Genetics 371B, Autumn 2000

Answer key, exam 3 YELLOW VERSION: 11 December

The short-tailed voles of the Canadian north have tails that measure 7 cm. In contrast, voles of the same species in Siberia have tails that are 11 cm long. A cross between a Canadian vole and a Siberian vole always gives F₁ progeny with 9 cm tails. F₂ progeny from crosses between these F₁ animals show a range of tail lengths -- most have tails that are 9 cm long, while the shortest tails are 5 cm long and the longest tails are 13 cm long. Of 1061 F₂ progeny, four had 5 cm tails and four had 13 cm tails.

Fraction of progeny showing one extreme of the range of phenotypes = (1/4)ⁿ where n = number of genes contributing to the trait

Therefore, (1/4)ⁿ = 4/1061

n = log(4/1061) / log(1/4) = 4.0

i.e., the number of genes involved in determination of tail length = 4

4 genes = 8 possible alleles (assuming 2 alleles per gene)

The range of tail lengths = (13 - 5) cm = 8 cm

i.e., each allele can contribute 8 cm / 8 alleles = 1 cm to tail length

• Only one tail length is seen in each parental strain, so the parental (Canadian and Siberian) strains must each be true-breeding.
• The Canadian strain must have 2 additive alleles (at 1 cm per additive allele plus the base length of 1.0 inches) -- so its genotype must be (e.g.) AAbbccdd
  The Siberian strain has 6 additive alleles (6 cm inches longer than the base length)
• The F₁ progeny must be heterozygous at 4 loci (or we wouldn't have got the observed range of phenotypes). Therefore, the Siberian parent must have additive alleles at loci where the Canadian strain did not have additive alleles -- i.e., if the Canadian strain is AAbbccdd, the Siberian strain must be aaBBCCDD, giving AaBbCcDd in the F₁ -- four additive alleles in the F₁, giving a tail length of 9 cm.

The sample size is small, so the value of 4/1061 may turn out to be an underestimate or overestimate of the real fraction. Also, with more progeny examined, animals with tails < 5 cm or > 13 cm may be found.

Reason 2:

Tail length may be subject to environmental influences

(Other reasons exist also -- for instance, we don't know that there are only 2 alleles per gene.)

You are studying seizures using a mouse model. You have a true-breeding strain of mice that has frequent, spontaneous seizures. You cross these to another true-breeding strain that does not have seizures. The F₁ all have seizures, and a single mating of two F₁ animals produces 2 normal mice and 7 mice with seizures.

Seizure is dominant, normal is recessive. The cross is aa (normal) x AA (seizures) giving Aa (seizing) F₁

and 3 A_ (seizures) : 1 aa (normal) in F₂.

The 7:2 observed ratio is approximately the expected 3:1 ratio.

The F₁ cross is AaBb x AaBb giving:

9 A_B_ (seizures)

3 A_bb (seizures)

3 aaB_ (normal)

1 aabb (seizures)

-- a 13:3 ratio, which is approximately the 7:2 ratio observed (7:2 = 14:4, which is close to 13:3). (Other ratios -- such as 15:1 -- are okay, provided the explanation makes sense.)

This would be a case of suppression (homozygous recessive bb suppresses the homozygous recessive aa phenotype).

The Great Northwestern Dragonfly develops in a manner similar to Drosophila , with the following difference: while Drosophila has one set of wings and one set of halteres (flight balancers) in the segment immediately posterior to the wing segment, the dragonfly has two sets of wings. The dragonfly larva has seven segments, shown here with their future adult fates.

A loss-of-function mutation in a homeotic (selector) gene that normally would act in the 4th segment would give wings instead of halteres in that segment (4th segment takes on a more anterior fate).

A French group working on dragonfly development has discovered three maternal genes expressed in the developing egg: fromage, pamplemousse (ppm), and chien. The expression pattern of these three genes in wild type and homozygous null mutant embryos is shown. Based on these data, suggest how these protein gradients are set up in wild type embryos, specifying the function of fromage and chien. Your model should explain the mutant phenotypes shown. (8 pts)

