well-expressed genes. While the addition of rapamycin has minimal effects on transcription in an untagged parental control strain, Pol II occupancy at coding regions of all genes tested is reduced to very low levels upon depletion of any GTF (Figure 1C and Figure 1—figure supplement 1). To extend these results to genome scale, we performed Pol II ChIP-seq analysis on the same samples to which a known amount of S. pombe chromatin was added as an internal control for immunoprecipitation and data normalization. In all cases, depletion of any GTF drastically reduced transcription to near-background levels as determined by metagene (Figure 2A) or individual gene (Figure 2B) analyses. In contrast, as will be discussed later, depletion of Taf1 results in a modest decrease in transcription. Furthermore, upon TBP depletion, TBP and Pol II occupancies decrease in a kinetically similar manner (Figure 2C), indicating that loss of TBP results in an immediate cessation of transcriptional initiation.

All GTFs are generally required for ongoing Pol II transcription.

(A) Mean Pol II occupancy averaged over 453 well-transcribed genes (metagene analysis) in strains depleted (+rap) for the indicated factor and in the parental (WT) strain (±rap). Partial reduction is observed only for the TAF1-depleted strain. (B) Pol II occupancy at individual genes (the same set of 453 genes ordered from top to bottom by expression level in WT) in strains depleted for the indicated factor. For each gene, the log2 change in Pol II occupancy after depletion is indicated according to the red/blue scale. (C) TBP and Pol II occupancies at the indicated promoters in the TBP-depletion strain at various times after rapamycin treatment. Error bars represent the standard error of at least three independent experiments.

In the above experiments, genes are expressed at steady-state levels prior to depletion of the GTF. To address the effect of GTF depletion on inducible transcription, we first depleted cells of an individual GTF and then analyzed the rapid transcriptional activation response to heat shock. In accord with drastic transcriptional effects described on non-inducible genes, induction of HSP12 (Figure 3A) and other heat shock genes (Figure 3—figure supplement 1A) is very strongly decreased, although not completely eliminated, for all GTFs (but not Taf1).

All GTFs are required for transcriptional induction upon heat shock.

(A) Mean Pol II occupancy at the HSP12 coding region (ORF) and promoter in strains depleted (or not) for the indicated factor and then induced for 15 min by shifting to 39°C. (B) FRB-tagged GTF:Pol II occupancy ratio at the induced HSP12 promoter in cells pretreated or not with rapamycin to deplete the indicated factors.

The residual levels of transcription observed upon GTF depletion could reflect GTF-independent transcription or, more simply, incomplete depletion of the GTF. In this regard, the anchor-away system works somewhat less efficiently in stress conditions (Petrenko et al., 2017). As discussed and shown elsewhere for TBP (Petrenko et al., 2017), incomplete depletion of an essential GTF will reduce (but not eliminate) transcription and its occupancy at the promoter, but it will not affect the nature of the PIC and hence the GTF:Pol II occupancy ratio. Indeed, the GTF:Pol II occupancy ratios for all cases of GTF depletion are comparable to that observed prior to depletion (Figure 3B and Figure 3—figure supplement 1B). Thus, the residual levels of transcription are due to incomplete depletion, indicating that all GTFs are required for Pol II transcription in vivo.

Depletion of any GTF prevents association of TBP with the promoter, suggesting that partial PICs do not exist at appreciable levels in vivo

As GTFs are components of the PIC, it is expected that their essential role in Pol II transcription reflects their requirement for a functional PIC. However, as partial PICs lacking various GTFs are stable in vitro (Buratowski et al., 1989), it is unclear whether such partial PICs are stable in vivo. To address this issue, we examined the effect of GTF depletion on PIC levels by measuring TBP occupancy at the promoter. For all GTFs, TBP occupancy levels at both continually expressed (Figure 4A and Figure 4—figure supplement 1A) and heat-shock inducible (Figure 4B and Figure 4—figure supplement 1B) genes are drastically reduced upon depletion. In addition, the TBP:Pol II and TBP:GTF occupancy ratios for all cases of GTF depletion are comparable to that observed prior to depletion and in the parental strain (Figure 4C and Figure 4—figure supplement 1C). Thus, each GTF is required for a stable PIC in vivo. As TBP is the only GTF that can independently and stably bind to DNA (Buratowski et al., 1989), these results indicate that, unlike the situation in vitro, partial PICs containing GTF subsets are very unstable in vivo.

All GTFs are required for TBP occupancy, and hence PIC stability/formation.

(A) TBP occupancy at the CCW12 and TEF1 promoters in strains depleted (+rap) or not (-rap) for the indicated factor. (B) TBP occupancy at the HSP12 and HSP82 promoters in strains depleted (+rap) or not (-rap) for the indicated factor and subject to a heat shock. (C) TBP:Pol II and TBP:FRB-tagged GTF occupancy ratios at the induced HSP12 promoter in cells pretreated or not with rapamycin to deplete the indicated factors. (D) Pol II and TFIIB occupancies at the indicated promoters previously reported to have partial PICs (Zanton and Pugh, 2006) under normal (blue) and heat shock induction (red).

Our results are in apparent contrast to a previous report claiming the existence of partial PICs in response to a mild heat shock (37°C) based on altered GTF:GTF occupancy ratios (Zanton and Pugh, 2006). However, in that report, the altered occupancy ratios for the vast majority of the genes with apparent partial PICs are very modest. To address this more directly, we measured the TFIIB and Pol II occupancies at several genes including those analyzed by Zanton and Pugh (2006). We did not observe increased TFIIB:Pol II occupancy ratios in response to heat shock (39°C) at any of these genes (Figure 4D). Thus, our results suggest that partial PICs do not exist at appreciable levels in vivo, although their existence cannot be completely excluded.

The PIC is not stable in the cells depleted of uracil

In vitro, the PIC is extremely stable in the absence of nucleotide triphosphates, and indeed is defined by its ability to initiate transcription upon addition of these precursors. We attempted to mimic this situation in vivo by analyzing PIC levels and transcription under conditions in which ura3 mutant cells were starved of uracil. As removal of uracil from the medium does not immediately eliminate intracellular uracil (because ura3 cells require and hence contain uracil for growth prior to the removal), we examined TBP and Pol II occupancy levels at various times after removal of uracil (Figure 5 and Figure 5—figure supplement 1A–C). As uracil depletion causes metabolic mayhem (Brauer et al., 2008), we examined genes that are typically inhibited (Figure 5A and Figure 5—figure supplement 1B), induced (Figure 5B and Figure 5—figure supplement 1C), or unaffected (Figure 5C) by metabolic stress.

A general loss of PICs in cells depleted for uracil.

(A) Pol II (ORF and promoter) and TBP occupancies at promoters (PMA1 and RPL28) of genes inhibited by metabolic stress in cells depleted for uracil or leucine for various times (color scale). (B) Similar analysis for HSP82, a gene induced by various metabolic stresses. (C) Similar analysis for genes TDH2, YRA1, and ACT1, genes unchanged upon metabolic stress. (D) Average TBP/Pol II ratio at all tested promoters at various times after depletion of leucine (blue) or uracil (red). See Figure 5—figure supplement 1D for individual tested promoters.

Upon removal of uracil from the medium, cells grow at near-normal rates for about 2 hr (using up the intracellular uracil) and then show decreased growth with cessation at about 4 hr (Figure 5—figure supplement 1A). With the exception of heat shock genes, TBP and Pol II occupancies at all promoters tested decrease over time, and PIC and transcription (Pol II occupancy at coding regions) levels are extremely low after 4 hr. Interestingly, the average TBP:Pol II occupancy ratio near the promoter decreases at intermediate times (15–60 min) of uracil depletion (Figure 5D and Figure 5—figure supplement 1D), as would be expected from Pol II buildup caused by reduced elongation due to decreased (but non-zero) UTP levels. At the heat shock genes, TBP and Pol II occupancies sharply increase at early times, presumably due to the stress response to uracil depletion, but they drop to virtually undetectable levels after 2–4 hr. In addition, TBP occupancy at Pol III and Pol I promoters is also extremely low upon uracil depletion.

The drastic drop in PIC levels upon uracil removal is unlikely to be due to growth arrest per se, because depletion of Kin28 (Wong et al., 2014) or Taf1 (see below) only modestly reduces TBP and Pol II occupancy, even though cell growth is blocked. To address whether PIC instability is due to metabolic limitation per se, we performed a similar experiment depleting cells of leucine (the strain is also a leu2 auxotroph). Leucine-depleted cells show a similar growth pattern as uracil-depleted cells with growth cessation at 4 hr. In contrast to the uracil-depleted cells, leucine-depleted cells do not show a drop in TBP and Pol II occupancy at all genes, although growth-inhibited genes are affected (Figure 5A–C, Figure 5—figure supplement 1). In addition, leucine-depleted cells do not show early induction of PIC levels at heat shock genes, presumably because they do not undergo the same stress response, nor do they show increased Pol II:TBP ratios at intermediate times of leucine depletion (Figure 5D and Figure 5—figure supplement 1D). The differences between uracil- and leucine-depleted cells are consistent with transcriptional profiling experiments in auxotrophic cells grown in chemostats at various concentrations of the required metabolites (Brauer et al., 2008). Thus, our results suggest that the drastic and general drop in PIC levels upon uracil removal is due to the absence of UTP precursors.

Depletion of FACT strongly reduces PIC levels

As depletion of uracil, and consequently UTP, blocks transcriptional elongation, we considered the possibility that other factors involved in the elongation process might also affect PIC levels. We therefore analyzed PIC levels and transcription upon depletion of FACT, a histone chaperone complex that travels with elongating Pol II in vivo (Mason and Struhl, 2003) and is important for elongation through chromatin templates in vitro (Orphanides et al., 1998). FACT does not directly associate with promoters, but rather associates (directly or indirectly) with the elongation machinery after Pol II escapes from the promoter (Mason and Struhl, 2003). In accord with previous results (Pathak et al., 2018), depletion of FACT strongly reduces Pol II occupancy throughout the genome (Figure 6). More interestingly, TFIIB occupancy at essentially all promoters is reduced to a comparable extent, indicating that FACT is important for PIC formation or stability. As FACT is not a component of the PIC, these observations suggest that some aspect of FACT-dependent elongation is important for PIC levels. As such, these observations are consistent with the drastic decrease in PIC levels upon uracil depletion, but the mechanism by which FACT affects PIC levels is unknown.

Depletion of Spt16 subunit of FACT reduces transcription and PIC formation.

Mean Pol II occupancy and TFIIB occupancy averaged over 453 well-transcribed genes and promoters before and after Spt16 depletion.

Taf1 is selectively important, but not required for transcription

The role of TFIID is controversial with respect to whether it is selectively (Moqtaderi et al., 1996 Kuras et al., 2000 Li et al., 2000 Basehoar et al., 2004) or generally (Warfield et al., 2017) required for transcription. Genome-scale analysis of Pol II occupancy in TAF1-depleted cells grown in SC medium reveals an overall

2 fold decreased in transcription (Figure 2A,B). Furthermore, transcription is significantly more affected at TATA-lacking vs. TATA-containing genes as observed by metagene (Figure 7A) or individual gene analyses (Figure 7B).

Taf1 depletion selectively affects TFIID-dependent genes.

(A) Mean Pol II occupancy averaged over 453 well-transcribed ‘SAGA’ or ‘TFIID’ genes in parental or Taf1-depleted strains. (B) Pol II (C) TBP, TFIIB, and TFIIA, and (D) Taf1 occupancies at ‘SAGA’ (PMA1 and CCW12) or ‘TFIID’ (ACT1, TEF1, RPL28, RPS14B) coding regions in the Taf1-depletion strain treated or untreated with rapamycin. (E) TBP:TAF1 occupancy ratios in the Taf1-depletion strain treated or untreated with rapamycin. (F) Log2 change in Pol II occupancy (measured at +200–400 from the TSS) for the indicated gene classes upon Taf1 depletion. ‘Active’ genes are the top 10% of transcribed genes, broken up into ‘SAGA’ and ‘TFIID’-dependent categories. Ribosomes include only the ribosomal genes from the TFIID-dependent category. ‘TFIID active, no ribosomes’ are the remainder of the top 10% transcribed genes in the TFIID-dependent category after ribosomal genes have been removed. ‘All’ genes are the entire set of

5000 s. cerevisiae genes, broken up into ‘SAGA’ and ‘TFIID’-dependent categories.

To exclude the possibility that these modest transcriptional effects are due to incomplete TAF1 depletion, we analyzed TBP, TFIIA, TFIIB (Figure 7C) and Taf1 (Figure 7D) occupancies at several promoters. Importantly, when Taf1 is depleted and no longer detected, TBP, TFIIA, and TFIIB strongly associate with promoters, but the effects depend on whether the promoter is ‘SAGA- or TFIID-dependent’. At ‘SAGA-dependent’ promoters (PMA1, CCW12), Taf1 levels are low, and Taf1 depletion has marginal effects on GTF occupancy. As a consequence, the TBP:Taf1 occupancy ratio at these promoters are far above those occurring in non-depleted cells (Figure 7E). In contrast, at ‘TFIID-dependent’ promoters with relatively high levels of Taf1 occupancy (RPL28 and RPS14B), GTF occupancy decreases upon Taf1 depletion, and the TBP:Taf1 occupancy ratio is only slightly affected (Figure 7E). In contrast and as shown above (Figure 2C), depletion of any individual GTF does not significantly alter the GTF:TBP or GTF:Pol II ratio. As expected, the TBP:TFIIA and TBP:TFIIB occupancy rations are similar in non-depleted and TAF1-depleted cells (Figure 7—figure supplement 1A). Thus, TAF1 behaves differently than all GTFs and hence is selectively important but not required for Pol II transcription.

Our results are in apparent contrast to those of a recent study that claimed that Taf-depletion (via auxin-induced degradation) resulted in ‘similar transcription decreases for genes in the Taf-depleted, Taf-enriched, TATA-containing, and TATA-less gene classes’ (Warfield et al., 2017). However, the conclusion of this other study was based on analysis of all

5000 yeast genes, most of which are transcribed at low or even background levels that prevent accurate measurements. We performed a similar analysis of all

5000 genes using the Taf1-depletion data presented here and confirmed a very modest difference between SAGA and TFIID-dependent promoters (Figure 7F see ‘all SAGA’ and ‘all TFIID’). Our analysis involves Pol II occupancy between +200 to+400 from the transcription start site, which avoids complications between transcriptional initiation and elongation. Although (Warfield et al., 2017) analyzed the region between 0 and +100, analysis of our Taf1-depletion data in this region also shows that the distinction between ‘all SAGA’ and ‘all TFIID’ genes is virtually non-existent (Figure 7—figure supplement 1B).

When we restrict the analysis to more actively transcribed genes, for which Pol II occupancy levels are well above the background (i.e. the top 453 as shown in Figure 2A,B and Figure 7A), the distinction between ‘active’ SAGA and TFIID genes is clear (Figure 7F). In addition, among TFIID-affected genes, ribosomal protein genes are even more strongly affected by Taf1 depletion (Figure 7F). In accord with this result, analysis of the same set of actively transcribed genes upon auxin-mediated depletion of Taf11 or Taf13- (Warfield et al., 2017) shows a clear distinction between SAGA and TFIID genes (Figure 7—figure supplement 2) and comparable to the Taf1-depletion data presented here. Thus, our experimental results are completely consistent with those of Warfield et al. (2017), and they clearly demonstrate that TFIID-specific Tafs are selectively important for transcription of TATA-less genes.


Carrying on

As we heard at the outset, the (protein) enzyme that transcribes DNA into RNA is RNA polymerase. 12 The enzyme certainly does not work alone, however, and its task is by no means automatic. To begin with, its critical interactions with various elements of the pre-initiation complex help determine whether and exactly where transcription will begin. Then, after those “decisions” have been made, RNA polymerase moves along the double helix transcribing the sequence of genetic “letters” into the complementary sequence of an RNA.

Throughout this productive journey, which is called elongation, the RNA polymerase still keeps good and necessary company. Certain molecular co-activators modify it during its transit of a gene’s sequence, and these modifications not only enable transcription elongation to begin, but also provide binding sites for yet other proteins that will cooperate throughout the transcription journey. The collective interaction here, as in the activities discussed above, can vary in many details from one context to another — all in order to contribute to a meaningful narrative that could hardly repeat itself in exactly the same way.

The table below offers some perspective on the number and variety of protein factors influencing elongation. You need not puzzle over the details. A quick browse of this incomplete listing (as of 2013) will give you at least an inkling of the kind of intricate complexity the cell must organize in order to carry out transcriptional elongation. As always, it is important to realize that each of the factors listed here enters the picture out of its own world of regulation. At the molecular level of the organism we are always looking at ever-widening circles of interaction, without limit. It’s just a question of how narrowly we choose to focus our attention — and how much of the context we consequently block from view.

Table 14.1. Don’t Read This Table! (Just feel it.) Some factors regulating RNA polymerase elongation (copied from Kwak and Lis 2013).

