Answers to Problems added to the Problem Book in 2015.
This is what you need if you have the 20th edition, re-revised (2014). If you have the 21st edition (2015), you do not need this page. All these answers (& the corresponding problems) have been added.
Answer to 7-19, parts B & C.
B. (a) would have the largest
absolute value. One explanation: Breaking a met-tRNA bond provides enough
energy to drive synthesis of a peptide bond.
Another explanation: it takes 2 so called high energy bonds (from one ATP) to synthesize met-tRNA. It takes only one high energy bond (from met-tRNA) to synthesize a peptide bond. Therefore hydrolysis, the reverse reaction, will release more energy from breaking a met-tRNA bond than from breaking a peptide bond. (Note that it is synthesis of met-tRNA, not hydrolysis, that uses up ATP.) Making a single peptide bond is often said to take 5 high energy bonds, but that includes all the energy needed to move the mRNA & tRNA one codon relative to the ribosome. The energy for movement is not involved here. Only the energy released by hydrolysis (not synthesis) of met-tRNA or hydrolysis of a peptide bond.
C-1. You expect all anticodons
given except AUC. (See note at * below.) In this problem you are given
anticodons, not codons; you are expected to look up (not memorize) both the code
table and the wobble rules. AUC is the anticodon for stop; there is no stop tRNA.
The other 3 match up to codons for AA.
*Note: it is possible that there is no tRNA with the anticodon UAG, because a tRNA with the anticodon UAI is all you need to read all 3 ile codons. If you explained this, that was fine.
C-2. CUG. CUI can pair up with more codons, but not all 3 codons specify the same amino acid – 2 code for asp and one codes for glu. So a tRNA with the anticodon CUI would not translate the mRNA accurately. CUG can pair up with all the codons for asp, but not with any of the codons for glu.
Answers to 7R-3 & 4. (These answers are on pages 183-184 in the 21st ed.)
A-1. If transcription occurs to the right, the 5’ end of the transcript will be 5’ CAUU 3’.
A-2. If transcription occurs to the left, the 5’ end of the transcript will be 3’ CCUC 5’ or 5’CUCC 3’.
Transcription to the right
will use the bottom strand (complement to strand shown) as template, so the top
(the strand shown) will be the sense strand; transcription to the left will do
B-1. For gene A, the sense strand of the DNA is (the complement).
B-2. The corresponding protein will have the amino acid (tyr).
Explanation of C:
C-1. Look at the bottom of handout 11-3; put the promoter to the right of the origin.
C-2. The primer on the 5’ end of a linear strand can be removed, but it cannot be replaced.
D-2. (neither strand).
Explanation of D:
D-1. GAAC can be used as primer to start synthesis to the right, using the bottom strand as template; you need the complement to ATGA (anti-parallel) to act as the ‘other’ primer.
D-2. There is no discontinuous synthesis or need for ligase in PCR.
A-1. In Mutant 1, the codon (must be changed). See below for possibilities.
A-2. In Mutant 2, the codon in the mRNA could be changed to GUA which is a (mis-sense).
A-3. In Mutant 3, the codon in the mRNA could be changed to CGA AND this is a (substitution)
Explanations for A:
A-1. Even though peptide Q is not changed in mutant 1, there must be a change in the DNA & mRNA (of one base) , because the problem says so. So this must be a ‘silent’ mutation -- you must change one leu codon to another. If you change CUA to another leu codon, you still have leu in the protein. If you check the code table, you will see there are 6 leu codons. Since one and only one base must be changed in the DNA, you can change CUA to CUG, CUC, CUU or UUA and still code for leu. (This is an example of degeneracy, not wobble. ) Note that in this case, you can change either the first or third base without changing the AA encoded.
A-2. The only val codon one base away from CUA is GUA.
A-3. There are 6 arg codons, but the only arg codon one base away from CUA is CGA. (It takes 2 base changes to convert CUA to AGA or any other arg codon.)
B. Mutant (3) is most likely to be drug sensitive. (Note that you are not expected to memorize the AA structures. They would be provided on an exam. However, given the structures, you are expected to be able to identify the side chains as polar, nonpolar, charged, etc.)
Explanation for B:
Mutation 1 is silent (no change of peptide Q).
Mutation 2 switches one hydrophobic (nonpolar) side chain for another, similar one.
Mutation 3 replaces a nonpolar side chain with a polar (charged) one. This is a big difference, unlike the previous cases, and is therefore much more likely to cause a change in the structure (and therefore function) of the peptide.
Answer to 8-16
A-1. # pairs = (12) = N. A tetraploid contains 4 sets of chromosomes in its
somatic cells, and will produce gametes with 2 sets -- half the number in a
somatic cell. In other words, a tetraploid is 4N and its gametes will be 2N.
Reminder: N = number of types of chromosomes = number of different types in one
Diploids have N pairs; in other words, they are 2N. Note that A-1 asks for the number of pairs of chromosomes at pro I, not for the total number of chromosomes.
