Answer key, practice problems 2000--week 7

1.
The number of generations needed to give the final number of cells can be deduced from the expression:
2n = final # of cells (where n = number of generations)
n = log(final # of cells)/log(2)
(e.g., if the final # of cells = 16, 2n = 16; n = 4
For a tumor with 109 cells, n = log(109)/log(2) = 29.9 or 30 generations
The number of cell divisions = (final # of cells - 1).

[Note that the number of cell divisions is not the same as the number of cell generations. In the first generation, there is one cell dividing, so there is one division. In the second generation, there are two cells dividing, so there are two divisions in this generation; in the third generation, there are 4 cells dividing, so there are 4 divisions. So after three generations, there are 8 cells--but it took 7 divisions total to get to 8 cells -- (1 division in the 1st generation) + (2 divisions in the 2nd generation) + (4 divisions in the 3rd generation).]

2. The tumor was derived from a single cell that had one X chromosome inactivated; since X inactivation is stably propagated through mitosis, all daughters of that cell have the same inactive X.

3. (a)

i. Most mutations will result in loss of function. Keeping that in mind --

  • Gene 1: the resulting protein (if any) will be unable to activate the transcription pathway. Therefore, most mutations will be non-cancer alleles.
  • Gene 2: the resulting protein will be unable to bind Protein 3; the released Protein 3 is potentially able to enter the nucleus and activate transcription. Therefore, most mutations will be cancer-promoting alleles.
  • Gene 3: the resulting protein (if any) will be unable to activate transcription. Therefore, most mutations will be non-cancer alleles.

ii and iii. Loss-of-function alleles generally result in products with reduced or no function; gain-of-function alleles give inappropriate activity (the product is still active, but may be active at inappropriate times or locations)

  • Gene 1: It would take a gain-of-function mutation to promote cancer (not any mutation will do; only those mutations that cause inappropriate activity of the protein will be capable of promoting cancer). The allele would be dominant (all it would take is one inappropriately active form of the gene).
  • Gene 2: As described above, a loss-of-function allele would be cancer-promoting; the allele would be recessive (the normal, tumor-suppressing allele would still be functioning)
  • Gene 3: As with Gene 1, it would take a gain-of-function mutation to promote cancer; the allele would be dominant.

Genes 1 and 3 are proto-oncogenes; Gene 2 is a tumor suppressor gene.
(b)

Proto-oncogene: two alleles are present, and mutation of either one is sufficient. Therefore, the chance of an oncogenic mutation is 2 x 10-5

For the tumor suppressor, both alleles have to be mutated, so the chance is 10-5 x 10-5 = 10-10

4. (a) E2F protein should still be in the cytoplasm, because there is still a normal copy of the Rb gene. (Assumes that the Rb protein is normally in excess.)
(b) There is no Rb protein being made, so E2F protein should be in the nucleus.
(c) E2F protein should still be in the cytoplasm, because there is still a normal copy of the Rb gene.
(d) Although there is no longer Rb protein, there is no E2F either, so the cell is not in increased danger through the E2F pathway. However, it remains possible that Rb has other targets besides E2F, so there could potentially be increased risk of cancer through other pathways.
(e) Since the mutant form blocks transcription (by permanently sequestering E2F in the cytoplasm), it is not expected to increase the risk of cancer.

5. (a) Since there is no pause in the cell cycle, the cell will probably enter S phase with uncorrected base adducts (methyl groups); there will probably be an increase in the mutation rate, with decreased viability and/or an increased risk of cancer.
(b) If the function of gene xyz is to give the cell time to repair the damage, then we should be able to compensate for the loss of gene xyz by artificially causing a delay in entry into S. Since we have complete control over Rb, we could delay the phosphorylation of Rb, which would delay the entry of E2F into the nucleus and therefore delay S phase. If the hypothesis is correct, MMS-treated mutant cells with an this Rb-mediated delayed S phase should show increased viability/reduced mutation rate compared to the non-delayed mutant.

6. (a) Since they both cause the appearance of his+ colonies, they both seem to be capable of causing mutations. Substance X reverts a missense mutation, so it appears to be capable of causing base substitutions; substance Y reverts a frame-shift mutation, so it is capable of causing frame-shift mutations.
(b) Substance X becomes mutagenic upon exposure to liver extract, so it would probably cause mutations in the body after being exposed to liver enzymes in the digestive tract. Substance Y appears to be rendered harmless by liver extract, so it might not be mutagenic in the body. Therefore, substance X would likely be worse for human consumption. However, since both are capable of promoting mutations (either before or after treatment with liver extract), neither compound seems safe for human consumption.

7. (a) The black cat has to be homozygous for the black fur allele (XrXr). Since she produced calico (XRXr) kittens, the father must have been orange (XRY). All of his female progeny would inherit the orange fur allele and be calico, so the black kitten must be male (XrY), getting the black fur allele from the mother and the Y chromosome from the father.
(b) Since the calico cat (XRXr) has a female black kitten (XrXr), the father must be contributing a black fur allele; he must be a black cat (XrY).

8. (a)
  1. Constance: If no Barr bodies are formed (i.e., no X chromosome is inactivated), XX fetuses will have an excess of active X-linked genes and therefore die.
  2. Male fetuses are not expected to have inactivated X chromosomes anyway, so they are not affected.
  3. Big Bertha: If all X chromosomes become inactivated, all X-linked genes will be silenced, and the fetus fails to develop.
(b)

The defect in Constance's aborted fetuses might be in the X inactivation center--if the XIC is missing, for instance, the cell will be able to detect only one X (the one with the normal XIC) and therefore not inactivate any X chromosome. (Production of non-functional Xist RNA by one X chromosome wouldn't explain why the other X fails to be inactivated.)

A variety of defects could explain Big Bertha's phenotype. For instance, she might have a mutated XIST gene such that the XIST RNA loses specificity and binds to any X chromosome. Alternatively, the defect could be in the gene that codes for the hypothetical "protector" protein that blocks inactivation of one X chromosome per cell. If this hypothetical gene is knocked out, no X chromosome would be protected from inactivation.
(c)

Lethality in this case will not be specific to females or males--but lethality might be higher in female fetuses than in males. (If X* represents an X that has lost its XIC because of the translocation and 14;X represents the translocation form of chr 14, then females inheriting X* and 14;X from Chimney will die because they will inactivate 14;X in half their cells. In contrast, males inheriting X* and 14;X will not inactivate 14;X --that being the only XIC in the cell-- and will therefore survive. This interpretation assumes that both copies of chromosome 14 must be active for viability.)
Note that these must all be germline mutations, otherwise wouldn't expect the females (Constance and her buddies) to be alive. Alternatively, the phenotypes might not be fully penetrant.