Class Factor name Function Related factors and notes
GAGA factor GAF Generates nucleosome-free region and promoter structure for pausing NURF
General Transcription Factors TFIID Generates promoter structure for pausing
TFIIF Increases elongation rate Near promoters
TFIIS Rescues backtracked RNA polymerase II RNA polymerase III
Pausing factors NELF Stabilizes RNA polymerase II pausing
DSIF Stabilizes RNA polymerase II pausing and facilitates elongation
Positive elongation factor P-TEFb Phosphorylates NELF, DSIF, and RNA polymerase II CTD for pause release
Processivity factors Elongin Increases elongation rate
ELL Increases elongation rate AFF4
SEC Contains P-TEFb and ELL Mediator, PAF
Activator c-Myc Directly recruits P-TEFb
NF-&kappaB Directly recruits P-TEFb
Coactivator BRD4 Recruits P-TEFb
Mediator Recruits P-TEFb via SEC
Capping machinery CE Facilitates P-TEFb recruitment, counters NELF/DSIF
RNMT Methylates RNA 5’ end to complete capping Myc
Premature termination factors DCP2 Decaps nascent RNA for XRN2 digestion Dcp1a/Edc3
Microprocessor Cleaves hairpin structure for XRN2 digestion Tat, Senx
XRN2 Torpedoes RNA polymerase II with RNA 5’-3’ exonucleation
TTF2 Releases RNA polymerase II from DNA
Gdown1 GDOWN1 Antitermination and stabilizes paused RNA polymerase II TFIIF, Mediator
Histone chaperone FACT H2A-H2B eviction and chaperone Tracks with RNA polymerase II
NAP1 H2A-H2B chaperone RSC, CHD
SPT6 H3-H4 chaperone Tracks with RNA polymerase II
ASF1 H3-H4 chaperone H3K56ac
Chromatin remodeler RSC SWI/SNF remodeling in gene body H3K14ac
CHD1 Maintains gene body nucleosome organization FACT, DSIF
NURF ISWI remodeling at promoter GAGA factor
Poly(ADP-ribose) polymerase PARP Transcription independent nucleosome loss Tip60
Polymerase-associated factor complex PAF Loading dock for elongation factors SEC, FACT
Histone tail modifiers MOF Acetylates H4K16 and recruits Brd4 H3S10ph, 14-3-3
TIP60 Acetylates H2AK5 and activates PARP
Elongator Acetylates H3 and facilitates nucleosomal elongation Also in cytoplasm
Rpd3C (Eaf3) Deacetylates and inhibits spurious initiation in gene body H3K36me3
SET1 Methylates H3K4 MLL/COMPASS
SET2 Methylates H3K36 and regulates acetylation-deacetylation cycle Rpd3C
PIM1 Phosphorylates H3S10 and recruits 14-3-3 and MOF
RNF20/40 Monoubiquitinates H2BK123 and facilitates nucleosomal DNA unwrapping UbcH6, PAF

I will mention here only one aspect of this cooperation of multiple factors. Transcription is an essentially rhythmical performance, with various sorts of pauses along the way. (Again, dynamic sculpture, or dance!) One pause of great significance occurs after RNA polymerase has just begun transcribing DNA but before it has fully separated from the pre-initiation complex. The factors that influence whether transcription will continue at this point — or remain paused for an extended period — play a large role in the regulation of gene expression.

But once that first pause is ended, the elongation journey often continues to be marked by a series of further, generally briefer pauses. These have to do, at least in part, with the need to disengage DNA from its intimate mutual embrace with certain constituents of chromatin (histone complexes, about which we will learn more below). The polymerase has various assistants to aid in this disengagement, which may involve disassembly of the protein complexes. Typical of chromatin in general, these complexes are rich repositories of regulatory information, so they will need to be reassembled behind the transcribing complex, and the remarkably nuanced meanings embodied in their composition and structure will somehow have to be preserved, reestablished, or modified.

So the rhythm of pauses depends, at least in part, on the polymerase’s helper molecules and on the positioning of certain protein complexes along the double helix, both of which will vary from one gene to another and even from one time to another. All this, and not just the so-called genetic code as such, shapes the functional significance of the DNA sequence within its chromosomal context. As we will see shortly, different versions of a protein may be produced, depending on the timing of the pauses.


Chromosomes without the airbrushing

Since Francis Crick and James Watson’s elucidation of the structure of DNA in 1953, biologists have been “in denial”, according to Nature columnist, Philip Ball. “That beautiful double helix, with its genetic information written into the spiral staircase of paired nucleic-acid bases, offers such an elegant picture of the chemical principles of life and inheritance that everyone fell for it”.

The spiraling image Ball refers to has become a dominant icon of the modern era, channeling the imagination along the alluring lines of its own geometric perfection. Yet its ubiquity and influence is matched only by its falsehood. For “when we come face to face with DNA in the cell”, writes Ball, “it’s like meeting a movie star whose airbrushed publicity photos don’t look at all like the real thing. You would barely recognise Crick and Watson’s perfectly-formed molecule in the tangled, twisted and bent spaghetti that is stuffed inside the nuclei of our cells”. 17

“Stuffed” probably isn’t the right word, given the functional sophistication of the ever-shifting sculptural forms whose activity in the nucleus “speaks” with such effective eloquence and on so many different levels. But let us see what we can of this “twisted and bent spaghetti”.

First look. The functioning of chromatin, the substance of chromosomes, is a matter, not of the serene, immaterial logic of an immortal code, but of a life-sustaining narrative performance. Think, for example, of the drama of mitotic cell division: chromosomes, previously replicated, become condensed some fifty-fold, are organized on the mitotic spindle through a delicate balance of tensions, and then are properly apportioned into the two daughter cells. Think, that is, what it might take to accomplish this.

But at every phase of the cell cycle the drama continues, and likewise through every stage of cell differentiation during an organism’s development. Similarly during illness, or high levels of emotional stress, or extreme exercise, or wound healing, or the passage from day to night and from night to day — in all the circumstances of life our chromosomes are gesturing their functions. That is, their movements and spatial configurations are integral to their functional roles within specific contexts. They “speak” gesturally, not merely through a passive and unvarying, computer-like configuration of hardware “logic gates”.

The performance we are looking at includes profound transformations of the double helix. For example, millions of “letters” in the DNA sequence are chemically altered, temporarily or for the longer term, by a process known as DNA methylation. By changing the local physical properties of the double helix, this modification “is observed to either inhibit or facilitate [DNA] strand separation, depending on methylation level and sequence context”. 18 This has a direct effect on gene expression, since strand separation is essential for the work of the enzyme that transcribes DNA.

Moreover, DNA methylation is considered a primary instrument of cellular memory, yet even where methylation levels are more or less stably sustained, the constancy is achieved through continual exchange of methyl groups. The average half-life of a particular methyl group on any given letter is measured in hours. 19 So even cellular “memory”, it would appear, is a function of dynamics rather than inert structures. In general, when it comes to chromosomes and DNA, performance, not static structure, is the fundamental reality.

The coming and going of DNA-associated proteins such as transcription factors is even more dramatic. Here the binding (to DNA or chromatin) and subsequent release can involve the cooperative or antagonistic activity of many molecules, and the temporal and spatial pattern of this activity is often what proves decisive for gene expression. Referring to “time-dependent events” as the important “‘fourth dimension’ of transcriptional regulation”, Bryan Turner, a geneticist at the University of Birmingham in the UK, notes that “studies of site-specific [transcription] factor binding in individual living cells have revealed residence times on chromatin measured in seconds”. 20

All this makes for an extraordinarily complex and dynamic picture, particularly when you consider this: looking at a single gene (Gal1 in baker’s yeast) at a particular moment, researchers recently identified more than fifty proteins residing at its locus. 21

In sum, there is “a surprisingly high mobility of proteins in the nucleus”. 22 Oscillations, rhythms, critical rates, continual exchange of subunits — these are now known to be common features of many molecular complexes or “structures” once thought to be more or less static. The architecture of the cell nucleus sometimes appears to be compounded more of standing waves than of rigid structures. Things get done by means of a startlingly indeterminate fluidity, not the huffing and puffing of imaginary “molecular machines”.

“It was amusing”, writes cell biologist Thoru Pederson of the University of Massachusetts Medical School, “to recall the incredulity expressed by some that &hellip chromosomes, relatively giant structures, are moving” — and doing so in the normal course of their business, even when the cell is not dividing. And yet, he continues, what else should we have expected? 23

Is it such a surprise, after all, to find that the processes of life are &hellip living processes? The idea that the essence of living things lay in a crystalline genetic code never had anything going for it apart from the mind’s preference for neat logical necessity (abstract and self-contained “explanation”) over observation and the reading of gestural narrative.

Second look. Focusing our vision more narrowly on one element of chromatin, we can home in on the millions of nucleosomes. I began my description of gene expression by mentioning the packing of chromosomes into the cell nucleus, and (in Box 2) said a little about the role of the nucleosome in this packing and in gene regulation generally. The nucleosomal core particle we should now note, “is much more flexible than the crystal structure [which is the basis for images like Figures 2 and 3] might lead us to believe”, and our current understanding of it “does not lend itself to simplifying generalisations”. 24

In particular, nucleosomes are partially or wholly disassembled during gene transcription, and then reassembled after the transcribing enzyme (RNA polymerase) has passed by. However, the rate of this assembly and disassembly differs with different genes, and has been proposed as one of the factors regulating gene expression.

Beyond that, individual histones of the nucleosome (there are eight in a fully constituted core particle) can come and go at an almost alarming rate — with an average exchange rate of just a few minutes for many nucleosomes, especially those associated with active genes. And the histones exchanged in this way can be different histones — histone variants — with each variant exerting its own sort of influence on gene expression and chromatin dynamics.

Like DNA methylation, histone variants are associated with the formation of cellular memory. (The members of a cell lineage, beginning with a stem cell, must proceed through a series of cell divisions on the way, for example, to a pancreatic islet cell destination. Each cell along the path needs to “remember” the trajectory along which it has been traveling — a trajectory leading toward islet cells rather than, say, muscle cells.) The puzzling discordance between the instability of histone variants, on the one hand, and their role in memory formation, on the other, is, according to University of Georgia molecular biologist Richard Meagher, “extreme”. 25

At least it is if we imagine cellular memory to be like a text safely stowed away somewhere. But the organism seems to have a much more complex and dynamic notion of memory. It is well to keep in mind that, whatever we may mean by a cell’s “remembering” where it is coming from, it must also be continually departing from the ways of its ancestors in order to reach where it is getting to. The story is through and through one of activity and transformation. The character of that activity within its larger active context, and not any fixed structure, may be the more essential bearer of identity and meaning.

One of the many aspects of the relation between the nucleosomal core particle and the encircling DNA has to do with the electrical attraction or repulsion between each local region of the core particle and the associated DNA. Also significant is which part of the DNA double helix at any point faces outward from the histones and which part faces inward. These factors directly affect the accessibility of any given DNA sequence to transcription factors and other regulatory molecules.

The nucleosome, we can fairly say, is a ceaselessly transforming matrix and organizational hub whose structure and pattern of activity is never exactly duplicated anywhere in the genome, nor ever sustained unchanged from one minute to the next. It is where the infinitely ramified interface between the larger cell and its DNA comes to its most focal expression. And that expression turns out to be livingly nuanced activity, dynamic beyond what anyone imagined.

Third look. We will burrow still further down into chromatin. I present below a modestly enlarged view of Figure 2, showing a ribbon diagram of a nucleosome — a histone core particle with the DNA double helix (in purple) wrapped around it a couple of times. You will note a number of squiggly “pig’s tails” extending outward from the core histones. These are the thin, flexible, and mobile histone tails. There are hundreds of distinct chemical modifications of these tails (referred to as post-translational modifications, or PTMs), and the countless resulting patterns within any given nucleosome or group of nucleosomes are intimately bound up with the expression of genes. In fact, there is virtually nothing relating to gene regulation, DNA replication, chromatin structure and dynamics, or the overall functional organization of the nucleus that is not correlated in one way or another with patterns of histone tail modifications.

Figure 4. An enlarged view of Figure 2, showing a ribbon drawing of a nucleosome.

Learning about these tails, we are — albeit across a considerable conceptual divide — easily reminded of both the “sensory” functions of insect antennae and the “motor” functions of limbs. On one hand, the tails are receivers of signals coming from all quarters and provide a context where the integrated significance of the signals can be “read off” (to use the standard phrase) by the innumerable gene-regulatory proteins that are sensitive to them. In this way, histone tail modifications, or combinations of them, “recruit” (again the standard phrase) various proteins that either restructure chromatin in one way or another, or more directly regulate the expression of particular genes.

There is in fact a galaxy of proteins that interact with single modifications, or with groups of them, or with the asymmetrically modified tails of a histone pair, or with a histone modification in proximity to a site of DNA methylation. Every such protein acts out of its own world of biochemical genesis, modification, and protein regulation, and together these proteins tell a great part of the story of gene regulation. Unfortunately, I can say almost nothing about them here.

Apart from their “sensory” function, the mobile tails can insinuate themselves into one of the grooves of the double helix, thereby loosening the DNA from the nucleosomal core particle (and making it more available for transcription), or else binding it more tightly. In both cases, one way this is accomplished is by altering the electrical balance between histone and DNA.

But a tail can also interact with the globular core of its own or a different histone, and can even find an affinity with an altogether different nucleosome, thereby having a potentially huge effect on the packing of the local chromatin. This, again, can make genes either more or less accessible for transcription and various forms of regulation.

Perhaps you can now see why the members of one research team, writing about histone tail modifications, find themselves reflecting upon

the incredibly intricate nature of the chromatin landscape and resultant interactions. The biological consequences of [interactions between histone tail modifications and regulatory proteins] are highly context dependent, relying on the combinatorial readout of the spatially and temporally fluctuating local [chromatin] environment and leading to a highly fine-tuned [regulation] of particular genomic sites. 26

Magnifying our view one last time, we will home in a single histone tail modification. I choose the one called ubiquitination simply because its gene regulatory roles do not seem quite as extensive as those performed by some of the other tail modifications — or, at least, its roles have not been as extensively documented. This makes their description here a little more manageable.

Monoubiquitination is the “attachment” (a poor word) of a single ubiquitin chemical group to a lysine amino acid of a protein. In the case of the histone tails, this can be done at more than one lysine, but we will look only at the monoubiquitination of lysine 120 on the tail of the histone known as H2B, all of which can be referred to as H2BK120ub1 (abbreviated here as H2Bub1).

So what is the significance of this modification at a single histone tail location? Here’s one summary:

H2Bub1 takes part in almost every molecular process associated with chromatin biology. H2Bub1 has been shown to regulate transcription initiation and elongation, DNA damage response and repair, DNA replication, nucleosome positioning, RNA processing and export, chromatin segregation and maintenance of chromatin boundaries. Given the large number of molecular processes regulated by H2Bub1, it is not surprising that H2Bub1 plays a vital role in some of the most fundamental biological processes that occur within multicellular organisms. [Loss of an enzyme responsible for ubiquitination] results in very early embryonic lethality. Furthermore, aberrant H2Bub1 levels can affect cell cycle progression, apoptosis [“programmed cell death”], stem cell differentiation, development, viral infection outcome and tumorigenesis. 27

Of course, H2Bub1 does nothing “in general” specific results are always context-dependent. For example, blocking this modification in a particular human cell line was found to upregulate some genes, downregulate others, and leave a great many unchanged. Under some circumstances, H2Bub1 is particularly needed for the transcription of relatively long genes. And the modification also plays an important role in histone “crosstalk”, helping to regulate other crucial modifications within the same or on different histones.

A search for “effector” molecules that, singly or cooperatively, associate and interact with (“read”) the H2Bub1 modification led to the identification of more than ninety proteins, many with known functions in gene regulation consistent with those of H2Bub1. This points us to what could be our “fifth look”, analyzing one or more of those proteins, the modifications they undergo, and the larger regulatory world in which they are caught up. But there would be no end of this, whereas there most definitely is an end to readers’ patience.

It remains to mention only that, with ubiquitination, as with so many other molecular biological investigations, researchers are vexed by “a need to establish causality more unequivocally” 28 — a need that never seems to be fully satisfied as understanding grows. The search for unambiguous causes is a fruitless one. 29 The kind of causes being looked for don’t exist in organisms.


How does RNA polymerase II CTD bind to the RNA modification proteins if the tail is flexible? - Biology

1) Garrod hypothesized that ʺinborn errors of metabolismʺ such as alkaptonuria occur
because
A) genes dictate the production of specific enzymes, and affected individuals have
genetic defects that cause them to lack certain enzymes.
B) enzymes are made of DNA, and affected individuals lack DNA polymerase.
C) many metabolic enzymes use DNA as a cofactor, and affected individuals have
mutations that prevent their enzymes from interacting efficiently with DNA.
D) certain metabolic reactions are carried out by ribozymes, and affected individuals lack
key splicing factors.
E) metabolic enzymes require vitamin cofactors, and affected individuals have
significant nutritional deficiencies.

According to Beadle and Tatumʹs hypothesis, how many genes are necessary for this
pathway?
A) 0
B) 1
C) 2
D) 3
E) It cannot be determined from the pathway.

