371B Autumn 2000 -- PS4 Answer Key

Genetics 371B, Autumn 2000

Answer key, Problem set 4

1.	Remember that the question being asked is whether the given genotype increases the chance of cancer -- not whether the cell in question is itself a cancer cell. To answer the question, ask yourself in each case: Does changing the genotype from homozygous normal to the given genotype increase the risk of cancer?
	(a)	Usually, it's the lack of tumor suppressor gene expression that is associated with an increased cancer risk. Therefore, increased expression of a tumor suppressor gene is not expected to increase the risk of cancer.
	(b)	This is a gain of function mutation in a protooncogene; since protooncogenes are generally in pathways that promote cell proliferation, such a mutation is expected to increase the risk of cancer (the mutant form is dominant).
	(c)	Failure to express a protooncogene is not expected to increase the risk of cancer.
	(d)	Since tumor suppressor genes keep cell proliferation pathways in check, failure to express such a gene will increase the risk of cancer.
	(e)	Again, loss of function of a protooncogene is not expected to increase the risk of cancer.
	(f)	While the mutation is recessive, being heterozygous for the mutation increases the risk of cancer, because all it would take is mutation of the remaining allele (or a mitotic recombination event) for complete loss of the tumor suppressor gene. Therefore, this genotype does carry an increased risk of cancer.

(a)

In each cross, one parent is homozygous for one mutation and the other parent is homozygous for another mutation (which may or may not be in the same gene as in the first parent, and in fact, could even be the very same mutation -- we just don't know at this point). If the mutations are in different genes, the cross can be represented as aaBB x AAbb, and the progeny will be AaBb, heterozygous at both loci, and will therefore show a wild type phenotype (assuming that the mutant phenotype is recessive). If both mutations are in the same gene, the cross can be represented as a₁a₁ x a₂a₂, giving a₁a₂ progeny. Since these progeny lack wild type alleles of gene A, they will not show wild type phenotype. Therefore, to identify the mutations that belong in the same complementation group, one just identifies pairwise crosses that fail to give wild type progeny. For example, mutations r1 and r3 fail to complement, so those mutations are members of the same complementation group.

By applying this logic, the three complementation groups are:

r1 and r3
r2, r5, and r6
r4, r7, and r8.

Again, in order to get complementation (i.e., to get wild type phenotype when crossing mutations in two separate genes) the mutant phenotype has to be recessive -- if the mutant phenotype is dominant, we wouldn't see the wild type phenotype even in the heterozygotes. Therefore, none of these mutations had a dominant phenotype.

(b)

Note: Your answer is not expected to be this long! I am jut spelling out the logic explicitly here.

We can come up with a working hypothesis (i.e., a hypothesis that we'll work with, and modify or discard if need be) by looking at the various phenotypes.

First, the normal phenotype has brown and white coat color; there is no black coat. Therefore, the presence of black in a mutant suggests (but does not prove) the following: black color is normally converted to something else (such as brown or white), but that conversion has failed in the mutant. Since the mutation that gives black coats is dd, the wild type gene D is presumably (in this hypothesis) responsible for converting black to something else:

D
black ---------> (something else)

Furthermore, since D is epistatic to the other genes (i.e., in the absence of gene D, we get a black coat phenotype, regardless of the genotype of the other two genes), we can hypothesize that the action of gene D must be requied early in the pathway, before the action of genes B and E.

What is that "something else" that brown is converted to? The simplest hypothesis is that that something else is either white or brown. Let us consider each of these two possibilities:

Possibility 1: Black is converted to white.

In this case, we'd have to hypothesize further that gene B is required to convert white to brown pigment -- since the phenotype of bb mutants is fully white (i.e., no brown). What is role of gene E in this hypothesis? Since the wild type phenotype is brown and white patches, we'd have to say that the role of gene E in this hypothesis is either to block the action of gene B in some patches (so that those patches remain white, and are not converted to brown), or to reverse the effect of gene B in some patches (brown is converted back to white in those patches):

However this hypothesis predicts that B should be epistatic to E -- i.e., in the absence of wild type gene B, the phenotype should be fully white, regardless of whether gene E is wild type or mutant. However, looking at the given phenotypes, we see that this prediction is NOT met -- B is NOT epistatic to E (i.e., the phenotype of bb is not the same as bbee). Therefore, this hypothesis is NOT CONSISTENT with the data.

Possibility 2: Black is converted to brown.

Here, we'd hypothesize that gene E is required to convert brown pigment to white -- since the phenotype of ee mutants is fully brown (i.e., no white). What is role of gene B in this hypothesis? Since the wild type phenotype is brown and white patches, we'd have to say that the role of gene B in this hypothesis is either to block the action of gene E in some patches (so that those patches remain brown, and are not converted to white), or to reverse the effect of gene E in some patches (white is converted back to brown in those patches):

This hypothesis predicts that E should be epistatic to B -- i.e., in the absence of wild type gene E, the phenotype should be fully brown, regardless of whether gene B is wild type or mutant. This is exactly what we do see -- i.e., the phenotype of ee is the same that of bbee -- fully brown. Therefore, this hypothesis IS CONSISTENT with the data; one of the these pathways must be correct:

CORRECT PATHWAY:

This pathway is also consistent with the cross described: the F₂ progeny from the F₁ cross (DdEe x DdEe) will be:

9 D_E_ : 3 D_ee : 3 ddE_ : 1 ddee
wildtype brown black black

giving a 9 wild type : 4 black : 3 brown ratio, which is exactly what is seen.