A-2. C = (5). By definition, C is the DNA content of one set of unreplicated chromosomes. C is not the total DNA content of whatever cell you are looking at – C does not depend on what stage of the cell cycle or life cycle the cell is in. C is constant for the species (here, lilies) but individual cells can contain different amounts of DNA – C, 2C, 4C, etc. Tetraploid gametes are 2N, which means they contain 2 sets of unreplicated chromosomes or 2C of DNA.
A-3. (somatic cell of a diploid in G1) and (zygote of a diploid line before S). 10 pg corresponds to the amount of DNA in 2 sets of unreplicated chromosomes = 2C. You’ll find 10 pg of DNA in any cell that is 2N, but has not yet replicated its DNA. So you’ll find 10 pg in a diploid cell in G1 (before S) or a zygote (before S).
Explanation of incorrect answers to A-3:
What rules out the other answers? After S, or in G-2, diploid cells contain 4C's worth of DNA. So does a germ cell of a diploid in pro I. In all these cases, there are 2N chromosomes, but 2 chromatids per chromosome, so the DNA content is 20 pg. Gametes and spores are haploid, so they contain only 5 pg of DNA. (A germinating spore in G2, after S, will have 2C's worth of DNA or 10 pg. However, when people speak of 'spores' they usually mean the haploid products of meiosis, before they start to germinate.)
B-1. The # of chromosomes/cell should be12.
B-2. The # of chromatids per chromosome should be 2.
After first division, you have one set of chromosomes, or N chromosomes, per cell, but each chromosome has 2 chromatids. Homologs have separated at first division, but sister chromatids will only separate at 2nd division, and 2nd div. hasn't happened yet. Note that the question asks for the # of chromatids per chromosome, not the # of chromosomes or chromatids per cell.
Answer to new 12R-5
You do not have any DNA in your in vitro system. Do you have any mRNA? (See note.)
A. Preparation (1) must come from batch L.
B. You need to do neither – the level of lactose doesn't matter.
A. Polysomes or polyribosomes are ribosomes attached to mRNA, so the polysome fraction contains mRNA, not just ribosomes, and is the only source of mRNA. That’s why the polysomes have to come from the induced cells (batch L). The remaining preparations can come from anywhere, as they will provide the same factors for translation in any case. (Even if the supernatant contains inducer or repressor protein it makes no difference – there is no DNA to induce.)
B. There is no DNA in this system, and no transcription. Although the lac operon is inducible, adding lactose makes no difference at all, because there is no DNA to induce. You have taken all the parts needed for translation (protein synthesis) and mixed them in a test tube. You do not have the parts needed for transcription. Whatever protein you get depends on whatever mRNA you already have (in the polysomes) – since there is no DNA, you can’t make more mRNA.
Note: This is an in vitro or test tube system for synthesizing protein. Once you make the preparations and mix them in a test tube, there is no possibility of induction. There is no DNA, so how the lac operon works is no longer relevant, as you are only measuring translation from pre-existing mRNA, not transcription or induction.
Answer to updated 13-3, part B.
B. (iv) & (viii) should work. The alleles themselves both lack a Dde site so they give the same # and size of pieces after Dde treatment or PCR. So one approach is to tell the alleles apart by their products. The products have about the same molecular weight but differ in charge because the R group of lys is positive and the R group of val is neutral. So HbS and HbC will have different mobilities during gel electrophoresis w/o SDS = (iv). The 2 PCR products can be distinguished by hybridization to allele-specific probes (= viii). That is, you can synthesize probes that will hybridize (under stringent conditions) to one sequence (one allele), but not to a different sequence (or allele) that is only one nucleotide different. (These days, analysis of the DNA is more common than analysis of the protein.)
Answer to new 13-16 & 13-17. (These answers are on pages 227 -228 in the 21st edition.)
A. The cDNA should hybridize to the sense strand of gene G & to the primary transcript of gene G.
Explanation. Both the primary transcript and the sense strand of the gene are complimentary to the cDNA. Both have introns, but the cDNA will hybridize to the parts with the exons, and the parts of the DNA that don’t hybridize will be looped out. To get full credit you had to figure out what was complementary to the cDNA and what was the same; you also had to take into account that both the sense strand and the primary transcript will hybridize to the cDNA even though they have introns, and won’t match up perfectly with the cDNA.
We know the cDNA is complementary to the mRNA. Anything that is complementary to the mRNA must be complementary to the sense strand, and the same as (not complementary to) the transcribed or template strand. Another way to see this – the primary transcript and the sense strand of the DNA are both the same as the mRNA, except they have introns. So both will contain sequences that are complementary to the cDNA.