• Since fromage and chien are maternal, and their gradients seem to form independently of each other (as seen in the two mutants), their mRNAs must be sequestered at the anterior and posterior ends, respectively, so that translation of those mRNAs and subsequent diffusion of the proteins gives the observed gradients of fromage and chein protein.
• In the absence of fromage, the ppm gradient extends further to the anterior than in wild type. Therefore, fromage must block synthesis of ppm
• In the absence of chien, ppm is not made. Therefore, chien must promote ppm synthesis. Furthermore, the synthesis of ppm is proportional to the concentration of chien.
• A subtle point -- since ppm is also a maternal gene, the mRNAs must be made from maternal cells surrounding the egg and dumped into the egg. Therefore, the gradient of ppm synthesis is likely NOT the result of differential levels of ppm transcription across the egg, but rather, differential levels of ppm mRNA translation -- chien promotes ppm mRNA translation, and fromage blocks it.

bateau is a gap gene.

(22 pts total)

An improbable pair of operons in the mythical bacterium Confusus inhabitabilis is outlined below:

E, F, G, H, J, and K are genes coding for proteins; P and O are the promoter and operator for each operon. G and K are known to be short-lived proteins. The sugar mannose induces transcription of the mannose operon, while melibiose induces transcription of the melibiose operon.

1. G is an activator of the melibiose operon and K is a repressor of the mannose operon (4 pts)
   g is a loss-of-function mutation of an activator, causing failure of HJK transcription
   k is a gain-of-function mutation of a repressor, so EFG transcription is always repressed
2. G is a repressor of the melibiose operon and K is an activator of the mannose operon(4 pts)
   g is a gain-of-function allele that makes a super-repressor, causing constitutive repression of HJK transcription
   k is a loss-of-function mutation of an activator, so EFG transcription does not occur

Mannose activates the activator protein K, promoting transcription of EFG.

Melibiose inactivates the repressor G, allowing transcription of HJK (by relieving repression of transcription of HJK)

According to Model (ii), K is an activator of EFG transcription; activation of K by mannose induces transcription of EFG. Upon induction of EFG transcription, the level of repressor G increases, so there is increased repression of HJK. The level of activator K therefore drops, and transcription of EFG drops correspondingly. However, in this model, G is a repressor of HJK transcription; melibiose inactivates the repressor, so as long as melibiose is present, the repressor should be inactive, and HJK transcription should occur. Thus, the observation (of transient induction of EFG transcription and sustained induction of HJK transcription) is consistent with model (ii) at least at a simplistic level.

According to Model (i), K is a repressor of EFG transcription; inactivation of K by mannose allows transcription of EFG. In this model, as long as mannose is present, K will remain inactive, allowing continuing transcription of EFG; in contrast, induction of HJK by melibiose should be transient (by the same logic as described above). The observations are not consistent with these predictions.

In a certain remote island nation (Island #1), 91% of the inhabitants are normal (phenotype T), while the remainder show an autosomal recessive trait (t).

Frequency of recessive phenotype = frequency of genotype TT = P_tt = q² = 0.04

Frequency of allele t = q = 0.2.

Frequency of allele T = p = (1 - 0.2) = 0.8.

Frequency of carriers = frequency of genotype Tt = 2pq = 2 x 0.8 x 0.8 = 0.32.

It turns out the the population is not in Hardy-Weinberg conditions after all, as the fitness of the two phenotypes is not the same: w_{T_} = 1, and w_tt = 1/9. Starting with the frequencies given above, predict the frequencies of alleles T and t in the next generation. Show your work! (6 pts)

In the next generation:

Abundance of homozygous dominant = (p²)(w_TT) = (0.8)² x 1 = 0.64.

Abundance of heterozygotes = (2pq)(w_Tt) = 2 x 0.8 x 0.2 x 1.0 = 0.32.

Abundance of homozygous recessive = (q²)(w_tt) = (0.2)² x (1/4) = 0.01.

New total of all genotypes = 0.64 + 0.32 + 0.01 = 0.97.

Therefore, frequency of allele T in this new generation = p' = (0.64 + (0.32/2)) / 0.97 = 0.825.

and frequency of allele t = q' = 1 - p' = 1 - 0.825 = 0.175.

while the frequency of carriers = 2pq = 2 x (1/1000) x (999/1000) = 1998/1,000,000.

Thus, while individuals showing the recessive trait are present at a frequency of 1 in a million, people carrying the recessive allele are present at almost 2000 times that frequency. Therefore, people showing the trait are only a tiny fraction of the people carrying the allele, so preventing people showing the trait from having children will have only a miniscule effect on the frequency of the recessive allele.