A strain lacking gene G can be rescued by any intermediate; therefore, the function of G must be to synthesize the very first intermediate from the precursor:

G
Precursor ---> (first intermediate)

Looking at gene E, we see that the only intermediate that does not rescue the mutation is #2; therefore, gene E must be required to convert #2 to one of the other intermediates. Furthermore, intermediates #1 and #3 do rescue this mutation, so they must be "downstream" of gene E. This observation also indicates that intermediate #2 must be the first intermediate in the pathway (because this is the only one that does not rescue mutant E):

G E
Precursor ---> supplement #2 ---> (second intermediate)

By the same logic, supplement #3 does rescue a defect in gene H, and that is the only supplement that rescues, so supplement #3 must be the last intermediate in the pathway, and gene H is required for its synthesis; also, since a mutation in gene F is not rescued by any intermediate, itmust be required to convert the last intermediate (#3) into tryptophan. So the complete pathway is:

G E H F
Precursor ---> #2 ---> #1 ---> #3 ---> Tryptophan

(a)

These mutations are polar mutations; the genes are in the order:

promoter --- gene B --- gene D --- gene A

Transcription begins at the promoter and extends rightwardto make a polycistronic mRNA, such that a mutation that causes premature termination of translation in gene B also results in failure to translate D and A, etc.

(b)

CONSISTENT: in the "haploid", a promoter mutation would prevent efficient transcription, giving the always-low phenotype; this mutation will not affect the ability of a wildtype operon to function normally, as seen in the partial diploid.
CONSISTENT: this mutation is predicted to be a loss-of-function mutation. Failure to activate transcription would explain the phenotype of the "haploid" r mutation; in the heterozygote, the wild type phenotype will be seen.
NOT CONSISTENT. A null allele of the repressor is expected to allow constitutive transcription in the haploid, always giving a spore wall phenotype. (ALTERNATIVE INTERPRETATION: the data can also be interpreted in the strict sense, as just looking for induction vs. non-induction. In that case, you can make the case that transcription levels will be always high in this mutant, but with no induction -- consistent with the observation. This answer is acceptable, provided your answer explains your thinking.)
NOT CONSISTENT. The phenotype of the partial diploid is not consistent with this hypothesis -- a gain-of-function mutation in a repressor is expected to behave in a dominant fashion, and that is not what is seen.
NOT CONSISTENT: Again, the phenotype of the partial diploid is not consistent with this hypothesis (and, depending on your interpretation of "failure to upregulate", neither is the phenotype of the "haploid").

(c)

The goal here is to come up with ONE GENOTYPE such that if hypothesis (i) is correct, we should get a phenotype for that genotype, and if hypothesis (ii) is correct, we get a different phenotype -- so that by looking at the phenotype, we can tell which hypothesis is correct.

The major difference in the expected behavior of a promoter vs. an activator is that a promoter acts in cis -- it only affects the genes that it is located upstream of -- whereas an activator is expected to behave in trans -- it does not matter where the activator gene is, it can affect its target anywhere in the genome. We can make use of that difference to distinguish between these two hypotheses. For example, a partial diploid of the genotype:

r BDA/R bda

will distinguish between the two hypotheses (b, d, and a being mutant alleles of B, D and A)--

if hypothesis (i) is correct, the partial diploid genotype given above would not give functional BDA gene products under any condition. According to this hypothesis, R is the promoter, and r is a non-functional promoter; the wildtype promoter drives transcription of the mutant bda genes (so no product is made), and the wild type alleles of B, D and A are under control of a non-functional promoter, so they are not transcribed -- and therefore, no functional B,D, A proteins are made.
if hypothesis (ii) is correct, the phenotype of this strain will be like wild type -- i.e., there will be induction of BDA transcription under nitrogen starvation. That is because in this hypothesis, R is an activator, and can act in trans -- so the wild type activator protein can activate transcription of the wild type BDA genes. (Remember that this is an "either/or" situation -- R is either the promoter or the activator, and we are tyring to figure out which it is. If R is the activator, then the promoter is not mutated in this strain.)

This one is just to get you thinking about selections and screens. In this particular instance, a selection is trickier than one might think initially. One could come up with various cruel and unusual schemes (e.g., surviving attacks by mouse-hungry cats), but they probably wouldn't be attempted in real labs. Instead, we could use some scheme where laser- and non-laser-mice could be separated based on the lasering itself. For example, we could set up a bin with passages leading out of the bin. The passages are branched and have laser-activated, fast-swinging doors (or trapdoors) so that mice that shoot lasers go one way (and, for instance, end up in a bin with lots of cheese and sunflower seeds, so they mice stop being angry) and non-laser mice go another way. We'd have to calibrate the laser-activated doors to respond only to a suitable laser strength, and we'd need some way of aggravating the mice (cat sounds?). Lastly, mice selected this way would have to go through a secondary screen, or perhaps be subjected to the same treatment again, to get rid of false positives. The main feature in this scheme is that we don't examine each mouse on an individual basis -- the mice are all dumped into the initial bin, and ones we want end up in one bin at the end, while ones we don't want are excluded. There are all sorts of other problems, of course -- for example, we have to keep the mice all happy, or they could end up killing each other.