Genetics 371B, Autumn 2000

Problem set 1 -- based on lectures 1-8

You have been trying to train Great Northwestern rabbits to do a pole-vaulting routine whenever they hear a dog whistle. After much experimentation, you realize that the ability to perform pole-vaulting routines is determined by an autosomal gene in this species of rabbit when you blow the whistle, the rabbits (depending on their genotype) either do the pole-vault or just ignore you and continue to chew thoughtfully on their carrots.

The results of a number of crosses with vaulters and non-vaulters are shown:

Which phenotype is dominant ? Explain your logic briefly (only one or two sentences needed!).

(b)

During discussion of your results with other members of the PVRS (Pole-Vaulting Rabbits Society), you find out that a mutation in an unrelated autosomal gene can cause deafness in rabbits, so that they cannot hear dog whistles. The ability to hear the whistle is dominant; inability to hear is recessive. The following cross was reported: a homozygous vaulter ("Anabelle") was crossed to a homozygous non-vaulter ("Mesquite"), and the resulting F₁ progeny were all non-vaulters. Give the genotype of the Anabelle and both possible genotypes of the non-vaulter parent (Mesquite). Use P and p for the dominant and recessive alleles with respect to the vaulting trait, and H and h with respect to hearing. (Remember what the test is: you blow the dog whistle and see if the rabbits do the pole-vault routine.)

David's dad and his two younger siblings (a sister, Samantha, and a brother, Matt, who is the youngest) are afflicted with a common genetic disorder that causes mild breathing problems, while David and his mom are unaffected. Samantha's and Matt's spouses are also unaffected, but both couples have affected children the older of Samantha's two sons is affected, as is the older of Matt's two daughters (the younger child in each case in not affected). David's only daughter, who is unaffected, is married to her cousin (Samantha's first son). David's wife is known to be homozygous normal.

Draw the pedigree as described; be sure to do all the appropriate numbering. Indicate in the pedigree who David is.

Assume that the disease is autosomal recessive (showing complete penetrance and expressivity). Show all of the genotypes as completely as possible given the information that you have. What is the probability that David's first grandchild will have this disorder?

Now assume that the disease is autosomal dominant (showing complete penetrance and expressivity). Show all of the genotypes as completely as possible. What is the probability that David's first grandchild will have this disorder?

Hemophilia and red-green colorblindness are both X-linked recessive traits in humans (X^h = hemophilia allele, X^H is normal; X^g = colorblindness allele, X^G is normal). A young couple have had a hemophilic son with Klinefelter syndrome (XXY male). They would like to have a second child, but are concerned about the risk of hemophilia. Their genetic counselor (who knows the genotypes of both parents with respect to hemophilia) tells them that there is a 50% chance that their next child will have hemophilia also regardless of whether it is a son or a daughter (assuming that the next child will be a normal XX or XY child).

What are the genotypes and phenotypes of the parents with respect to hemophilia? Explain BRIEFLY.

What combinations of egg and sperm genotypes (with respect to the sex chromosomes and hemophilia) could have given rise to the XXY son?

In humans, the index finger can be longer than the fourth (ring) finger. Long index fingers is dominant in women but recessive in men.

What is the probability of a daughter with long index fingers being born to heterozygous parents?

The amount of DNA in a haploid yeast nucleus is about 0.015 picograms (0.15 x 10^-13 gram).

(i) a diploid yeast cell in the G1 phase of the cell cycle

(ii) a haploid cell in G2

(iii) a cell in metaphase I of meiosis

(iv) a cell at the end of meiosis II