A mutation results in a defective enzyme A. Which of the following would be a
consequence of that mutation?
A) an accumulation of A and no production of B and C
B) an accumulation of A and B and no production of C
C) an accumulation of B and no production of A and C
D) an accumulation of B and C and no production of A
E) an accumulation of C and no production of A and B

If A, B, and C are all required for growth, a strain that is mutant for the gene encoding
enzyme A would be able to grow on which of the following media?
A) minimal medium
B) minimal medium supplemented with nutrient ʺAʺ only
C) minimal medium supplemented with nutrient ʺBʺ only
D) minimal medium supplemented with nutrient ʺCʺ only
E) minimal medium supplemented with nutrients ʺAʺ and ʺCʺ

If A, B, and C are all required for growth, a strain mutant for the gene encoding enzyme B
would be capable of growing on which of the following media?
A) minimal medium
B) minimal medium supplemented with ʺAʺ only
C) minimal medium supplemented with ʺBʺ only
D) minimal medium supplemented with ʺCʺ only
E) minimal medium supplemented with nutrients ʺAʺ and ʺBʺ

The nitrogenous base adenine is found in all members of which group?
A) proteins, triglycerides, and testosterone
B) proteins, ATP, and DNA
C) ATP, RNA, and DNA
D) alpha glucose, ATP, and DNA
E) proteins, carbohydrates, and ATP

Using RNA as a template for protein synthesis instead of translating proteins directly from
the DNA is advantageous for the cell because
A) RNA is much more stable than DNA.
B) RNA acts as an expendable copy of the genetic material.
C) only one mRNA molecule can be transcribed from a single gene, lowering the
potential rate of gene expression.
D) tRNA, rRNA and others are not transcribed.
E) mRNA molecules are subject to mutation but DNA is not

If proteins were composed of only 12 different kinds of amino acids, what would be the
smallest possible codon size in a genetic system with four different nucleotides?
A) 1
B) 2
C) 3
D) 4
E) 12

The enzyme polynucleotide phosphorylase randomly assembles nucleotides into a
polynucleotide polymer. You add polynucleotide phosphorylase to a solution of adenosine
triphosphate and guanosine triphosphate. How many artificial mRNA 3 nucleotide codons
would be possible?
A) 3
B) 4
C) 8
D) 16
E) 64

A particular triplet of bases in the template strand of DNA is 5ʹ AGT 3ʹ. The corresponding
codon for the mRNA transcribed is
A) 3ʹ UCA 5ʹ.
B) 3ʹ UGA 5ʹ.
C) 5ʹ TCA 3ʹ.
D) 3ʹACU 5ʹ.
E) either UCA or TCA, depending on wobble in the first base.

A possible sequence of nucleotides in the template strand of DNA that would code for the
polypeptide sequence phe-leu-ile-val would be
A) 5ʹ TTG-CTA-CAG-TAG 3ʹ.
B) 3ʹ AAC-GAC-GUC-AUA 5ʹ.
C) 5ʹ AUG-CTG-CAG-TAT 3ʹ.
D) 3ʹ AAA-AAT-ATA-ACA 5ʹ.
E) 3ʹ AAA-GAA-TAA-CAA 5ʹ.

What amino acid sequence will be generated, based on the following mRNA codon
sequence?
5ʹ AUG-UCU-UCG-UUA-UCC-UUG 3ʹ
A) met-arg-glu-arg-glu-arg
B) met-glu-arg-arg-gln-leu
C) met-ser-leu-ser-leu-ser
D) met-ser-ser-leu-ser-leu
E) met-leu-phe-arg-glu-glu

A peptide has the sequence NH2-phe-pro-lys-gly-phe-pro-COOH. Which of the
following sequences in the coding strand of the DNA could code for this peptide?
A) 3ʹ UUU-CCC-AAA-GGG-UUU-CCC
B) 3ʹ AUG-AAA-GGG-TTT-CCC-AAA-GGG
C) 5ʹ TTT-CCC-AAA-GGG-TTT-CCC
D) 5ʹ GGG-AAA-TTT-AAA-CCC-ACT-GGG
E) 5ʹ ACT-TAC-CAT-AAA-CAT-TAC-UGA

What is the sequence of a peptide based on the following mRNA sequence?
5ʹ . . . UUUUCUUAUUGUCUU 3ʹ
A) leu-cys-tyr-ser-phe
B) cyc-phe-tyr-cys-leu
C) phe-leu-ile-met-val
D) leu-pro-asp-lys-gly
E) phe-ser-tyr-cys-leu

The genetic code is essentially the same for all organisms. From this, one can logically
assume all of the following except
A) a gene from an organism could theoretically be expressed by any other organism.
B) all organisms have a common ancestor.
C) DNA was the first genetic material.
D) the same codons in different organisms usually translate into the same amino acids.
E) different organisms have the same number of different types of amino acids.

The ʺuniversalʺ genetic code is now known to have exceptions. Evidence for this could be
found if which of the following is true?
A) If UGA, usually a stop codon, is found to code for an amino acid such as tryptophan
(usually coded for by UGG only).
B) If one stop codon, such as UGA, is found to have a different effect on translation than
another stop codon, such as UAA.
C) If prokaryotic organisms are able to translate a eukaryotic mRNA and produce the
same polypeptide.
D) If several codons are found to translate to the same amino acid, such as serine.
E) If a single mRNA molecule is found to translate to more than one polypeptide when
there are two or more AUG sites.

Which of the following nucleotide triplets best represents a codon?
A) a triplet separated spatially from other triplets
B) a triplet that has no corresponding amino acid
C) a triplet at the opposite end of tRNA from the attachment site of the amino acid
D) a triplet in the same reading frame as an upstream AUG
E) a sequence in tRNA at the 3ʹ end

Which of the following is true for both prokaryotic and eukaryotic gene expression?
A) After transcription, a 3ʹ poly-A tail and a 5ʹ cap are added to mRNA.
B) Translation of mRNA can begin before transcription is complete.
C) RNA polymerase binds to the promoter region to begin transcription.
D) mRNA is synthesized in the 3ʹ → 5ʹ direction.
E) The mRNA transcript is the exact complement of the gene from which it was copied.

In which of the following actions does RNA polymerase differ from DNA polymerase?
A) RNA polymerase uses RNA as a template, and DNA polymerase uses a DNA
template.
B) RNA polymerase binds to single-stranded DNA, and DNA polymerase binds to
double-stranded DNA.
C) RNA polymerase is much more accurate than DNA polymerase.
D) RNA polymerase can initiate RNA synthesis, but DNA polymerase requires a primer
to initiate DNA synthesis.
E) RNA polymerase does not need to separate the two strands of DNA in order to
synthesize an RNA copy, whereas DNA polymerase must unwind the double helix
before it can replicate the DNA.

Which of the following statements best describes the termination of transcription in
prokaryotes?
A) RNA polymerase transcribes through the polyadenylation signal, causing proteins to
associate with the transcript and cut it free from the polymerase.
B) RNA polymerase transcribes through the terminator sequence, causing the
polymerase to fall off the DNA and release the transcript.
C) RNA polymerase transcribes through an intron, and the snRNPs cause the
polymerase to let go of the transcript.
D) Once transcription has initiated, RNA polymerase transcribes until it reaches the end
of the chromosome.
E) RNA polymerase transcribes through a stop codon, causing the polymerase to stop
advancing through the gene and release the mRNA.

RNA polymerase moves in which direction along the DNA?
A) 3ʹ → 5ʹ along the template strand
B) 3ʹ → 5ʹ along the coding (sense) strand
C) 5ʹ → 3ʹ along the template strand
D) 3ʹ → 5ʹ along the coding strand
E) 5ʹ → 3ʹ along the double-stranded DNA

RNA polymerase in a prokaryote is composed of several subunits. Most of these subunits
are the same for the transcription of any gene, but one, known as sigma, varies
considerably. Which of the following is the most probable advantage for the organism of
such sigma switching?
A) It might allow the transcription process to vary from one cell to another.
B) It might allow the polymerase to recognize different promoters under certain
environmental conditions.
C) It could allow the polymerase to react differently to each stop codon.
D) It could allow ribosomal subunits to assemble at faster rates.
E) It could alter the rate of translation and of exon splicing.

Which of these is the function of a poly (A) signal sequence?
A) It adds the poly (A) tail to the 3ʹ end of the mRNA.
B) It codes for a sequence in eukaryotic transcripts that signals enzymatic cleavage

10
—35 nucleotides away.
C) It allows the 3ʹ end of the mRNA to attach to the ribosome.
D) It is a sequence that codes for the hydrolysis of the RNA polymerase.
E) It adds a 7-methylguanosine cap to the 3ʹ end of the mRNA.

In eukaryotes there are several different types of RNA polymerase. Which type is involved
in transcription of mRNA for a globin protein?
A) ligase
B) RNA polymerase I
C) RNA polymerase II
D) RNA polymerase III
E) primase

Transcription in eukaryotes requires which of the following in addition to RNA
polymerase?
A) the protein product of the promoter
B) start and stop codons
C) ribosomes and tRNA
D) several transcription factors (TFs)
E) aminoacyl synthetase

A part of the promoter, called the TATA box, is said to be highly conserved in evolution.
Which might this illustrate?
A) The sequence evolves very rapidly.
B) The sequence does not mutate.
C) Any mutation in the sequence is selected against.
D) The sequence is found in many but not all promoters.
E) The sequence is transcribed at the start of every gene.

The TATA sequence is found only several nucleotides away from the start site of
transcription. This most probably relates to which of the following?
A) the number of hydrogen bonds between A and T in DNA
B) the triplet nature of the codon
C) the ability of this sequence to bind to the start site
D) the supercoiling of the DNA near the start site
E) the 3-dimensional shape of a DNA molecule

Which of the following help(s) to stabilize mRNA by inhibiting its degradation?
A) TATA box
B) spliceosomes
C) 5ʹ cap and poly (A) tail
D) introns
E) RNA polymerase

What is a ribozyme?
A) an enzyme that uses RNA as a substrate
B) an RNA with enzymatic activity
C) an enzyme that catalyzes the association between the large and small ribosomal
subunits
D) an enzyme that synthesizes RNA as part of the transcription process
E) an enzyme that synthesizes RNA primers during DNA replication

What are the coding segments of a stretch of eukaryotic DNA called?
A) introns
B) exons
C) codons
D) replicons
E) transposons

A transcription unit that is 8,000 nucleotides long may use 1,200 nucleotides to make a
protein consisting of approximately 400 amino acids. This is best explained by the fact that
A) many noncoding stretches of nucleotides are present in mRNA.
B) there is redundancy and ambiguity in the genetic code.
C) many nucleotides are needed to code for each amino acid.
D) nucleotides break off and are lost during the transcription process.
E) there are termination exons near the beginning of mRNA.

Once transcribed, eukaryotic mRNA typically undergoes substantial alteration that
includes
A) union with ribosomes.
B) fusion into circular forms known as plasmids.
C) linkage to histone molecules.
D) excision of introns.
E) fusion with other newly transcribed mRNA.

Introns are significant to biological evolution because
A) their presence allows exons to be shuffled.
B) they protect the mRNA from degeneration.
C) they are translated into essential amino acids.
D) they maintain the genetic code by preventing incorrect DNA base pairings.
E) they correct enzymatic alterations of DNA bases.

A mutation in which of the following parts of a gene is likely to be most damaging to a cell?
A) intron
B) exon
C) 5ʹ UTR
D) 3ʹ UTR
E) All would be equally damaging.

Which of the following is (are) true of snRNPs?
A) They are made up of both DNA and RNA.
B) They bind to splice sites at each end of the exon.
C) They join together to form a large structure called the spliceosome.
D) They act only in the cytosol.
E) They attach introns to exons in the correct order.

During splicing, which molecular component of the spliceosome catalyzes the excision
reaction?
A) protein
B) DNA
C) RNA
D) lipid
E) sugar

Alternative RNA splicing
A) is a mechanism for increasing the rate of transcription.
B) can allow the production of proteins of different sizes from a single mRNA.
C) can allow the production of similar proteins from different RNAs.
D) increases the rate of transcription.
E) is due to the presence or absence of particular snRNPs

In the structural organization of many eukaryotic genes, individual exons may be related to
which of the following?
A) the sequence of the intron that immediately precedes each exon
B) the number of polypeptides making up the functional protein
C) the various domains of the polypeptide product
D) the number of restriction enzyme cutting sites
E) the number of start sites for transcription

Each eukaryotic mRNA, even after post-transcriptional modification, includes 5ʹ and 3ʹ
UTRs. Which are these?
A) the cap and tail at each end of the mRNA
B) the untranslated regions at either end of the coding sequence
C) the U attachment sites for the tRNAs
D) the U translation sites that signal the beginning of translation
E) the U — A pairs that are found in high frequency at the ends

In an experimental situation, a student researcher inserts an mRNA molecule into a
eukaryotic cell after he has removed its 5ʹ cap and poly(A) tail. Which of the following
would you expect him to find?
A) The mRNA could not exit the nucleus to be translated.
B) The cell recognizes the absence of the tail and polyadenylates the mRNA.
C) The molecule is digested by restriction enzymes in the nucleus.
D) The molecule is digested by exonucleases since it is no longer protected at the 5ʹ end.
E) The molecule attaches to a ribosome and is translated, but more slowly.

A particular triplet of bases in the coding sequence of DNA is AAA. The anticodon on the
tRNA that binds the mRNA codon is
A) TTT.
B) UUA.
C) UUU.
D) AAA.
E) either UAA or TAA, depending on first base wobble.

Accuracy in the translation of mRNA into the primary structure of a polypeptide depends
on specificity in the
A) binding of ribosomes to mRNA.
B) shape of the A and P sites of ribosomes.
C) bonding of the anticodon to the codon.
D) attachment of amino acids to tRNAs.
E) both C and D

A part of an mRNA molecule with the following sequence is being read by a ribosome: 5ʹ CCG-ACG 3ʹ (mRNA). The following charged transfer RNA molecules (with their anticodons shown in the 3ʹ to 5ʹ direction). The dipeptide that will form will be
A) cysteine-alanine.
B) proline-threonine.
C) glycine-cysteine.
D) alanine-alanine.
E) threonine-glycine.

What type of bonding is responsible for maintaining the shape of the tRNA molecule?
A) covalent bonding between sulfur atoms
B) ionic bonding between phosphates
C) hydrogen bonding between base pairs
D) van der Waals interactions between hydrogen atoms
E) peptide bonding between amino acids

Figure 17.4 represents tRNA that recognizes and binds a particular amino acid (in this
instance, phenylalanine). Which codon on the mRNA strand codes for this amino acid?
A) UGG
B) GUG
C) GUA
D) UUC
E) CAU

The tRNA shown in Figure 17.4 has its 3ʹ end projecting beyond its 5ʹ end. What will occur
at this 3ʹ end?
A) The codon and anticodon complement one another.
B) The amino acid binds covalently.
C) The excess nucleotides (ACCA) will be cleaved off at the ribosome.
D) The small and large subunits of the ribosome will attach to it.
E) The 5ʹ cap of the mRNA will become covalently bound.

A mutant bacterial cell has a defective aminoacyl synthetase that attaches a lysine to tRNAs
with the anticodon AAA instead of a phenylalanine. The consequence of this for the cell
will be that
A) none of the proteins in the cell will contain phenylalanine.
B) proteins in the cell will include lysine instead of phenylalanine at amino acid
positions specified by the codon UUU.
C) the cell will compensate for the defect by attaching phenylalanine to tRNAs with
lysine-specifying anticodons.
D) the ribosome will skip a codon every time a UUU is encountered.
E) None of the above will occur the cell will recognize the error and destroy the tRNA.

There are 61 mRNA codons that specify an amino acid, but only 45 tRNAs. This is best
explained by the fact that
A) some tRNAs have anticodons that recognize four or more different codons.
B) the rules for base pairing between the third base of a codon and tRNA are flexible.
C) many codons are never used, so the tRNAs that recognize them are dispensable.
D) the DNA codes for all 61 tRNAs but some are then destroyed.
E) competitive exclusion forces some tRNAs to be destroyed by nucleases.


MCAT Biology Review

What process is shown by the following:

What process is shown by the following: Also called a nucleoside phosphate, it is composed of a five carbon sugar, a nitrogen base, and an inorganic phosphate

At the 3rd and 5th carbons designated the 3' and 5' positions
  • DNA polymerase can add free nucleotides only to the 3' end of the template strand. This results in elongation of the newly forming strand in a 5'-3' direction.
  • Semi-discontinuous
  • Leading
  • Lagging
  • approximately 18-22 hours
  • New DNA is synthesized and partitioned equally so that division can then occur.

  • Interphase (G1 [also called gap phase], S [also called the synthesis phase], G2 [also called the growth phase]) and
  • Mitosis (prophase, metaphase, anaphase and telophase.
  • During prophase, the chromosomes condense to the point that they are visible under the light microscope as an X-shaped structure. Pairs of centrioles migrate away from each other (to opposite poles of the cell) while microtubules appear in between forming a spindle. Other microtubules emanating from the centrioles give a radiating star-like appearance thus they are called asters. Therefore, centrioles form the core of the Microtubule Organizing Centers (MTOC). Simultaneously, the diffuse nuclear chromatin condenses into the visible chromosomes which consist of two identical sister chromatids. The area of constriction where the two chromatids are attached is the centromere. Ultimately, the nuclear envelope disappears at the end of prophase.
The area of constriction where the two chromatids are attached it the centromere and the ends are referred to as the telomere
  • This is when the centromeres line up along the equatorial plate. At or near the centromeres are the kinetochores which are proteins that face the spindle poles (asters). Microtubules, from the spindle, attach to the kinetochores of each chromosome. The kinetochore fibers help to align and maintain the chromosomes on the metaphase plate.
  • Sister chromatids are pulled apart (as a result of the splitting of the centromere) such that each migrates to opposite poles being guided by spindle microtubules. With the separation of the sister chromatids, each chromatid is called a daughter chromosome. Cytokinesis begins during the last part of anaphase.

