Genetics 371B, Autumn 1999

Answer key, Problem set 5

1.

From the first two crosses (yellow x blue, yellow x red), and the 3:1 F2 ratio seen in them, it looks like a simple allelic series -- one gene with red, blue and yellow alleles, red and blue both being dominant over yellow. However, the third and fourth crosses (yellow x purple, red x blue) reveal that two genes must be involved -- the F2 ratios in each case are in sixteenths, indicating that these must be dihybrid crosses.

How do we tell which gene is doing what? Recall the components of the 9:3:3:1 ratio in F2 -- if the genes are R/r and B/b, the F2 ratio is 9/16 R_B_, 3/16 R_bb, 3/16 rrB_, and 1/16 rrbb. Therefore, B_R_ must be purple. Similarly, having dominant alleles at only one of two genes gives red (R_bb), while blue is seen if there's a dominant allele at the other locus only (rrB_). Yellow comes from being fully homozygous recesive (rrbb).

The crosses were:

Parents F1 F2
rrbb (yellow) x rrBB (blue) rrBb (all blue) rrB_ (blue), rrbb (yellow) in 3:1 ratio
rrbb (yellow) x RRbb (red) R_bb (all red) R_bb (red), rrbb (yellow) in 3:1 ratio
rrbb (yellow) x RRBB (purple) R_B_ (all purple) R_B_ (purple), R_bb (red), rrB_ (blue), rrbb (yellow) in 9:3:3:1 ratio
rrBB (blue) x RRbb (red) R_B_ (all purple) R_B_ (purple), R_bb (red), rrB_ (blue), rrbb (yellow) in 9:3:3:1 ratio

The action of the two genes together gives purple -- a case of complementation. (It may look like co-dominance, but remember that co-dominance refers to alleles of the same gene, while there are two separate genes at play here.)

The pathway of color production can be represented as:

2. (a) The screen is straightforward -- just look at each mutagenized firefly and pick out the ones that either shine continuously and the ones that don't shine at all. A selection may take more gadgetry -- for instance, one could make the fireflies walk through a long tube (one fly at a time) and set up some kind of photactivated device such that that darkness in the tube causes a trapdoor to open, allowing the fly to fall through -- so flies that don't shine continuously wouldn't make it to the end of tube. The converse setup (trapdoor opening if there is light) would select for fireflies that never shine. Lots of other schemes are possible, of course.
(b) She must have done a complementation test -- crossing homozygous mutant 1 with homozygous mutant 2. The progeny must have been wildtype, indicating that the two mutations were in separate genes (therwise the progeny wouldn't get any normal alleles). The mutant phenotypes must be recessive, otherwise the progeny wouldn't show the normal phenotype.

3. (a)

Mutations in Group A genes should complement Group B and Group C mutations, but not each other; Group B mutations should complement A and C but not each other, etc. --

Mutation 1 2 3 4 5 6 7 8 9 10
1 - + - + - + + + + +
2 - + + + - + - + +
3 - + - + + + + +
4 - + + - + + +
5 - + + + + +
6 - + - + +
7 - + + +
8 - + +
9 - -
10 -

(b)

Protocatechuic acid allows growth of group D mutations only, while dihydroshikimic acid allows growth of C and D, and shikimic acid allows growth of group A, C, and D mutations. Therefore, the first intermediate in the pathway must be protocatechuic acid; the second is dihydroshikimic acid, and the third is shikimic acid. Group D mutations are rescued by all supplements; Gene D must therefore act first in the pathway, converting a precursor into protocatechuic acid. Group C genes (rescued by 3 of the supplements) is next, converting protocatechuic acid into dihydroshikimic acid; then A converts dihydroshikimic acid into shikimic acid. Group B is not rescued by any of the supplements except tryptophan itself. Therefore, it must act last, converting shikimic acid into tryptophan. The pathway can be represented as:
(c)
(i)
Genes A and B are missing, so only tryptophan will rescue.
(ii)
Genes A and C are missing, so shikimic acid and tryptophan will rescue; the others will not.
(iii)
Genes B and C are missing, so only tryptophan will rescue.