B. Of the pieces generated by Rest5, 1 piece contains no exons.
The mRNA contains only exons, no introns. So the cDNA probe has sequences complementary to exons, but not to introns. Therefore, the probe hybridizes only to pieces of DNA that contain exonic sequences. If a piece is entirely derived from an intron, it won’t have any sections complementary to the cDNA probe, and the probe won’t hybridize to it. So the exon-less piece will be there, but it won’t be fluorescent. (It’s in the gel, but it doesn’t bind the fluorescent probe.) There must be two restrictions sites in an intron, not one, to produce a piece of DNA with no exons. The solution here is that one piece of DNA has no introns, and two sites for Rest 5, not that there is one restriction site in an intron.
Note that UTRs are encoded in exons – exons code for both translated and untranslated regions of the mRNA. DNA does not encode separately for exons and UTRs.
C. >900 in the mRNA and much larger in the transcribed part of the gene.
The protein is approximately 300 AA long, so you need about 900 translatable codons. However, mRNA should not be about 900 bases/nucleotides – it should be longer, because it will contain a 5’ UTR and a 3’ UTR. The primary transcript should be considerably longer than the mRNA, because the primary transcript will include all the introns, and the average intron is much longer than the average exon.
13-17. A. #6. Note the question asks about the sites at the ends of the introns, not the sites at the ends of the exons. That’s why the answer is #6, not #7. In kidney cells, exon 6 is excised along with introns 5 & 6 as one big intron. As a result, exon 5 is connected to exon 7. The splicing out of the big intron connects the 3’ end of exon 5 (which is the donor site at the 5’ end of intron 5) to the 5’ end of exon 7 (which is the acceptor site at the 3’ end of intron 6). Donor and acceptor sites for RNA splicing are defined by the ends of the introns, not by the ends of the exons. (For an explanation, it was sufficient to draw exons 5 to 7 with accompanying introns, and show where the splicing occurs.)
Answers: B-1. 160 AA; B-2. the same length; B-3. Smaller.
B-1. 30 nt corresponds to 10 codons, or 10 additional amino acids. Therefore the protein in eye cells will be 10 AA longer than the protein found in kidney cells, which is 150 AA long.
B-2. The primary transcript is the same length in the two cell types; it’s the way it’s spliced that’s different in kidney cells and eye cells. After splicing, 30 additional nt are included in the mRNA in eye cells.
B-3. In kidney cells, one large intron is spliced out as a lariat. The big intron or lariat includes introns 5 & 6 and exon 6. (See A.) In eye cells, the two introns, 5 & 6, are removed as separate (smaller) lariats. The individual lariats are smaller in eye cells, but the number of lariats is smaller in kidney cells.
Answers: C-1. both; C-2. eye cells; C-3. the mRNA (but see below); C-4. Either probe with mRNA from eye cells.
C-1. Probe 5 hybridizes to exon 5 in the mRNA. This exon is found in mRNA from both cell types.
C-2. Probe 6 hybridizes to exon 6 in the mRNA. This exon is only found in mRNA from eye cells. This exon is transcribed in both cell types, but spliced out of the primary transcript in kidney cells. (See above.)
C-3. The mRNA is separated by size on a gel first. Then the probe is added to visualize the location of the mRNA. The size of the probe does not affect the position of the band – the probe sticks and forms a visible band wherever the mRNA happens to be, and that depends on the size of the mRNA.
Some students chose ‘neither’ because they felt that a RNA this size was too long to code for a protein that contains only 160 amino acids. Therefore they reasoned that the RNA in the gel or blot must be the primary transcript, not the mRNA. Nine hundred nt may seem on the long side, but eukaryotic mRNAs often contain very long UTRs, and the problem says it is messenger RNA.
C-4. Eye cell mRNA with either probe. (See diagram below.). There are 4 possible cases here – two different mRNAs with each of two different probes. Here are the patterns of bands you will get on the blots:
Kidney mRNA Eye Cell mRNA
Probe 5 Probe 6 Probe 5 Probe 6
A B C D
You are looking for two cases that give the same pattern of labeled bands. You can see from the results above that C and D give the same pattern, but none of the other pairs do. If you want to get the same pattern in different cases, you have to use the same mRNA both times. That’s because the position of the band depends on the size (length) of the mRNA, not on the size of the probe.
The bands in A and C are in different places because the two mRNAs (from the two cell types) are of a different length. Whatever probe you use, if you compare blots of the two mRNAs with the same probe (A vs C or B vs D) you won’t get the same pattern.
Since it is the size of the mRNA that determines the position of the band, you need to use two different probes with the same mRNA (A vs B or C vs D), not two different mRNAs with the same probe. If you use kidney cell mRNA (A vs B), you’ll get one band with probe 5 and no bands with probe 6. If you use eye cell mRNA (C vs D), you’ll get one band with either probe, and both bands will be in the same spot (since the position of the band depends on the length of the mRNA, which is identical in both cases). The two different probes will bind to different parts of the mRNA, but the position of the band will not change.In this problem, many students compared the wrong blots – they compared the bands you get when using probe 5 with the two different mRNAs (A vs C) instead of comparing the two different mRNAs with the same probe (A vs B and C vs D).