The daughter chromosomes are positioned at opposite poles of the cell and the kinetochore fibers disappear. New membranes (a nuclear membrane) form around the daughter nuclei nucleoli reappear the chromosomes uncoil (decondense) and become less distinct and are no longer visible by light microscopy and finally, cytokinesis (cell separation) occurs.

  • The last of the sequence of phases of the cell cycle. It is the phase of the cell cycle in which the cell spends the majority of its time and performs the majority of its purposes including preparation for cell division. In preparation for cell division, it increases its size and makes a copy of its DNA, which is made during the S phase. Interphase does not describe a cell that is merely resting but is rather an active preparation for cell division.
  • adenine (A) - a purine
  • cytosine (C) - a pyrimidine
  • guanine (G) - a purine
  • thymine (T) - a pyrimidine
  • Memory tool :
  • Pure Ag (silver element symbol) --> the purines are Adenine and Guanine
  • 1. The Ori is recognized on the DNA strand.
  • 2. The DNA strands are separated via helicase.
  • 3. A replication bubble is formed, with two replication forks.
  • 4. RNA primers are laid down by primase.
  • 5. DNA polymerases read the template strand (in the 3' to 5' direction) and synthesizes a new complementary strand (in the 5' to 3' direction). This can occur in a continuous way (as in the leading strand) or via a semidiscontinuous mechanism (as in the lagging strand).
  • 6. Okazaki fragments of the lagging strand are joined by DNA ligase.
  • 7. The RNA primers are removed and replaced with DNA (this is done by DNA polymerases).
  • 1. two ribonucleoside triphosphates (or nucleotides) approach the DNA template
  • 2. temporary hydrogen bonds form between RNA nucleotide residues and DNA nucleotide residues to form a DNA-RNA hybrid
  • 3. RNA polymerase catalyzes the formation of a phosphodiester bond between the two nucleotide residues, and a pyrophosphate is released
  • 4. additional ribonucleoside triphosphates are added to the growing chain, and additional phosphodiester bonds are formed
  • 5. when the growing chain is about nine nucleotide residues in length, the sigma factor detaches from RNA polymerase, and initiation is complete.
  • local unwinding of the DNA double helix
  • positioning of two nucleotide units, bound within ribonucleoside triphosphates at the start site, according to the base-pairing rules and
  • the formation of a phosphodiester bond between the 3' end of the first nucleotide residue and the 5' end of the second
  • 1. the ribosome recognizes some protein of the mRNA molecule and binds it
  • 2. the ribosome reads the mRNA molecule, three nucleotides at a time (these three nucleotides are called a codon)
  • 3. when read by the ribosome, each codon on the mRNA orders that some particular amino acid be brought to the ribosome
  • 4. molecules of tRNA within the cytosol bring amino acids to the ribosome, as instructed by mRNA and
  • 5. amino acids brought to the ribosome, one by one, in a sequence ordered by the mRNA molecule, form peptide bonds to form a polypeptide.
  • It is the binding of an amino acid to its tRNA to produce the corresponding aminoacyl-tRNA and is an endergonic (meaning ΔG is positive) reaction.
  • Amino Acid + ATP + tRNA ----->
  • ------>Aminoacyl-tRNA + AMP + PPi
  • C. In summary, the steps of translation initiation are:
  • 1. The 30S subunit binds the 5' end of the mRNA transcript.
  • 2. The UAC antocodon of fMET-tRNA Met recognizes and binds the AUG start codon on a molecule of mRNA
  • 3. The large ribosomal subunit joins the initiation complex to form the 70S ribosome.
  • 1. a mRNA codon is exposed at the A site of the ribosome.
  • 2. An aminoacyl-tRNA, whose anticodon is complementary to the codon, arrives at the A site and forms hydrogen bonds with the codon. This step requires the hydrolysis of one phosphate from GTP.
  • 3. The enzyme peptidyl transferase transfers the amino acid in the P site from its tRNA carrier to the amino terminus of the aminoacyl-tRNA in the A site. A peptide bond is formed between the two amino acids. The bond between the amino acid and the tRNA in the P site is broken.
  • 4. The ribosome moves, or translocates, a distance of three nucleotides along the mRNA molecule in the 5' to 3' direction. Translocation moves the growing polypeptide anchored to a tRNA molecule to the P site of the ribosome, the old tRNA to the E site and exposes the next mRNA codon at the A site. This step requires the hydrolysis of one GTP.
  • UAG, UGA, UAA - stop codons that upon appearing at the ribosomal A site will end translation elongation
  • Memory tool : "U Are Gone, U Go Away, U Are Away"
  • Bacterial ribosomes comprise 30S and 50S subunits. Together, they make a 70S ribosome.
  • The eukaryotic ribosome comprises a 40S and 60S subunit which make an 80S ribosome. Eukaryotic ribosomes are synthesized in the nucleolus, which is also the site of rRNA synthesis.
  • Because prokaryotic ribosomes are different from eukaryotic ribosomes, many antibiotic drugs have been developed that target the prokaryotic ribosome.
  • In prokaryotes, mRNA is made 5' to 3' via transcription. Once the 5' end of the transcript has been made by RNA polymerase, it can start to undergo translation. In prokaryotes, translation and transcription occur simultaneously RNA polymerase can be building the 3' end of the mRNA as ribosomes are already reading the 5' end of the mRNA.
  • In eukaryotes, such synchrony is impossible because the two processes occur in different cellular compartments. Transcription takes place in the nucleus and translation in the cytoplasm. Also, mRNA processing must take place between transcription and translation in eukaryotes.
  • 1. They do not grow by increasing in size.
  • 2. They can not carry out independent metabolism.
  • 3. They do not respond to external stimuli.
  • 4. They have no cellular structure.
  • A virus attaches to a specific receptor on a cell.
  • Some viruses may now enter the cell others will simply inject their nucleic acid.
  • Either way, viral molecules induce the metabolic machinery of the host cell to produce more viruses.
  • The new viral particles may now exit the cell by lysing (bursting). The preceding is deemed lytic or virulent.
  • Some viruses lie latent for long periods of time without lysing the host cell. These are called lysogenic or temperate viruses.
  • Cocci : spherical or sometimes elliptical
  • Bacilli : rod-shaped or cylindrical
  • Spirilli : helical or spiral
  • Two identical DNA molecules migrate to opposite ends of a cell as a transverse wall forms, dividing the cell in two. The cells can now separate and enlarge to the original size.
  • Under ideal conditions, a bacterium can undergo fission every 10-20 minutes producing over 10 30 progeny in a day and a half.
  • b=B*2 n where :
  • b is the number of bacteria at the end of the time interval
  • B is the number of bacteria at the beginning of the time interval
  • n is the number of generations
  • Thus if starting with 2 bacteria and follow for 3 generations then:
  • b=B*2 n =2*2 3 =2*8=16
  • note : bacterial doubling time is a relatively popular question type
  • Aerobic : refers to metabolism in the presence of oxygen
  • Anaerobic : refers to metabolism in the absence of oxygen (i.e. fermentation)
  • Proteins with two different configurations, each with different biological properties
  • Important regulators of transcription
  • (a) Simple proteins which contain only amino acids like digestive enzymes ribonuclease, trypsin and chymotrypsin.
  • (b) Complex proteins which contain amino acids and a non-amino acid cofactor. Thus the complete enzyme is called a holoenzyme and is made up of a protein portion (apoenzyme) and a cofactor.
  • Holoenzyme = Apoenzyme + Cofactor
  • A metal : For example, Zinc is a cofactor for the enzymes carbonic anhydrase and carboxypeptidase.
  • An organic molecule : Such as pyridoxal phosphate or biotin. Cofactors such as biotin, which are covalently linked to the enzyme are called prosthetic groups or ligands.
  • Their specificity is linked to the concept of an active site.
  • An active site is a cluster of amino acids within the tertiary (i.e. 3-dimensional) configuration of the enzyme where the actual catalytic events occurs.
  • The active site is often similar to a pocket or groove with properties (chemical or structural) that accommodate the intended substrate with high specificity.
  • Enzymes exhibit saturation kinetics.
  • The mechanism that lies largely with feedback inhibition - when the product of the enzyme catalyzed reaction returns (feeds back) to prevent or inhibit further reactions between the enzyme and its substrate
  • Reversible : these generally interact non-covalently and virtually instantaneously with an enzyme.
  • Irreversible : these usually react covalently to render the enzyme inactive.
  • Induction : enhancement of its synthesis
  • Repression : a decrease in its biosynthesis
  • Covalent modification : Phosphorylation of specific serine residues by protein kinases increases or decreases catalytic activity depending upon the enzyme. Proteolytic cleavage of proenzymes (e.g., chymotrypsinogen, trypsinogen, protease and clotting factors) converts an inactive form to an active form (e.g., chymotrypsin, trypsin, etc.)
  • Environment : especially pH and temperature (most enzymes exhibit optimal activity at a pH in the range of 6.5-7.5)
  • Non-covalent or allosteric mechanisms : Isocitrate dehydrogenase is an enzyme in the Kreb's Tricarboxylic Acid Cycle, which is activated by ADP, which is not a substrate or substrate analogue. It is postulated to bind a site distinct from the active site called the allosteric site.
  • Some enzymes fail to behave by simple saturation kinetics. In such cases, a phenomenon called positive co-operativity is explained in which binding of one substrate or ligand makes it easier for the second to bind.
  • Medium term:
  • glycogen (broken down to glucose)
  • Long term:
  • fat (broken down to fatty acids)
  • protein - as a last resort (amino acids)
  • fatty acids and amino acids can enter the Kreb's cycle
  • Short term:
  • ATP (1 GTP is approx. equal to 1 ATP)
  • Glycolysis
  • Kreb's Citric Acid Cycle
  • The electron transport chain (ETC)
  • Oxidative phosphorylation
  • Glucose + 2ADP + 2NAD + + 2Pi
  • 2Pyruvate + 2ATP + 2NADH + 2H +
  • ADP : adenosine diphosphate
  • NAD : nicotinamide adenine dinucleotide
  • Pi : inorganic phosphate
  • Aerobic : pyruvate is converted to Acetyl CoA which will enter the Kreb's Cycle followed by the oxidative phosphorylation producing a total or 38 ATP per molecule of glucose (i.e. 2 pyruvate)
  • Anaerobic : pyruvate is quickly reduced by NADH to lactic acid using the enzyme lactate dehydrogenase. A net of only 2 ATP is produced per molecule of glucose (this process is called fermentation)
  • It is the breaking apart of glucose.
  • Insulin helps to get glucose across the cell into the cytoplasm.
  • It wants to keep it in the cell, so it gets a big phosphate group.
  • It then changes to a fructose with a phosphate.
  • It then gets another phosphate group. By having TWO phosphate groups on a single molecule, the molecule becomes unstable and breaks apart because the two phosphate groups repel.
  • It goes through lysis to become 2 trioses (two phosphate trioses)
  • These transform to 2 pyruvate. It is the critical factor to determine whether things will proceed with or without oxygen.
  • (At this point 2 ATP has been produced - some energy is released - and get some reducing equivalents - 2 NADH + H which feed into the Electron Transport system to create more ATP)
  • If no oxygen (anaerobic, fermentation), mammals get lactic acid production.
  • If oxygen, then the pyruvate will enter the mitochondrian (the powerhouse of the cell) and produces 2 Acetyl CoA (Acetyl coenzyme A) that enters into the Kreb's cycle.
  • 1) glucose → 2 acetyl CoA → 2 turns around the TCA Cycle (Kreb's cycle)
  • 2) 2 CO2per turn is generated as a waste product which will eventually be blown off in the lungs
  • 3) one GTP per turn is produced by substrate level phosphorylation one GTP is equivalent to one ATP (GTP + ADP → GDP + ATP)
  • 4) reducing equivalents are hydrogens which are carried by NAD + (→NADH + H + ) three times (producing 3 ATP) per turn and FAD (→FADH2) once per turn (producing 2 ATP) these reducing equivalents will eventually be oxidized to produce ATP (oxidative phosphorylation) and eventually produce H2O as a waste product (the last step in ETC)
  • 5) the hydrogens (H) which are reducing equivalents are not protons (H + ). Often the reducing equivalents are simply called electrons.

1 NADH produces how many ATP molecules while 1 FADH2 produces how many?

What is the cost of getting two molecules of NADH generated in the cytoplasm to enter the mitochondrion?

What is the net yield of ATP for eukaryotes?

The net yield for eukaryotes is 36 ATP

  • (a) 3 molecules of NADH
  • (b) 1 molecule of FADH2, and
  • (c) 1 molecule of GTP
  • Two molecules of CO2 are also produced.
  • Two turns, producing:
  • (a) 4 molecules of CO2,
  • (b) 6 molecules of NADH,
  • (c) 2 molecules of FADH2, and
  • (d) 2 molecules of GTP
  • The right atrium's
  • where the process begins,
  • Where the C02 blood enters the heart
  • Through the tricuspid valve
  • to the right ventricle
  • The pulmonary artery and lungs.
  • Once inside the lungs it dumps its carbon dioxide
  • And picks up its oxygen supply
  • Then it's back to the heart through the pulmonary vein
  • Through the atrium and left ventricle.
  • The aortic valve's where the blood leaves the heart
  • Then it's channeled to the rest of the bod
  • The arteries, arterioles, and capillaries too
  • Bring the oxygenated blood to the cells
  • The tissues and the cells trade off waste and CO2
  • Which is carried through the venules and the veins
  • Through the larger vena cava to the atrium and lungs
  • And we're back to where we started in the heart.
  • Edema is caused by disruption of the hydrostatic/oncotic pressure dynamics and/or compromise of the vessel membrane integrity. Either condition can arise in both capillaries and lymph vessels. When lymph vessels are affected, normal drainage of tissue fluid is disturbed. An increase in hydrostatic pressure forces fluid out of capillaries the capillaries then fail to reabsorb the fluid in adequate amounts.
  • Dimished capillary oncotic pressure reduces the blood's tendency to draw fluid back from the interstitium into the capillary. Insufficient plasma concentrations of albumin reduces the blood's ability to draw water inward across the capillary membrane, and edema results. Disruption of capillary wall integrity, precipitating loss of plasma proteins, reduces the ability of the vessels to hold and resorb fluid, causing edema
  • They contain a concentration of lymphocytes, which fight infection.
  • Note : Lymphocytes also circulate freely in the blood stream.
  • Lymphocytes are a class of white blood cell (leukocyte).
  • They are broadly divided into two classes- T lymphocytes (T cells) and B lymphocytes (B cells).
  • T cells are the basis of cell-mediated immunity.
  • They have their origins in the bone marrow but mature in the thymus.
  • They originate in the bone marrow.
  • They participate in humoral immunity by producing antibodies, which belong to a class of proteins called immunoglobulins.
  • Mechanical : mechanical digestion begins with the shredding and grinding of food into small pieces by chewing (mastication). It continues with vigorous churning in the stomach.
  • Chemical : chemical digestion occurs by means of enzymes produced by several different organs associated with the alimentary canal. Enzymes break down food into absorbable molecules.
  • Salivary amylase
  • It initiates the digestion of starch, hydrolyzing glycosidic bonds to produce component sugars
  • Mouth : mastication it secretes amylase-containing saliva, which begins the digestion of starch.
  • Esophagus : transports food from the mouth to the stomach through the synchronized muscular contractions called peristalsis.
  • Stomach : initiates protein digestion (and performs mechanical digestion). The parietal cells within the walls of the stomach secrete hydrochloric acid, rending the lumen highly acidic (pH between 1.5-2.5). The chief cells of the stomach secrete pepsinogen, which is converted to its active form-pepsin-by hydrochloric acid. Pepsin breaks down proteins.
  • They are called zymogens.
  • To be activated, zymogens normally must be cleaved by an enzyme. (For example, among the zymogens released into the intestine is trypsinogen. The pancreatic enzyme trypsinogen is activated by an enzyme in the duodenum called enterokinase (also called enteropeptidase). The activation process produces trypsin, an active protein-degrading enzyme. Trypsin cleaves a number of other zymogens into their active forms.)
  • Pancreatic amylase : (chemically identical to salivary amylase) continues the digestion of carbohydrates, which was initiated in the mouth.
  • Pancreatic lipase : serves in the enzymatic breakdown of fats (lipids).
  • Trypsin and chymotrypsin : the two most important proteolytic, or protein-digesting, enzymes in the gastrointestinal tract. These two enzymes break peptide bonds, reducing large proteins into small chains composed of only a few amino acids.
  • Active in the duodenum, it is produced in the liver and stored in the gall bladder.
  • It is a complex mixture of water, electrolytes, cholesterol, bilirubin, steroid hormones, and several other substances.
  • Containing no enzymes, it acts as an emulsifier, helping to separate large globules of fat molecules into smaller lobules in order to increase the surface area available for the action of lipase.
  • It enters the midsection of the duodenum via the common bile duct.
  • Bile production
  • Plays a significant role in carbohydrate metabolism (for example, converting glucose to a storage form, glycogen), converts amino acids to keto acids and urea, and processes toxins.
  • Degradation of senescent (aged) erythrocytes.
  • The resorption of large amounts of water from its lumen
  • By passing through the ileocecal valve
  • Major component of the skeleton, providing an anchor for muscular contraction
  • Structural support and protection for organs and nerves
  • Red blood cells and platelets are formed in bone marrow
  • A storage depot for calcium, phosphate and other ions of biological significance (takes up and releases as needs)
  • Maintain a variety of ion concentrations within acceptable limits
  • Dynamic connective tissue a living tissue
  • Matrix (which is composed of organic, like Type I collagen and amorphous ground substance (glycosaminoglycans and proteins), and inorganic, like calcium and phosphorous, substances in about a 1:1 ratio) and cells
  • Hydroxyapatite
  • Osteoblasts : located on the inner surfaces of bone tissue, synthesize Type I collagen and other organic components of the matrix, build and nourish bone
  • Osteocytes : occupy minute spaces (lacunae) within the bony matrix, are simply osteoblasts with greatly reduced synthetic capacity, are responsible for maintaining the matrix, build and nourish bone
  • Osteoclast : also known as a multinucleated giant cell, promotes ongoing breakdown, resorption, and remodeling of bone.
  • The mineral composition of bone is largely a crystalline substance called hydroxyapatite. Dissolution of the mineral component would make the bone softer and more flexible.
  • (Collagen, composed of protein chains, can be correctly presumed to impart some flexibility to the bone.)
  • The removal of the organic component of the bone matrix results in an inflexible, hard, brittle substance, subject to destruction under force.
  • Compact : the outer, dense portion
  • Spongy bone : the inner spongy-looking area (due to its many small, marrow-filled cavities)
  • Despite differences in their gross appearances, compact bone and spongy bone are similarly constituted. Each consists of matrix and cells.
  • Red : the site of red blood cell and platelet production and some immune cell development and maturation
  • Yellow : filled with adipocytes (fat cells)
  • In addition to the osteoblasts, osteocytes, and osteoclasts, other cell species inhabit the marrow.
  • Newborn : all marrow is red
  • Adult : red marrow is confined primarily to flat bones, such as ribs, clavicles, pelvic bones, and skull bones. Under stress of blood loss or poor oxygen supply, however, yellow marrow may be transformed into red marrow to increase red blood cell production.
  • Haversian system, or osteon.
  • Haversian systems exist to distribute nutrients throughout compact bone. Many Haversian systems together give compact bone its strength.
  • At the center of each Haversian system is a canal known as a Haversian canal.
  • A Haversian canal runs the length of a Haversian system.
  • This canal carries blood vessels and nerves, and its filled with loose connective tissue.
  • Shallow indentations mark the surfaces of the lamellae
  • Spicules
  • Because of their thinness, the spicules within spongy bone are able to absorb nutrients directly from the marrow contained within their cavities, and so do not require Haversian systems for nutrient delivery
  • Compact bone and spongy bone are similar in that:
  • They have the same chemical and structural composition
  • both are composed of matrix and bone cells
  • both are hard and resistant to bending or compression
  • The essential difference between compact bone and spongy bone:
  • The manner in which the components are arranged
  • Compact bone is so densely arranged that the canals of the Haversian systems are required to convey nutrients to its cells
  • The hard substance of spongy bone is laid out in thin, bubble-like form that precludes the need for special nutrient delivery channels

