Answer key, practice problems 2000--week 5
| 1. | (a) |
The primers that will work are #s (ii) and (iv). Primer (ii) anneals to the bottom strand, near the left end, and primes "rightward" synthesis, while primer (iv) anneals to the top strand near the right end, and primes "leftward" synthesis. [Remember that DNA polymerase adds to the 3' end of the primer. So primer # (i), for instance, will hybridize to the top strand, but the 3' end of the primer will be "pointing" away from the target region--leftward, off the left end of the DNA.] |
| (b) | The notation indicates the genotype of the individual with respect to this polymorphic site. The person is heterozygous for this locus; one homolog has 3 (CA) repeats at this particular location, while the other homolog has 7 repeats -- | |
Homolog 1:
Homolog 2:
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| 2. | (i) | Candidate 3 does not have any alleles in common with the lottery winner, and so can be eliminated from the list of potential fathers. |
| (ii) | For each of the three polymorphic loci, Candidate 1 and Candidate
2 each have one allele in common with the young woman. However,
for each of the three polymorphic loci, one of the alleles that
the woman has is less frequent in the population than the other.
Therefore, the candidate who shares the less common allele with the woman is more likely to be her father. Candidate 1 has 22, 28, and 16-repeat alleles
in common with the woman at loci 1, 2, and 3, respectively, while
Candidate 2 has 28, 25, and 35-repeat alleles in common with the
woman. Because Candidate 1's alleles are much less frequent in
the population, it is much less probable that he has those alleles
just by chance--he has a stronger claim of direct relationship
with the woman.
However, we cannot with certainty conclude that he is the real father -- all the data are saying is that if we pulled a random person out of the population, it is improbable (but not impossible) that their genotype would match the woman's genotype just by chance. One way of firming up the conclusion would be to test more polymorphic sites. |
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| (iii) | If Candidate 1 is indeed the father, then the woman's 28-repeat allele at Locus 1, 25 repeat at Locus 2, and 36-repeat allele at Locus 3 must all have come from her mother. So the mother must have these three alleles; we have no information on her other allele at each of the three loci. |
| 3. | (a) | The Lod score graph tells us that the pedigree data favor a map distance of 5 cM between Gene 1 and PS1; a map distance of 15 cM between Gene 1 and PS2; a map distance of 10 cM between Gene 1 and PS3, etc. A map that is consistent with these interpretations is:
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| (b) | The assumption being made is that the apex of the curves (where the evidence in favor of linkage is statistically strongest) match the actual map distance between the loci... a reasonable assumption. Although a researcher would be well aware that the assumption might not hold, he or she would probably begin their search for the gene at that location (i.e., at that distance from the polymorphic site). |
| 4. | No, the stategy would not be expected to work–the disease phenotype is dominant, so the the wildtype allele is recessive. Introduction of the wildtype allele into the mutant cell will not reverse the disease phenotype. |
| 5. | (a) |
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| (b) | The inappropriate expression of a gene often results in a "gain of function" – all it takes is one copy of the mutated gene to give a phenotype, so the mutant phenotype is often dominant. An example is the AY allele of the agouti gene. This mutant form of the allele has a promoter mutation, causing inappropriate expression of the gene -- and one copy of the mutant phenotype is sufficient to give the yellow coat color phenotype. But clearly, it is not an absolute rule that inappropriate expression will give a dominant phenotype, because the lethality caused by the same AY allele is recessive. | |
| (c) | The small, truncated peptide may often be completely non-functional and innocuous, in which case we can expect the mutant phenotype to be recessive. However, there may instances where the truncated peptide interferes with the functioning of the full-length protein made from the normal allele. In such cases, the presence of a single mutant allele can block the function of the normal allele; the mutant is "dominant negative". |
| 6. | The experimental strategy was to detect lethal mutations on the X chromosome that was marked with the wildtype w+ allele. If crossovers were allowed, the mutation could be transferred to the other X chromosome in the female in the cross. By preenting crossovers, the lethal phenotype was forced to remain associated with the red-eye phenotype, and could therefore be detected as a deficit in red-eyed male progeny. |
| 7. | (a) | Microtubules form the spindle apparatus that segregates the chromatids at mitosis (and does the segregation in meiosis too, although we are just concerned with mitosis in this particular instance). Failure to form spindles will result in cells that reach metaphase but cannot proceed beyond that stage -- the cells are arrested at metaphase. |
| (b) | One interpretation is that by arresting cell cycle progression of the cells for period of time by treatment with nocodazole, the cells were given time to repair the DNA damage, and were therefore able to survive. In fact, this experiment was instrumental in the model that the function of checkpoints is to detect damge or incomplete processes and halt the cell cycle. |
| 8. | To mutate strain #1 so that it is not cuttable at Site B, all it takes is a single change of any type within the cut site (i.e., mutation of any one of the six base pairs in the sequence GGATCC will result in a non-cuttable sequence). In contrast, to change Site B in strain #2 to a cuttable site, one particular mutation has to occur at one particular location (GAATCC has to change to GGATCC) -- the requirement here is much more stringent, so it is not surprising that this event (the "reverse mutation" is much more rare than the cuttable --> non-cuttable change (the "forward mutation"). |