4. (a) Maternal effect mutations don't cause a lethal phenotype in the zygote provided the mother is not homozygous for the mutant. So, for instance, we could set up a cross Mm x Mm where m = mutant allele. The homozygous recessive mm progeny (which would survive, because the mothers had the normal allele) could then be examined for the second phenotype. To ensure that we are looking at the right animals (how do we know which progeny are mm?) we could use two morphological markers that are tightly linked to the gene of interest and flanking it -- e.g., a genotype A M B/a m b where a and b are marker alleles. If we pick out progeny that show phenotypes a and b, it is highly probably that they would be homozygous for allele m also.
(b)

For a zygotic gene mutation, we have to start with zygotes that have a dominant allele -- else they won't make it past embryogenesis -- which means we have to think of some way to cause loss of the dominant allele(s). So, for instance, we could start with Mm heterozygous zygotes, bu then we'd have to cause loss of heterozygosity in the animal. One way of doing that -- inducing mitotic recombination once the essential function has been completed. Again, we'd need some way of identifying the patches of cells that have become homozygous mm, so we'd have to include a marker gene. As we saw with the Jedi story in the previous problem set, the marker gene would have to be between M/m and the centromere, and tightly linked to M/m.

For your information: A different strategy, which has not been discussed in class, is to use a type of mutation called conditional mutation. Such mutations have the property that they allow wildtype or near-wildtype function of the allele under some conditions (called permissive condition) but under a different condition (non-permissive or restrictive) they don't function. Temperature-sensitive mutations are examples of this type of mutation. Generally, these mutations allow normal function at low temperature, but fail to function at elevated temperatures (presumably because the mutated protein is unstable at higher temperatures). Coat color is Siamese cats is temperature sensitive -- pigment is produced only at lower temperatures, which is why Siamese cats show pigmentation at their extremities -- the temperature is lower there. In this Drosophila problem, having a temperature-sensitive allele would allow testing of the second function -- we could let homozygous mutant embryos complete embryogenesis at the permissive temperature, then shift the embryos to the restrictive temperature to look at the second function of the gene.

5. (a)

NOTE: Your answer does not need to be this long!

Solving this problem is much like solving the question of whether GAL4 and GAL80 are activators or repressors of transcription. In this case, stress induces transcription of the hulk operon -- so stress either activates an activator of transcription, or inactivates a repressor of transcription. Velveeta stops transcription of the operon, so it either activates a repressor of transcription or inactivates an activator of transcription. The regulatory genes (activators/repressors) are m and am; we have to decide whether m and am are activators or repressors:

Just as we saw with GAL4 and GAL80 (see p.119 of lecture notes)... to tell whether a gene makes an activator or a repressor of transcription, we just have to ask -- does a mutation that causes constitutively low levels of transcription behave in dominant or recessive fashion? If it is dominant, then the gene product must be a repressor; if recessive, it must make an activator. Likewise, we can ask if constitutively high transcription is dominant or recessive. If dominant, the regulatory gene must code for an activator; if recessive, it must be a repressor.

Here, mutation m1 causes constitutively low hulk levels, and this phenotype is recessive (m1/m+ animals behave like wildtype). Therefore, gene m must encode an activator of transcription; m is itself activated by stress. This conclusion is confirmed by looking at mutation m2. This mutation causes constitutively high levels of hulk transcription (in the absence of Velveeta), and behaves in dominant fashion (m+/m2 animals behave like m2/m2)-- so, again, gene m must be an activator of transcription.

What about gene am? am1 animals continue to transcribe hulk even in the presence of Velveeta, and this phenotype is recessive -- so am must be a repressor of hulk transcription, and Velveeta activates am. Again, the other mutation confirms this conclusion -- am2 animals transcribe hulk at constitutively low levels, and this phenotype is dominant, consistent with am being a repressor.

Putting all these conclusions together, a model for hulk gene transcription that makes sense is as follows:

An alternative model is that am works not directly on hulk, but indirectly, through m:
(b) The phenotype of gene X deletion suggests that the gene product is normally required to maintain transcription of hulk at high levels. One way that might be accomplished is if gene X activates hulk transcription, perhaps by activating gene m. In this model, stress first activates hulk gene transcription, including gene X transcription. Thereafter, even if stress is withdrawn, gene X keeps transcription of hulk going until transcription is blocked by gene am.

.. Color