Which of the following applies to osteoclasts?
I. They are multinucleated cells.
II. They are a type of bone cell.
III. They deposit new bone during bone remodeling.

A. II only
B. I and II only
C. II and III only
D. I, II, and III

  • Joints allow for motion and flexibility.
  • Fibrous : composed of collagen fibers, designed to allow minimal movement
  • Synovial : consists of the approximated ends of two bones (nearly in contact, covered with smooth, tough articular cartilage), covered with a common synovial capsule made of fibrous tissue (encloses a sac of synovial fluid that acts as a lubricant, which, with the underlying articular cartilage, allows smooth movement of the joint, protecting the bones from damage that might otherwise result from friction) allow for the great range and extent of movement seen in the body examples include knees, hips, shoulders, and fingers
  • Cartilaginous : articular cartilage
  • Ligaments : keep bones attached to joints
  • Tendons : attach muscles to bones
  • skeletal:
  • Voluntary contraction - A skeletal muscle that traverses a joint will be among those responsible for bending that joint (like biceps brachii, traversing the elbow joint, which is antagonistic to the triceps).
  • cardiac:
  • operates involuntarily (sinoatrial node initiates contractions)
  • smooth:
  • Involuntarily controlled, the operative muscle of organs and systems such as blood vessels, stomach, intestines, skin, glands and ducts.
  • a long, multinucleated cell in which many striations are visible
  • commonly referred to as muscle "fibers" or myofibers
  • A group or bundle of skeletal muscle fibers is referred to as a fascicle

a segment of muscle fiber between two Z lines

  • |-----A Band-----|-I Band-|
  • ___________ ____________
  • -----⟨_____ >->->- (-- )-<-<-< ______⟩-----
  • -----⟨_____ >->->- (--) -<-<-<__ ____⟩-----
  • -----⟨_____ >->->- (--) -<-<-< ______⟩-----
  • Z line |-H-| Z line
  • Myosin filament (thick filaments) that have no connection to the Z lines - The length of myosin filament corresponds to the A band. Because the filament itself does not contract, the A band has a fixed length equal to that of the myosin strands. The center of the sarcomere, containing only myosin filaments with no overlapping actin filaments , is the H zone.
  • Actin filaments (thin filaments) that are anchored at one end to a Z line - A given length of thin filament that does not overlap with any thick filament is called the I band.
  • Myosin and Actin are proteins
  • Contraction is achieved by the sliding of actin and myosin filaments, each over the other
  • |-----A Band-----|-I Band-|
  • ___________ ____________
  • -----⟨_____ >->->- (--) -<-<-< ______⟩-----
  • -----⟨_____ >->->- (--) -<-<-< ______⟩-----
  • -----⟨_____ >->->- (--) -<-<-< ______⟩-----
  • Z line |-H-| Z line
  • Regularly spaced crossbridges extend from the myosin filaments to the actin filaments.
  • These crossbridges make the physical interaction possible
  • 1) the myosin head binds to the actin filament at a myosin binding site
  • 2) the myosin heads interact with the actin filaments in such a way as to draw them inward so that Z lines come closer together and the sarcomere shortens.
  • Membrane resting potential : the inside is negative relative to the outside
  • (1) Na + concentration is higher outside the cell than inside
  • (2) K + concentration is higher inside the cell than outside
  • (3) the cell's interior is electrically negative relative to its exterior
  • A wave of depolarization that spreads rapidly and extensively along the entirety of the cell membrane so that the whole of the cell is depolarized (a rapid influx of sodium ions results). The entire cell reverses its normal charge gradient and its inside becomes positive relative to its outside. After depolarization, it loses its increased permeability to sodium and temporarily experiences and increased permeability to potassium, restoring the original resting potential.
  • Often used to designate the combination of the two sequential events:
  • (1) rapid depolarization followed by
  • (2) repolarization
  • Tropomyosin : in a resting muscle, these molecules "mask," or "cover," those sites on the actin molecule with which myosin heads are prone to interact.
  • The troponin complex : Troponin has a tendency to bind calcium ions, if they are present, and then undergoes a change of shape and position to cause the tropomyosin molecule to "uncover" the actin sites with which myosin is prone to interact.
  • Therefore : If the myofiber's troponin complex is exposed to calcium ions, the actin and myosin interact at their crossbridges, the sarcomere shortens, and the fiber contracts.
  • (1) cell membrane is called the sarcolemma
  • (2) cytoplasm is called the sarcoplasm
  • (3) endoplasmic reticulum is called the sarcoplasmic reticulum , which stores calcium ions in a muscle (Ca 2+ ) and
  • (4) sarcolemma and sarcoplasmic reticulum are roughly continuous with each other.
  • A. A motor neuron that innervates the myofiber undergoes depolarization and an action potential , causing the motor neuron to release the neurotransmitter acetylcholine .
  • B. The neurotransmitter binds to receptors on the sarcolemma , which is continuous with the sarcoplasmic reticulum .
  • C. The neurotransmitter causes depolarization of both sarcolemma and sarcoplasmic reticulum .
  • D. The depolarization, if sufficient in magnitude, triggers an action potential, and the entire sarcolemma depolarizes.
  • E. The action potential causes calcium ions to move from the sarcoplasmic reticulum to the sarcoplasmic space , surrounding the fiber's actin and myosin filaments.
  • F. The calcium ions bind to troponin.
  • G. The binding of calcium ions to troponin causes tropomyosin to undergo change in shape and position.
  • H. The sites at which actin interacts with myosin are "uncovered." Myosin heads are thus able to contact actin filaments.
  • I. Actin and myosin interact as "sliding filaments."
  • A nerve impulse arrives at the neuromuscular junction, causing the release of acetylcholine.
  • Depolarization of muscle cell membranes is carried throughout the fiber by the T tubules.
  • The depolarization causes a release of calcium ions, which bind to troponin on the actin filaments.
  • This causes a conformational change that uncovers myosin binding sites on the actin filaments. Myosin crossbridges are then able to attach to the actin and flex, pulling the actin fiber along the myosin fiber. This causes the simultaneous shortening of sarcomeres all along the muscle cell.
  • To maintain-
  • (1) the resting potential that allows for depolarization and action potential
  • (2) the dynamic interaction of myosin and actin that underlies the physical shortening of the sarcomere, and
  • (3) the return of calcium ions from the sarcoplasm to the sarcoplasmic reticulum after contraction is complete
  • It embodies a high-energy phosphate bond.
  • Used to regenerate ATP from ADP and inorganic phosphate
  • Intercalated discs
  • The attachment points between adjacent muscle cells.
  • The attachments, gap junctions, allow a nearly unimpeded flow of ions from one fiber to the next (so the action potential created in one cardiac cell is easily and readily propagated to the next).
  • The whole of the cardiac muscle mass, the myocardium, is termed a syncytium (in which rapid intercellular transmission of action potentials promotes a particularly organized and synchronous mode of contraction)
  • Tropomyosin and troponin complex are absent.
  • Calcium ions induce contraction by some device different than that which operates in the sarcomeres of skeletal and cardiac muscle.
  • Calcium ions bind to a calcium-binding protein called calmodulin.
  • The calcium-calmodulin complex then interacts with a protein called myosin light chain kinase (MLCK) which, in turn, directly phosphorylates the myosin head.
  • Phosphorylation of the myosin head then allows myosin to form a cross bridge with actin, whereupon contraction occurs
  • Like cardiac muscle, adjoining smooth muscle cells feature gap junctions so that the action potential is readily and rapidly transmitted from one cell to the next.
  • (1) Excretion of hydrophilic waste note that the liver is responsible for excreting hydrophobic or large waste products which cannot be filtered by the kidney.
  • (2) Maintain constant solute concentration.
  • (3) Maintain constant pH (approximately 7.4).
  • (4) Maintain constant fluid volume, which is important for blood pressure and cardiac output.
  • Renal cortex : an outer portion
  • Renal medulla : an inner portion (composed of wedge-shaped tissue structures called the renal pyramids, or medullary pyramids.)
  • The nephron (each kidney consists of more than a million)
  • Each consists of a renal corpuscle, continuous with a long "urinary pipeline," or renal tubule (a hollow tube surrounded by epithelium cells)
  • Glomerulus : a tuft of capillaries
  • A glomerular basement membrane
  • A surrounding Bowman's capsule : a double-walled cup formed as an enlargement of the proximal end of the renal tube.
  • This area of the nephron functions in blood filtration.
  • the proximal convoluted tubule (at the beginning in cortex)
  • the descending loop of Henle (extends into the pyramids)
  • the ascending loop of Henle (back up into the cortex)
  • the distal convoluted tubule
  • the collecting duct (back into the medullary pyramids, a single duct carries fluid away from numerous distal convoluted tubules.
  • Papillary ducts they empty into the funnel-shaped sections of the renal pelvis called calyces (singular : calyx)
  • From there, the ureter carries the wine away from the kidney to the urinary bladder, where it is stored until it passes from the body in micturition (urination).
  • via the renal artery, which branches off the abdominal aorta
  • the artery branches in to afferent ("approaching") arterioles which travel to individual renal corpuscles, where they branch profusely to form the glomerular capillaries inside a Bowman's capsule.
  • The inner wall (or visceral layer) is porous and permeable to plasma and other small blood constituents.
  • The exterior wall of the capsule (parietal layer) is neither porous nor permeable.
  • The space enclosed by the two walls-the Bowman's space- is the origin of the renal tubule.
  • The Descending Loop of Henle : The salt concentration of the cortical interstitial fluid-around the renal corpuscles, proximal convoluted tubules, and upper portion of the loop of Henle- is relatively low. As filtrate travels through the descending loop of Henle, it encounters increasingly higher salt concentrations in the surrounding interstitial environment. Because the wall of the descending limb of the loop of Henle is permeable to water but nearly impermeable to solutes, water passes out of the loop of Henle. Therefore, water is reabsorbed back into the body in the descending loop of Henle.
  • The Ascending Loop of Henle : the concentration gradient is reversed here (it decreases) but the ascending limb is impermeable to water and permeable to sodium. As the filtrate travels here, it becomes dilute once again.
  • the pancreas, the adrenal glands, the thyroid gland, the parathyroid glands, the ovaries, the testes, the hypothalamus, and the pituitary gland
  • Note : this is not an exhaustive list there are other organs (such as the thymus, heart, and kidney) that can also be classified as endocrine glands
  • First, polysaccharides that we eat are chemically broken down and then absorbed by the gastrointestinal tract.
  • Second, glucose can be produced by the liver and released into the blood.
  • The brain : the brain takes up glucose from the blood whether or not insulin is secreted.
  • The liver : the liver converts large quantities of the glucose it takes up to glycogen, a long carbohydrate polymer that serves as a storage form of glucose.
  • 1. increasing cellular uptake of glucose
  • 2 .promoting formation of glycogen from glucose in the liver
  • 3. reducing glucose concentration in the blood, and
  • 4. increasing protein and triglyceride synthesis
  • The levels of the minerals sodium and potassium in the body
  • Aldosterone
  • It acts primarily at the distal convoluted tubule of the kidney to promote sodium-potassium exchange (increasing the removal of three sodium ions from the renal filtrate for every two potassium ions transported into the filtrate and this promotes the movement of water form tubule to interstitium)
  • The effects are:
  • 1. increase urinary excretion of potassium,
  • 2. increase interstitial sodium concentration, and
  • 3. increase water conservation (as an effect of #2)
  • As a result, it is not surprising that aldosterone secretion is stimulated by high levels of extracellular potassium, low levels of extracellular sodium, and low fluid levels (blood volume).
  • Glucocorticoids affect : plasma glucose concentrations, increase blood glucose levels (especially in response to environmental stressors), strengthen cardiac muscle contractions, increase water retention, and have anti-inflammatory and anti-allergic activities.
  • Cortisol : released as part of the long-term stress response and affects most tissues in the body. It increases plasma glucose levels and inhibits immune activity.
  • Cortisol release is controlled by the hypothalamus and the anterior pituitary and causes negative feedback to both these areas of the brain.
  • catecholamines, mostly epinephrine (also known as adrenaline)
  • Epinephrine is an amino acid derivative that acts like a peptide hormone.
  • The thyroid is a flat gland located in the neck, in front of the larynx.
  • It synthesizes and secretes two hormones : calcitonin and the thyroid hormones (thyroxine- thyroid hormone T4, triiodothyronine- thyroid hormone T3)
  • A set of four small glands located on the posterior aspect of the thyroid.
  • The peptide hormone parathyroid hormone (PTH, also known as parathormone) exerts effects opposite those of calcitonin (which is produced in the parafollicular cells of the thyroid). It is secreted in response to low blood levels of calcium, and through actions on various organs, it increases levels of blood calcium. It acts to:
  • 1. increase blood resorption and consequent calcium release,
  • 2. increase intestinal calcium uptake, and
  • 3. promote calcium re-uptake at the kidney.
  • follicular phase (the first phase):
  • rapid growth of the ovarian follicule
  • the anterior pituitary gland secretes two hormones (follicle-stimulating hormone (FSH) and luteinizing hormone (LH)) that stimulate the growth of one follicle containing several ova, only one of which fully matures
  • the follicle itself is secretory, releasing the hormone estrogen as it develops
  • the follicular phase ranges from seven to twenty-one days
  • Increased ovarian estrogen release prevents maturation of more than one follicle at a time.
  • at the end of the follicular stage (around day 14), there is a surge in LH secretion from the anterior pituitary causing the release of the ovum from the enlarged follicle (ovulation), and it is swept into the Fallopian tube by the fimbriae where the ovum waits for fertilization by sperm
  • luteal phase (the final phase):
  • typically day 14 to day 28
  • after ovulation, the part of the ruptured follicle that remains in the ovary is referred to as the corpus luteum
  • the corpus luteum secretes estrogen and progesterone
  • Menses (the first phase):
  • shedding of the uterine lining
  • lasts between 4 and 7 days and occurs at the same time as the early follicular phase in the ovary
  • Proliferative phase:
  • occurs until day 14.
  • estrogen from the ovaries induces the proliferation of the endometrium
  • Secretory phase (final phase):
  • lasts for the final 14 days of the menstrual cycle
  • progesterone from the corpus luteum promotes the rapid thickening and vascularization of the uterine lining in preparation for the implantation of a fertilized ovum -
  • if the mature ovum is not fertilized, it will not implant into the uterine lining
  • at approximately 13 days after ovulation (day 27), the corpus luteum in the ovary degenerates and ceases to secrete estrogen and progesterone
  • the lack of progesterone causes the lining to slough off for about 5 days and then a new proliferative phase begins.
  • if the ovum is fertilized, the developing placenta begins to secrete human chorionic gonadotropin (hCG) preventing the corpus luteum from degenerating, allowing it to continue secreting progesterone
  • before the end of the first trimester of pregnancy, the placenta begins to secrete estrogen and progesterone and the corpus luteum degenerates and these hormones are secreted at continuously increasing levels throughout the pregnancy

FSH and LH are peptide hormones secreted from the anterior pituitary during the follicular phase to cause follicular development in the ovary. A surge of LH causes ovulation.

Estrogen is secreted by the follicular cells during the follicular phase and promotes proliferation of the endometrium.

Estrogen and progesterone are secreted from the corpus luteum during the luteal phase. They cause further development of the endometrium, in preparation for zygote implantation.

Human chorionic gonadotropin (hCG) is secreted from the developing placenta upon fertilization and implantation. It stimulates the corpus luteum to continue progesterone and estrogen secretion. In the absence of implantation and the consequent absence of placental hCG, the corpus luteum degenerates, and its hormonal secretion terminates. Without progesterone, the endometrial lining deteriorates. Menstrual bleeding follows. Also, in response to the post-luteal fall in circulating levels of estrogen and progesterone, FSH production rises, initiating a new proliferative phase.

Each testis contains specialized reproductive organs. What are they called and what do they contain?

What do the interstitial cells, situated among the twisted reproductive organs within the testes, secrete, and what stimulates this production? Give the function of this hormone.

  • seminiferous tubules
  • spermatogonia, the precursors of spermatozoa formation
  • testosterone
  • stimulated by pituitary LH (also called interstitial cell stimulating hormone, ICSH, in mature males)
  • --predominately male hormone, does not become plentiful until puberty, secreted into the bloodstream and comes into contact with all parts of the body --
  • It's principle role is to promote spermatogenesis, the division of the spermatogonia within the seminiferous tubules to produce haploid spermatozoa. It also promotes the development of secondary sex characteristics including deepening of the voice, growth of facial, axillary, and pubic hair, and enlargement of the penis and scrotum
  • The hypothalamus is a portion of the diencephalon of the forebrain.
  • It releases several hormones that control secretions from the pituitary gland.
  • I provides neural input and central control of the endocrine glands outside of the brain.
  • It serves as a high-level coordinating and regulating center for both the endocrine system and the autonomic nervous system.
  • It integrates a variety of information from the cerebral cortex and limbic system, and regulates output from the pituitary glands.
  • 1. Thyroid-stimulating hormone (TSH) : stimulates the thyroid gland to secrete thyroid hormone.
  • 2. Adrenocorticotropic hormone (ACTH) : stimulates the adrenal cortex to secrete cortisol.
  • 3. Luteinizing Hormone (LH) : stimulates the gonads (ovaries or testes) to promote sex hormone secretion and gamete production.
  • 4. Follicle Stimulating Hormone (FSH) : stimulates the gonads (ovaries or testes) to promote sex hormone secretion and gamete production.
  • 5. Growth hormone (GH) : influences the development of skeletal muscle, bone, and organs in infants and children. Without growth hormone, children fail to develop normally. GH is also known as somatotropin (STH).
  • 6. Prolactin : directly targets the female breasts, where it stimulates breast development and milk production.
  • 1. Antidiuretic hormone (ADH) : Discussed previously, in connection with the kidney. It is also known as vasopressin.
  • 2. Oxytocin : Released at childbirth (parturition), causing the uterus to contract and push the fetus through the birth canal.
  • 1. The hypothalamus and the pituitary glad (both anterior and posterior) are the higher regulatory organs of the endocrine system many of the hormones they secrete control other endocrine glands. Therefore, many functions of the endocrine system depend on instructions from the brain.
  • 2. Hormones can regulate other hormones and many endocrine secretions are part of a pathway that is master regulated. Hormones that control the release of other hormones are called tropic hormones.
  • Examples:
  • organ:
  • tropic hormone
  • controlled hormone
  • hypothalamus:
  • corticotropin-releasing hormone (CRH)
  • cortisol release
  • thyrotropin-releasing hormone (TRH)
  • thyroid hormone levels
  • gonadotropin-releasing hormone (GnRH)
  • sex hormones
  • (androgens,
  • estrogens,
  • and progesterone)
  • anterior pituitary:
  • adrenocorticotropic hormone (ACTH)
  • cortisol release
  • thyroid-stimulating hormone (TSH)
  • thyroid hormone levels
  • gonadotropins
  • (luteinizing hormone (LH) and
  • follicle stimulating hormone (FSH)
  • sex hormones
  • (androgens,
  • estrogens,
  • and progesterone)
  • 3. hormone levels are controlled by feedback regulation, especially negative feedback.
  • A. A change in physiological status in response to a hormone can feed back to the endocrine gland to stop hormone secretion.
  • B. Hormones can cause feedback regulation to the higher regulatory organs (hypothalamus and pituitary gland) that are controlling their release.
  • Dendrites : cytoplasmic extensions of the cell that act like antenna, or sensors: they receive stimuli.
  • Axon : nerve fiber that is a single, elongated cytoplasmic extension that transmits signals.
  • Synaptic terminals : the distal end of the axon bears these small extensions
  • Synaptic vesicles : contained within synaptic terminals
  • Neurotransmitters : contained within the synaptic vesicles, these molecules transmit chemical signals from one neuron to the next. In some situations, a neurotransmitter will exert and excitatory effect on a neuron, while in other, it will have an inhibitory effect.
  • presynaptic neuron
  • presynaptic membrane
  • synaptic cleft - the place into which the neurotransmitters are released from the synaptic vesicles to the postsynaptic membrane
  • postsynaptic membrane - The part of the cell membrane of a neuron or muscle fiber with which an axon terminal forms a synapse.
  • They are specific to the neurotransmitter.
  • Each postsynaptic neuron can receive signals from many presynaptic neurons (they express receptors for many types of neurotransmitters).
  • Note : Each neuron will release only ONE type of neurotransmitter from its presynaptic axon terminus.
  • 1. Acetylcholine : triggers muscle contraction and is degraded by the enzyme acetylcholinesterase .
  • 2. Epinephrine : (aka adrenaline) increases heart rate and blood pressure and decreases metabolic activity, such as that of the smooth muscle of the digestive system. It is oxidized and methylated to inactive metabolites by monoamine oxidase (MAO) and catechol-O-methyl transferase (COMT) , respectively.
  • The Nervous System
  • ↙ ↘
  • Peripheral Central
  • ↙↘ ↙ ↘
  • Somatic Autonomic Spinal cord* Brain*
  • ↙ ↘
  • Sympathetic Parasympathetic*
  • ♦ Spinal cord : white matter, gray matter
  • ♦ Brain : cerebral cortex, hypothalamus, thalamus, pons, cerebellum, medulla
  • ♦ Parasympathetic nervous system : vagus nerve
  • 1. Sensory receptor cells : register a given stimulus, such as smell or sound, and gather information.
  • 2. Sensory neurons (aka afferent neurons) : receive information from the sensory receptors and send it to the central nervous system in some cases (such as olfactory transduction), the receptor is a modified part of the sensory neuron itself.
  • 3. Interneurons (aka associative neurons) : In the CNS, one or more interneurons receive and process the information and generally function in relaying signals from neuron to neuron.
  • 4. Motor neurons (aka effector or efferent neurons) : transmit nerve impulses from the CNS to the target muscle, organ, or gland.
  • **Afferent nerves transmit nerve impulses to the CNS and efferent nerves conduct impulses from the CNS to the muscles or glands.
  • Brain : the anterior section of the neural tube
  • Spinal cord : the posterior portion of the neural tube
  • 1. layers of connective tissue (the meninges)
  • 2. bone (the spine and the skull)
  • 3. circulating cerebrospinal fluid (CSF) that acts as a liquid shock absorber
  • hindbrain : cerebellum, the pons, the medulla
  • midbrain : structures that govern visual auditory reflexes and coordinate information of posture and muscle tone
  • forebrain : the diencephalon (includes the thalamus, the hypothalamus, and the pituitary gland) and the telencephalon (includes the cerebrum, the limbic system and the basal nuclei)
  • ♦Composed of two hemispheres
  • ♢Divided by a longitudinal fissure
  • ♢Connected by the corpus callosum
  • -a thick bundle of axons
  • ♦The largest portion of the human brain
  • ♦Cerebral cortex
  • ♢Gray matter overlying the cerebrum
  • -contains neuronal cell bodies (or soma)
  • that conduct the highest of intellectual
  • functions
  • ♢Governs voluntary motor activity, functions
  • of language and cognition
  • ♢Divided into four pairs of lobes
  • -frontal
  • -parietal
  • -temporal
  • -occipital
  • Spinal cord:
  • ♢no subdivision
  • ♦simple spinal reflexes
  • ♦control of primitive processes
  • Medulla:
  • ♢Hindbrain
  • ♦controls autonomic processes (bp, heartbeat,
  • respiratory rate, vomiting)
  • Pons:
  • ♢Hindbrain
  • ♦some autonomic control
  • ♦controls anti-gravity posture and balance
  • Cerebellum:
  • ♢Hindbrain
  • ♦integrating center
  • ♦coordination of complex movement, balance,
  • and posture
  • Midbrain:
  • ♢no subdivision
  • ♦integration of visual and auditory info
  • ♦wakefulness
  • Thalamus:
  • ♢Forebrain diencephalon
  • ♦somatic/conscious sensation
  • Hypothalamus:
  • ♢Forebrain diencephalon
  • ♦homeostasis (ex. temperature)
  • ♦primitive emotions (ex. hunger, rage..)
  • Pituitary:
  • ♢Forebrain diencephalon
  • ♦homeostasis via hormone release
  • ♦controlled by the hypothalamus
  • ♦2 peptide hormones from posterior pituitary
  • ♦6 peptide hormones from anterior pituitary
  • Basal nuclei:
  • ♢Forebrain telencephalon
  • ♦regulate body movement
  • Limbic System:
  • ♢Forebrain telencephalon
  • ♦emotion
  • Cerebral cortex (in general)
  • ♢Forebrain telencephalon
  • ♦intelligence, communication, memory,
  • planning, reading, voluntary movement..
  • ♦left side controls speech and motor function
  • on the right side of body
  • ♦right side controls visual spatial reasoning
  • and music, and left side motor function
  • Frontal lobes:
  • ♢Forebrain telencephalon
  • ♦voluntary movement
  • ♦complex reasoning skills
  • ♦problem solving
  • Parietal lobes:
  • ♢Forebrain telencephalon
  • ♦general sensation (ex. touch, temperature..)
  • ♦gustation (taste)
  • Temporal lobes:
  • ♢Forebrain telencephalon
  • ♦auditory and olfactory sensation
  • ♦short-term memory
  • Occipital lobes:
  • ♢Forebrain telencephalon
  • ♦visual sensation and processing
  • Corpus callosum
  • ♢no subdivision
  • ♦connects the left and right cerebral
  • hemispheres
  • Simple motor reflexes
  • Interior : gray matter (composed of cell bodies of spinal cord neurons)
  • Exterior : composed of white matter (or myelinated spinal cord axons) -- so called for the pale appearance of myelin which insulates the axon
  • Afferent : receive information from sensory receptors for pain, touch, temperature and proprioception
  • Efferent : innervate skeletal muscle by releasing the neurotransmitter acetylcholine
  • **The somatic nervous system governs voluntary activities that we can consciously control.
  • The preganglionic neuron has its cell body in the brainstem or the spinal cord. It synapses with the postganglionic neuron and releases acetylcholine. The postganglionic neuron can release acetylcholine or norepinephrine to control the effector tissue.
  • Note that a ganglion is a cluster of nerve cell bodies in the PNS.
  • Part of the parasympathetic nervous system
  • It sends parasympathetic innervation to the thoracic and abdominal regions
  • One of the major effects of the vagus nerve innervation is slowing down the heart rate below the rate that automatically generated by the SA node.
  • System Sympathetic || Parasympathetic
  • General: Fight & flight || Rest & digest
  • Mobilize energy || store energy
  • Heart rate: Increased || Decreased
  • Pupils: Dilate || Constrict
  • Vision: Favors far vision || Favors near vision
  • GI Tract: Inhibit mobility || Stimulate mobility
  • Bladder: Inhibit || Stimulate
  • Bronchial Relaxed, || Constricted,
  • smooth muscle: open || closed, shallow
  • System: Sympathetic || Parasympathetic
  • 1. Thoracic & lumbar || Brainstem & sacral
  • spinal cord || spinal cord
  • 2. Short || Long
  • 3. Acetylcholine || Acetylcholine
  • 4. Long || Short
  • 5.Norepinephrine (most) || Acetycholine
  • Mechanoreceptors : responds to mechanical disturbances. These include stretch receptors, tactile receptors, proprioceptors, and auditory receptors.
  • Chemoreceptors : Responds to particular chemicals and register taste and smell
  • Thermoreceptors : stimulated by changes in temperature
  • Electromagnetic receptors : stimulated by electromagnetic waves (photoreceptors-rods and cones in the eye)
  • Nociceptors : pain receptors
  • The inner ear
  • interprets positional information required for maintaining equilibrium
  • a membranous labyrinth situated within the three semicircular canals
  • Movement of the head causes movement of fluid within the labyrinths and displacement of specialized hair cells (the crista) located in the ampulla at the base of the semicircular canals.
  • This crista displacement initiates sensory impulses conveyed via the vestibular nerve to centers in the cerebellum, midbrain, and cerebrum, where directional movement and position are interpreted.
  • External (or outer) ear:
  • Composed of the pinna, which funnels sound waves into the ear canal.
  • Middle ear:
  • Sound waves cause vibrations in the tympanic membrane, setting into motion the three auditory bones-- the malleus, incus, and stapes. The stapes movement is transmitted across the oval window into the inner ear.
  • Inner ear:
  • movement from across the oval window sets up vibrations in the fluid of the cochlea, which causes bending of auditory hair cells in the organ of Corti. The cochlear nerve and the vestibular nerve form the two branches of the acoustic nerve (8th cranial nerve).
  • Light enters the cornea
  • traverses the aqueous humor
  • passes through the pupil
  • proceeds through the lens
  • and the vitreous humor
  • until it reaches the light receptors of the retina.
  • Electrical signals are then transmitted via the optic nerve to visual centers in the brain.
  • located in the outer layer, both contain pigments, allowing them to absorb energy from light rays
  • rods : specialized to register dim light, rhodopsin is the rod's pigment
  • cones : specialized to register bright lights as well as color, subdivided into red-absorbing, blue-absorbing, and green-absorbing, opsin (similar to rhodopsin) mediates light reception for the cones
  • The skin
  • -maintains body temperature
  • -registers information from the environment
  • -provides a barrier against infection
  • Composed of stratified squamous epithelium
  • Layered, flat cell structure
  • Stratum corneum (external layer)
  • composed of many layers of dead cells containing the protein keratin
  • waterproof and provides resistance to invasion of the body by microorganisms
  • continuously renews itself by sloughing off cells, which are replaced by keratinized epithelial cells from deeper layers
  • Stratum germinativum (beneath the stratum corneum)
  • where skin cells replicate through mitosis and where keratin is produced
  • cells from the germinativum layer migrate upward to the surface, away from the capillary beds that nourish the skin. As they lose contact with capillaries, the cells die and form the layers of the corneum
  • Directly underlies the stratum germinativum of the epidermis
  • Contains the blood vessels, nerve endings, sebaceous glands (which secrete oils), and sweat glands.
  • The sweat glands secrete water and ions in response to high temperatures and sympathetic stimulation serving to maintain a stable body temperature and optimal balance of sodium and chloride ions in the body.
  • Primarily adipose tissue
  • Also called hypodermis
  • A protective, insulating layer of fat, or adipose tissue
  • 1) rRNA, made via transcription,
  • 2) tRNA, made via transcription,
  • 3) polypeptide, made via transcription and translation. The polypeptide for which a gene codes might represent a discrete functional protein or one subunit of a protein, which is functional when all of its subunits are fully assembled. As a general rule, one gene codes for one polypeptide, but remember there may be different forms of the polypeptide in eukaryotes due to alternative splicing. In most eukaryotic organisms, most genes code for peptides.
  • regulate the function of other DNA sequences that do code directly for polypeptides
  • in eukaryotes, there are also non-protein-coding DNA sequences called introns
  • some chromosomal DNA sequences serve no (presently) known function at all
  • only about 1 percent of the DNA found on a given human chromosome directly codes for polypeptide formation -- all genes contain chromosomal DNA sequences, but not all chromosomal DNA sequences constitute genes.
  • 1) Genes are composed of DNA on chromosomes and can code for one of three final gene products : rRNA, tRNA, or a polypeptide.
  • 2) The proteins formed via transcription and translation are encoded by DNA on the chromosomes.
  • 3) An organism's genetic traits are traceable, largely due to the proteins formed by its cells via the processes of transcription and translation.
  • Recall that homologous chromosomes line up on the metaphase plate, across from each other, in meiotic metaphase I. During anaphase I, homologous pairs of chromosomes separate so that each of the two daughter takes one member of each homologous pair. For every genetic locus on every chromosome of the diploid parent cell, the gamete obtains one allele. Because the homologous chromosomes are separated into separate daughter cells at the end of meiosis I, the cell is haploid from this point onward.
  • So, the law of segregation:
  • --every individual possess a pair of alleles for any particular trait and that each parent passes a randomly selected copy (allele) of only one of these to its offspring.
  • It reflects the phenomenon that at anaphase I, each homologous pair separates independently to the manner in which any other homologous pair separates.
  • --separate genes for separate traits are passed independently of one another from parents to offspring.
  • 1) Dominant alleles are denoted by a capital letter and recessive traits are denoted by a lower case letter.
  • 2) When an individual is either homozygous for the dominant allele (such as TT) or heterozygous (such as Tt), they express the dominant phenotype (such as T positive).
  • 3) When an individual is homozygous for the recessive allele (such as tt), they express the recessive trait (such as T negative).
  • If two individuals of known genotype mate, a Punnett square can be drawn to show possible combinations. Consider the trait of eye color and assume that it manifests via classical dominance the dominate B allele codes for brown eyes and the recessive b allele codes for blue eyes. If a mating occurs between a heterozygous individual and a homozygous recessive individual, the Punnett sequence would look like this:
  • Possible gametes
  • from Bb parent B b
  • Possible gametes b | Bb | bb |
  • from bb parent b | Bb | bb |
  • B = brown eye color allele
  • b = blue eye color allele
  • For the offspring, the Punnett square shows two possible genotypes : Bb and bb.
  • Further, it shows that in ideal statistical terms,
  • fifty percent of offspring will have Bb genotype
  • fifty percent of offspring will have bb genotype
  • fifty percent of offspring will be brown-eyed
  • fifty percent of offspring will be blue-eyed
  • 1) 100% Aa | 100% dominant A phenotype
  • 2) 50% AA | 100% dominant
  • 50% Aa | A phenotype
  • 3) 50% Aa | 50% dominant A phenotype
  • 50% aa | 50% recessive a phenotype
  • 4) 25% AA |
  • 50% Aa | 75% dominant A phenotype
  • 25% aa | 25% recessive a phenotype
  • This is also called a 1:2:1 ratio
  • 1) Genotypic = 25% of each (AaBb, Aabb, aaBb, aabb), Phenotypic = 25% of each (A and B, A and b, a and B, a and b
  • This cross is also called a testcross, because one of the individuals is homozygous recessive.
  • 2) Genotypic = It is too complicated to predict the genotypic ratios in a heterozygous dihybrid cross. Instead, break the cross down into the two separate genes, calculate the probabilities associated with each, and then multiply these two numbers, since you want to include the results from gene 1 and gene 2 in the final answer. Phenotypic = 9:3:3:1 ratio (9 offspring have the A and B phenotypes, 3 A and b, 3 a and B, 1 a and b)
  • B. is correct. If E = brown eyes and e = blue eyes, and H = brown hair and h = blond hair, the parental cross is EeHh x EeHh. Starting with the eye color locus, if the son has brown eyes, he must have one E allele. If he is able to have glue-eyed children, he must also have one e allele. Therefore, we want to find the probability of the son being Ee given the parental cross Ee x Ee for this locus. This is 1/2.
  • Next, analyze the hair color locus. He must have an H allele to have brown hair. Since this is the only information given, the second allele could be either h or H. Given the parental crossing of Hh x Hh, the probability of the son having Hh or HH is 3/4.
  • Finally, the probability that the child will be a boy is 1/2.
  • Since the child must be male, and have brown hair, and have brown eyes (but be able to have blue-eyed children), the rules of probabilities says to multiply the three results.
  • The overall probability:
  • 1/2 x 3/4 x 1/2 = 3/16.
  • Note : if the question have asked for probability of the offspring having brown hair or brown eyes, the results would have needed to be added, not multiplied.
  • Alternative alleles do not always (or even usually) interact to exhibit classic dominance. For some traits, alleles interact to produce an intermediate phenotype or a blended phenotype. In this case, the trait is said to exhibit incomplete dominance.
  • For example, if flower color in plants exhibits incomplete dominance, and plants with an RR genotype have red flowers, and plants with an rr genotype have white flowers, plants with an Rr genotype would have pink flowers.
  • Two different alleles for the same locus might express themselves not as an intermediate phenotype, but as two distinct phenotypes both present in a single individual. Alleles for the human blood groups (which determine your blood type) exhibit this form of interaction, known as co-dominance.
  • Human blood is commonly typed as A, B or O, and either positive or negative. Blood type is determined by the expression of antigens on the surface of red blood cells (also called erythrocytes). An antigen is a molecule that is recognized by an antibody.
  • The blood group (A, B, AB or O) is governed by three alleles designated I A , I B , and i at one locus, (1) the I A allele codes for an enzyme that adds the sugar galactosamine to the lipids on the surface of red blood. People with this allele express antigen A on their erythrocytes, (2) the I B allele codes for an enzyme that adds the sugar galactose to the lipids on the surface of red blood cells. People with this allele express antigen B on their erythrocytes, and (3) the i allele codes for a protein that does not add any sugar to the surface of red blood cells.
  • An individual of genotype (1) I A I A or I A i shows the addition of galactosamine to the surface lipids of the red blood cells (or expresses antigen A) this person has blood type A, (2) I B I B or I B i shows the addition of galactose to the surface lipids of the red blood cells (or expresses antigen B) this person has blood type B, (3) I A I B shows the addition of both galactose and galactosamine to the surface lipids of the red blood cells (or expresses both antigens A and B) this person has blood type AB, and (4) ii shows the addition of no sugar to the surface lipids of the red blood cells (expresses neither antigen A nor antigen B) this person has type O blood.
  • Note : the alleles I A and I B exhibit co-dominance. The individual who carries both alleles exhibits the trait tied to each. Neither allele is recessive in relation to the other. At the same time, however, the i allele, which does not code for the addition of any sugar on the red blood cell surface, is recessive in relation to both the I A and I B alleles. Only the genotype ii produces that phenotype.
  • Positive or negative blood type is determined by a separate gene called the Rh factor or antigen, which exhibits classical dominance. If an individual expresses antigen D, they have positive blood. If they do not express antigen D, they have negative blood.
  • A process in which genetic information on one chromosome is moved to a
  • chromosome that belongs to some other cell, or a
  • different chromosome within the same cell (often a homologous chromosome.)
  • Example : The human immunodeficiency virus (HIV) transfers a segment of its genome to a human chromosome, and the human cell thereby undergoes (severely pathologic) genetic recombination.
  • An instance of genetic recombination
  • During the synapsis phase of the first meiotic division, homologous chromosomes joined on the spindle apparatus undergo breakage and an exchange of genetic information. A physical "piece" breaks from each chromosome of the pair and "crosses over" to become integrated into its counterpart.
  • Mutation refers to the actual alteration of the DNA sequence on the chromosome.
  • Often mutation produces effects deleterious to the organism's function, adaptability, and survival. Much more rarely, it leads to improvement in the organism's adaptability. In some cases, mutation confers on the organism neither an advantage nor a disadvantage.
  • point mutation : when one nucleotide unit is substituted for another
  • frameshift mutation : when one or more nucleotides are added or deleted (note that the addition or deletion of a multiple of three nucleotides does not cause a frameshift mutation)
  • silent mutation
  • missense mutation (missense mutation tend to be more serious if the occur at an important location of the protein, such as the active site of an enzyme)
  • nonsense mutation

Refer to the genetic code. If a DNA triplet that codes for the mRNA codon CGC undergoes a point mutation, after which the triplet codes for the mRNA codon CGU, the result, in terms of polypeptide synthesis, will be:
2nd - G
1st - C || Arginine || C - 3rd
Arginine || U

A. the replacement of a histidine residue with a glutamine residue.
B. none.
C. the replacement of a valine residue with a leucine residue.
D. the shifting of the reading frame in which RNA polymerase and the ribosome assess mRNA codons.

  • They are passed onto offspring from the mother and only the mother.
  • This is because spermatozoom pass only twenty-three nuclear chromosomes to the zygote. The ovum also donates twenty-three nuclear chromosomes and all the other cellular components, including organelles. Since there is a very small and separate genome in the matrix of the mitochondria, all individuals inherit their mitochondrial genome (and any associated traits) from their mother. These traits are not recessive or dominant since there is only one copy it is either present or absent.
  • A male who carries a sex-linked (X-linked) recessive gene will be positive for the relevant phenotype because they only have one X chromosome. Males who are positive received the trait from their mother and will always pass the recessive gene to their daughter.
  • A female who is homozygous for a sex-linked recessive trait will be positive for the relevant phenotype, but who are heterozygous for a recessive sex-linked will be negative and will be called a carrier of the trait.
  • 1. Check for Y-linked and mitochondrial inheritance. Both of these have very distinct patterns that can be easily spotted.
  • 2. Check for skipped generations. Is there a pattern such as "affected grandma, unaffected dad, affected son"? If so, the trait is probably recessive. If not, it is likely dominant.
  • 3. Check the ratio of affected males and affected females in this family. If there are approximately equal numbers, it suggests the trait is autosomal. If there are more males affected, the trait is likely X-linked. It is very rare for a trait to affect females more than males on the MCAT

The Punnett square presented below indicates that:

A. the parents produced four offspring.
B. on average, mating of a heterozygous and a homozygous recessive parent yields equal numbers of heterozygous and homozygous recessive offspring.
C. on average, mating of heterozygous parents yields a 50-50 ratio of heterozygous and homozygous offspring.
D. the probable outcome of heterozygous-homozygous matings is determined by the paternal chromosomal make-up.

Which of the following is true regarding the pedigree shown below?

A. A trait illustrated in the pedigree is autosomal and sex-linked because both affected offspring and heterozygotes have affected or heterozygous fathers, and only generation III involves a carrier mother.
B. The pedigree is incorrect because genotype for the trait should skip generations in autosomal recessive inheritance.
C. Phenotypic expression of autosomal recessive traits in offspring requires that both parents be heterozygous, or that one parent is heterozygous and the other is homozygous recessive.
D. Phenotypic expression of autosomal recessive traits in parents must result in affected offspring.
  • 1) population size is very large
  • 2) mating is random
  • 3) mutation does not occur
  • 4) the population takes in no genes from other populations
  • 5) selection does not occur
  • Simple counting tells us that the total number of alleles in this small population is 8. There are 3T alleles and 5 t alleles.
  • The frequency of the dominant (T) allele:
  • 5/8 = 0.625
  • The frequency of the recessive (t) allele:
  • 3/8 = 0.375
  • 0.625 + 0.375 = 1
  • Hardy-Weinberg law:
  • p + q = 1
  • (p + q) 2 = 1 2
  • p 2 + 2pq +q 2 = 1
  • p 2 = frequency of homozygous dominant
  • q 2 = frequency of homozygous recessive
  • 2pq = frequency of heterozygous genotype
  • Frequency of the homozygous recessive genotype:
  • q 2 = 160,000/1,000,000 = 0.16
  • Frequency of the recessive allele:
  • q = √0.16 = 0.4
  • Frequency of the dominant allele:
  • p + (0.4) = 1 p=0.6
  • Frequency of the homozygous dominant genotype:
  • p 2 = (0.6) 2 =0.36
  • Frequency of the heterozygous genotype:
  • 2pq = 2(0.4)(0.6) = 0.48
  • Check the math:
  • 0.48 + 0.36 + 0.16 = 1.0

kingdom, phylum, class, order, family, genus, and species

King Phylum Courts Ordinary Farm GirlS


Water has the ability to stabilize temperature. Its ability to stabilize temperature comes from its high specific heat. According to Campbell Biology “the sp.

Furthermore, as catalase, a biological enzyme, denatures at high temperatures, it is essential for us to keep the catalase at lower temperatures. By utilizin.

First off, water has a very high boiling point, one hundred degrees Celsius, relative to its small molecular weight at roughly eighteen grams. Water requires.

Salting out Salting out is used for the purification process which is apply on the basis of protein solubility. It actually relies on the principle that muc.

Diastereo isomeers - It the configurational changes considered C2, C3, or C4 in glucose. Example: Mannose, galactose. Annomerism - It is the spatial configur.

The structural and molecular biology of type I galactosemia: Enzymology of galactose 1-phosphate uridylyltransferase. 2011, 63, 694-700. Figure 3. Mechanism .

This characteristic is very important as it contributes to the moderation of the earth’s temperature and the moderates the internal temperature of all living.

It is also called anabolism or biogenesis. It is a enzyme-catalyzed method in which substrates are transformed into complex products. In this process simple.

Salivary amylase works to hydrolyze the – 1, 4 –glycosidic linkages between the maltose and glucose monosaccharides in starch (Tracey et al. 2016).

Glycolysis takes place outside the mitochondria in the cytoplasm. Glycolysis breaks down a molecule of glucose which has six carbon, down into two molecules .


How does RNA polymerase II CTD bind to the RNA modification proteins if the tail is flexible? - Biology

) Garrod hypothesized that ʺinborn errors of metabolismʺ such as alkaptonuria occur because

  1. A) genes dictate the production of specific enzymes, and affected individuals have genetic defects that cause them to lack certain enzymes.
  2. B) enzymes are made of DNA, and affected individuals lack DNA polymerase.
  3. C) many metabolic enzymes use DNA as a cofactor, and affected individuals have mutations that prevent their enzymes from interacting efficiently with DNA.
  4. D) certain metabolic reactions are carried out by ribozymes, and affected individuals lack key splicing factors.
  5. E) metabolic enzymes require vitamin cofactors, and affected individuals have significant nutritional deficiencies.
  1. A) genes dictate the production of specific enzymes, and affected individuals have genetic defects that cause them to lack certain enzymes.

The following questions refer to Figure 17.1, a simple metabolic pathway:

2) According to Beadle and Tatumʹs hypothesis, how many genes are necessary for this pathway?

E) It cannot be determined from the pathway.

  1. A mutation results in a defective enzyme A. Which of the following would be a consequence of that mutation? A) an accumulation of A and no production of B and C B) an accumulation of A and B and no production of C C) an accumulation of B and no production of A and C D) an accumulation of B and C and no production of A E) an accumulation of C and no production of A and B
  1. 4) If A, B, and C are all required for growth, a strain that is mutant for the gene encoding enzyme A would be able to grow on which of the following media? A) minimal medium B) minimal medium supplemented with nutrient ʺAʺ only C) minimal medium supplemented with nutrient ʺBʺ only D) minimal medium supplemented with nutrient ʺCʺ only E) minimal medium supplemented with nutrients ʺAʺ and ʺCʺ
  1. If A, B, and C are all required for growth, a strain mutant for the gene encoding enzyme B would be capable of growing on which of the following media? A) minimal medium B) minimal medium supplemented with ʺAʺ only C) minimal medium supplemented with ʺBʺ only D) minimal medium supplemented with ʺCʺ only E) minimal medium supplemented with nutrients ʺAʺ and ʺBʺ
  1. The nitrogenous base adenine is found in all members of which group? A) proteins, triglycerides, and testosterone B) proteins, ATP, and DNA C) ATP, RNA, and DNA D) alpha glucose, ATP, and DNA E) proteins, carbohydrates, and ATP
  1. Using RNA as a template for protein synthesis instead of translating proteins directly from the DNA is advantageous for the cell because
    1. A) RNA is much more stable than DNA.
    2. B) RNA acts as an expendable copy of the genetic material.
    3. C) only one mRNA molecule can be transcribed from a single gene, lowering the potential rate of gene expression.
    4. D) tRNA, rRNA and others are not transcribed.
    5. E) mRNA molecules are subject to mutation but DNA is not.
    1. If proteins were composed of only 12 different kinds of amino acids, what would be the smallest possible codon size in a genetic system with four different nucleotides? A) 1 B) 2 C) 3 D) 4 E) 12
    1. The enzyme polynucleotide phosphorylase randomly assembles nucleotides into a polynucleotide polymer. You add polynucleotide phosphorylase to a solution of adenosine triphosphate and guanosine triphosphate. How many artificial mRNA 3 nucleotide codons would be possible? A) 3 B) 4 C) 8 D) 16 E) 64
    1. A particular triplet of bases in the template strand of DNA is 5ʹ AGT 3ʹ. The corresponding codon for the mRNA transcribed is A) 3ʹ UCA 5ʹ. B) 3ʹ UGA 5ʹ. C) 5ʹ TCA 3ʹ. D) 3ʹACU 5ʹ. E) either UCA or TCA, depending on wobble in the first base.

    A possible sequence of nucleotides in the template strand of DNA that would code for the polypeptide sequence phe-leu-ile-val would be

    A) 5ʹ TTG-CTA-CAG-TAG 3ʹ. B) 3ʹ AAC-GAC-GUC-AUA 5ʹ.

    C) 5ʹ AUG-CTG-CAG-TAT 3ʹ. D) 3ʹ AAA-AAT-ATA-ACA 5ʹ. E) 3ʹ AAA-GAA-TAA-CAA 5ʹ.

    1. 12) What amino acid sequence will be generated, based on the following mRNA codon sequence? 5ʹ AUG-UCU-UCG-UUA-UCC-UUG 3ʹ A) met-arg-glu-arg-glu-arg B) met-glu-arg-arg-gln-leu C) met-ser-leu-ser-leu-ser D) met-ser-ser-leu-ser-leu E) met-leu-phe-arg-glu-glu
    1. A peptide has the sequence NH2-phe-pro-lys-gly-phe-pro-COOH. Which of the following sequences in the coding strand of the DNA could code for this peptide? A) 3ʹ UUU-CCC-AAA-GGG-UUU-CCC B) 3ʹ AUG-AAA-GGG-TTT-CCC-AAA-GGG C) 5ʹ TTT-CCC-AAA-GGG-TTT-CCC D) 5ʹ GGG-AAA-TTT-AAA-CCC-ACT-GGG E) 5ʹ ACT-TAC-CAT-AAA-CAT-TAC-UGA
    1. What is the sequence of a peptide based on the following mRNA sequence? 5ʹ . . . UUUUCUUAUUGUCUU 3ʹ A) leu-cys-tyr-ser-phe B) cyc-phe-tyr-cys-leu C) phe-leu-ile-met-val D) leu-pro-asp-lys-gly E) phe-ser-tyr-cys-leu
    1. The genetic code is essentially the same for all organisms. From this, one can logically assume all of the following except A) a gene from an organism could theoretically be expressed by any other organism. B) all organisms have a common ancestor. C) DNA was the first genetic material. D) the same codons in different organisms usually translate into the same amino acids. E) different organisms have the same number of different types of amino acids.
    1. The ʺuniversalʺ genetic code is now known to have exceptions. Evidence for this could be found if which of the following is true?
      1. A) If UGA, usually a stop codon, is found to code for an amino acid such as tryptophan (usually coded for by UGG only).
      2. B) If one stop codon, such as UGA, is found to have a different effect on translation than another stop codon, such as UAA.
      3. C) If prokaryotic organisms are able to translate a eukaryotic mRNA and produce the same polypeptide.
      4. D) If several codons are found to translate to the same amino acid, such as serine.
      5. E) If a single mRNA molecule is found to translate to more than one polypeptide when there are two or more AUG sites.
      1. A) If UGA, usually a stop codon, is found to code for an amino acid such as tryptophan (usually coded for by UGG only).
      1. Which of the following nucleotide triplets best represents a codon? A) a triplet separated spatially from other triplets B) a triplet that has no corresponding amino acid C) a triplet at the opposite end of tRNA from the attachment site of the amino acid D) a triplet in the same reading frame as an upstream AUG E) a sequence in tRNA at the 3ʹ end
      1. Which of the following is true for both prokaryotic and eukaryotic gene expression? A) After transcription, a 3ʹ poly-A tail and a 5ʹ cap are added to mRNA. B) Translation of mRNA can begin before transcription is complete. C) RNA polymerase binds to the promoter region to begin transcription. D) mRNA is synthesized in the 3ʹ → 5ʹ direction. E) The mRNA transcript is the exact complement of the gene from which it was copied.
      1. In which of the following actions does RNA polymerase differ from DNA polymerase?
        1. A) RNA polymerase uses RNA as a template, and DNA polymerase uses a DNA template.
        2. B) RNA polymerase binds to single-stranded DNA, and DNA polymerase binds to double-stranded DNA.
        3. C) RNA polymerase is much more accurate than DNA polymerase.
        4. D) RNA polymerase can initiate RNA synthesis, but DNA polymerase requires a primer to initiate DNA synthesis.
        5. E) RNA polymerase does not need to separate the two strands of DNA in order to synthesize an RNA copy, whereas DNA polymerase must unwind the double helix before it can replicate the DNA.
        1. D) RNA polymerase can initiate RNA synthesis, but DNA polymerase requires a primer to initiate DNA synthesis.
        1. Which of the following statements best describes the termination of transcription in prokaryotes?
          1. A) RNA polymerase transcribes through the polyadenylation signal, causing proteins to associate with the transcript and cut it free from the polymerase.
          2. B) RNA polymerase transcribes through the terminator sequence, causing the polymerase to fall off the DNA and release the transcript.
          3. C) RNA polymerase transcribes through an intron, and the snRNPs cause the polymerase to let go of the transcript.
          4. D) Once transcription has initiated, RNA polymerase transcribes until it reaches the end of the chromosome.
          5. E) RNA polymerase transcribes through a stop codon, causing the polymerase to stop advancing through the gene and release the mRNA.
          1. B) RNA polymerase transcribes through the terminator sequence, causing the polymerase to fall off the DNA and release the transcript.
          1. RNA polymerase moves in which direction along the DNA? A) 3ʹ → 5ʹ along the template strand B) 3ʹ → 5ʹ along the coding (sense) strand C) 5ʹ → 3ʹ along the template strand D) 3ʹ → 5ʹ along the coding strand E) 5ʹ → 3ʹ along the double-stranded DNA
          1. RNA polymerase in a prokaryote is composed of several subunits. Most of these subunits are the same for the transcription of any gene, but one, known as sigma, varies considerably. Which of the following is the most probable advantage for the organism of such sigma switching?
            1. A) It might allow the transcription process to vary from one cell to another.
            2. B) It might allow the polymerase to recognize different promoters under certain environmental conditions.
            3. C) It could allow the polymerase to react differently to each stop codon.
            4. D) It could allow ribosomal subunits to assemble at faster rates.
            5. E) It could alter the rate of translation and of exon splicing.
            1. B) It might allow the polymerase to recognize different promoters under certain environmental conditions.
            1. Which of these is the function of a poly (A) signal sequence?
              1. A) It adds the poly (A) tail to the 3ʹ end of the mRNA.
              2. B) It codes for a sequence in eukaryotic transcripts that signals enzymatic cleavage
              1. 24) In eukaryotes there are several different types of RNA polymerase. Which type is involved in transcription of mRNA for a globin protein? A) ligase B) RNA polymerase I C) RNA polymerase II D) RNA polymerase III E) primase
              1. Transcription in eukaryotes requires which of the following in addition to RNA polymerase? A) the protein product of the promoter B) start and stop codons C) ribosomes and tRNA D) several transcription factors (TFs) E) aminoacyl synthetase
              1. A part of the promoter, called the TATA box, is said to be highly conserved in evolution. Which might this illustrate? A) The sequence evolves very rapidly. B) The sequence does not mutate. C) Any mutation in the sequence is selected against. D) The sequence is found in many but not all promoters. E) The sequence is transcribed at the start of every gene.
              1. The TATA sequence is found only several nucleotides away from the start site of transcription. This most probably relates to which of the following? A) the number of hydrogen bonds between A and T in DNA B) the triplet nature of the codon C) the ability of this sequence to bind to the start site D) the supercoiling of the DNA near the start site E) the 3-dimensional shape of a DNA molecule
              1. Which of the following help(s) to stabilize mRNA by inhibiting its degradation? A) TATA box B) spliceosomes C) 5ʹ cap and poly (A) tail D) introns E) RNA polymerase
              1. 29) What is a ribozyme?
                1. A) an enzyme that uses RNA as a substrate
                2. B) an RNA with enzymatic activity
                3. C) an enzyme that catalyzes the association between the large and small ribosomal subunits
                4. D) an enzyme that synthesizes RNA as part of the transcription process
                5. E) an enzyme that synthesizes RNA primers during DNA replication
                1. What are the coding segments of a stretch of eukaryotic DNA called? A) introns B) exons C) codons D) replicons E) transposons
                1. A transcription unit that is 8,000 nucleotides long may use 1,200 nucleotides to make a protein consisting of approximately 400 amino acids. This is best explained by the fact that A) many noncoding stretches of nucleotides are present in mRNA. B) there is redundancy and ambiguity in the genetic code. C) many nucleotides are needed to code for each amino acid. D) nucleotides break off and are lost during the transcription process. E) there are termination exons near the beginning of mRNA.
                1. Once transcribed, eukaryotic mRNA typically undergoes substantial alteration that includes A) union with ribosomes. B) fusion into circular forms known as plasmids. C) linkage to histone molecules. D) excision of introns. E) fusion with other newly transcribed mRNA.
                1. Introns are significant to biological evolution because A) their presence allows exons to be shuffled. B) they protect the mRNA from degeneration. C) they are translated into essential amino acids. D) they maintain the genetic code by preventing incorrect DNA base pairings. E) they correct enzymatic alterations of DNA bases.
                1. A mutation in which of the following parts of a gene is likely to be most damaging to a cell? A) intron B) exon C) 5ʹ UTR D) 3ʹ UTR E) All would be equally damaging.
                1. Which of the following is (are) true of snRNPs? A) They are made up of both DNA and RNA. B) They bind to splice sites at each end of the exon. C) They join together to form a large structure called the spliceosome. D) They act only in the cytosol. E) They attach introns to exons in the correct order.
                1. During splicing, which molecular component of the spliceosome catalyzes the excision reaction? A) protein B) DNA C) RNA D) lipid E) sugar
                1. Alternative RNA splicing A) is a mechanism for increasing the rate of transcription. B) can allow the production of proteins of different sizes from a single mRNA. C) can allow the production of similar proteins from different RNAs. D) increases the rate of transcription. E) is due to the presence or absence of particular snRNPs.
                1. In the structural organization of many eukaryotic genes, individual exons may be related to which of the following? A) the sequence of the intron that immediately precedes each exon B) the number of polypeptides making up the functional protein C) the various domains of the polypeptide product D) the number of restriction enzyme cutting sites E) the number of start sites for transcription
                1. Each eukaryotic mRNA, even after post-transcriptional modification, includes 5ʹ and 3ʹ UTRs. Which are these? A) the cap and tail at each end of the mRNA B) the untranslated regions at either end of the coding sequence C) the U attachment sites for the tRNAs D) the U translation sites that signal the beginning of translation E) the U — A pairs that are found in high frequency at the ends
                1. In an experimental situation, a student researcher inserts an mRNA molecule into a eukaryotic cell after he has removed its 5ʹ cap and poly(A) tail. Which of the following would you expect him to find? A) The mRNA could not exit the nucleus to be translated. B) The cell recognizes the absence of the tail and polyadenylates the mRNA. C) The molecule is digested by restriction enzymes in the nucleus. D) The molecule is digested by exonucleases since it is no longer protected at the 5ʹ end. E) The molecule attaches to a ribosome and is translated, but more slowly.
                1. A particular triplet of bases in the coding sequence of DNA is AAA. The anticodon on the tRNA that binds the mRNA codon is A) TTT. B) UUA. C) UUU. D) AAA. E) either UAA or TAA, depending on first base wobble.
                1. Accuracy in the translation of mRNA into the primary structure of a polypeptide depends on specificity in the A) binding of ribosomes to mRNA. B) shape of the A and P sites of ribosomes. C) bonding of the anticodon to the codon. D) attachment of amino acids to tRNAs. E) both C and D
                1. A part of an mRNA molecule with the following sequence is being read by a ribosome: 5ʹ CCG-ACG 3ʹ (mRNA). The following charged transfer RNA molecules (with their anticodons shown in the 3ʹ to 5ʹ direction) are available. Two of them can correctly match the mRNA so that a dipeptide can form.

                tRNA Anticodon Amino Acid

                The dipeptide that will form will be A) cysteine-alanine.

                B) proline-threonine. C) glycine-cysteine. D) alanine-alanine.

                What type of bonding is responsible for maintaining the shape of the tRNA molecule? A) covalent bonding between sulfur atoms

                B) ionic bonding between phosphates
                C) hydrogen bonding between base pairs
                D) van der Waals interactions between hydrogen atoms

                E) peptide bonding between amino acids

                C) hydrogen bonding between base pairs

                Figure 17.4 represents tRNA that recognizes and binds a particular amino acid (in this instance, phenylalanine). Which codon on the mRNA strand codes for this amino acid?

                A) UGG B) GUG C) GUA D) UUC E) CAU

                1. The tRNA shown in Figure 17.4 has its 3ʹ end projecting beyond its 5ʹ end. What will occur at this 3ʹ end? A) The codon and anticodon complement one another. B) The amino acid binds covalently. C) The excess nucleotides (ACCA) will be cleaved off at the ribosome. D) The small and large subunits of the ribosome will attach to it. E) The 5ʹ cap of the mRNA will become covalently bound.
                1. A mutant bacterial cell has a defective aminoacyl synthetase that attaches a lysine to tRNAs with the anticodon AAA instead of a phenylalanine. The consequence of this for the cell will be that
                  1. A) none of the proteins in the cell will contain phenylalanine.
                  2. B) proteins in the cell will include lysine instead of phenylalanine at amino acid positions specified by the codon UUU.
                  3. C) the cell will compensate for the defect by attaching phenylalanine to tRNAs with lysine-specifying anticodons.
                  4. D) the ribosome will skip a codon every time a UUU is encountered.
                  5. E) None of the above will occur the cell will recognize the error and destroy the tRNA.
                  1. B) proteins in the cell will include lysine instead of phenylalanine at amino acid positions specified by the codon UUU.
                  1. There are 61 mRNA codons that specify an amino acid, but only 45 tRNAs. This is best explained by the fact that A) some tRNAs have anticodons that recognize four or more different codons. B) the rules for base pairing between the third base of a codon and tRNA are flexible. C) many codons are never used, so the tRNAs that recognize them are dispensable. D) the DNA codes for all 61 tRNAs but some are then destroyed. E) competitive exclusion forces some tRNAs to be destroyed by nucleases.
                  1. From the following list, which is the first event in translation in eukaryotes? A) elongation of the polypeptide B) base pairing of activated methionine-tRNA to AUG of the messenger RNA C) the larger ribosomal subunit binds to smaller ribosomal subunits D) covalent bonding between the first two amino acids E) the small subunit of the ribosome recognizes and attaches to the 5ʹ cap of mRNA
                  1. Choose the answer that has these events of protein synthesis in the proper sequence. 1. An aminoacyl-tRNA binds to the A site. 2. A peptide bond forms between the new amino acid and a polypeptide chain. 3. tRNA leaves the P site, and the P site remains vacant. 4. A small ribosomal subunit binds with mRNA. 5. tRNA translocates to the P site. A) 1, 3, 2, 4, 5 B) 4, 1, 2, 5, 3 C) 5,4,3,2,1 D) 4, 1, 3, 2, 5 E) 2, 4, 5, 1, 3
                  1. As a ribosome translocates along an mRNA molecule by one codon, which of the following occurs? A) The tRNA that was in the A site moves into the P site. B) The tRNA that was in the P site moves into the A site. C) The tRNA that was in the A site moves to the E site and is released. D) The tRNA that was in the A site departs from the ribosome via a tunnel. E) The polypeptide enters the E site.
                  1. What are polyribosomes? A) groups of ribosomes reading a single mRNA simultaneously B) ribosomes containing more than two subunits C) multiple copies of ribosomes associated with giant chromosomes D) aggregations of vesicles containing ribosomal RNA E) ribosomes associated with more than one tRNA
                  1. Which of the following is a function of a signal peptide? A) to direct an mRNA molecule into the cisternal space of the ER B) to bind RNA polymerase to DNA and initiate transcription C) to terminate translation of the messenger RNA D) to translocate polypeptides across the ER membrane E) to signal the initiation of transcription
                  1. When translating secretory or membrane proteins, ribosomes are directed to the ER membrane by
                    1. A) a specific characteristic of the ribosome itself, which distinguishes free ribosomes from bound ribosomes.
                    2. B) a signal-recognition particle that brings ribosomes to a receptor protein in the ER membrane.
                    3. C) moving through a specialized channel of the nucleus.
                    4. D) a chemical signal given off by the ER.
                    5. E) a signal sequence of RNA that precedes the start codon of the message.
                    1. When does translation begin in prokaryotic cells? A) after a transcription initiation complex has been formed B) as soon as transcription has begun C) after the 5ʹ caps are converted to mRNA D) once the pre-mRNA has been converted to mRNA E) as soon as the DNA introns are removed from the template
                    1. When a tRNA molecule is shown twisted into an L shape, the form represented is A) its linear sequence. B) its 2-dimensional shape. C) its 3-dimensional shape. D) its microscopic image.

                    An experimenter has altered the 3ʹ end of the tRNA corresponding to the amino acid methionine in such a way as to remove the 3ʹ AC. Which of the following hypotheses describes the most likely result?

                    A) tRNA will not form a cloverleaf.
                    B) The nearby stem end will pair improperly.

                    C) The amino acid methionine will not bind.
                    D) The anticodon will not bind with the mRNA codon.

                    E) The aminoacylsynthetase will not be formed.

                    C) The amino acid methionine will not bind.

                    A transfer RNA (#1) attached to the amino acid lysine enters the ribosome. The lysine binds to the growing polypeptide on the other tRNA (#2) in the ribosome already.

                    1. 59) Which enzyme causes a covalent bond to attach lysine to the polypeptide? A) ATPase B) lysine synthetase C) RNA polymerase D) ligase E) peptidyl transferase

                    A transfer RNA (#1) attached to the amino acid lysine enters the ribosome. The lysine binds to the growing polypeptide on the other tRNA (#2) in the ribosome already.


                    Acknowledgements

                    We thank Saskia Hutten and Hilary Wunderlich for critically reading the manuscript and Manuela Neumann for providing figure material. DD is supported by the Deutsche Forschungsgemeinschaft (DFG, German Research Foundation) within the framework of the Munich Cluster for Systems Neurology (EXC 1010 SyNergy) and Emmy Noether program DO 1804/1-1 and the Junior Researcher Fund of the Ludwig-Maximilians-Universität München. HE is supported by a PhD fellowship of the Hans-and-Ilse Breuer Foundation.


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