Genetics 371B, Autumn 1999 Problem set 4 -- based on lectures 15-21

Due Wed Nov. 24 at the start of class

Reminder: You are welcome to work in groups, but you must write out the answers on your own, IN YOUR OWN WORDS. Credit will NOT be given for answers that are identical.

1.	In Drosophila, two loci Hs and D are normally 20 cM apart on the long arm of chromosome 2. A genetic map of the region is shown below where loci a-f represent polymorphic markers between the Hs and D genes. The number of kilobase pairs between markers is also shown below each map. Certain fly strains carry a paracentric inversion with breakpoints just outside of the polymorphic markers a and d (as shown below). Other fly strains carry a 100 kb deletion which includes polymorphic markers e and f (as shown below; the dotted line represents the missing portion). Assuming that recombination frequency is proportional to the physical distance in this portion of the chromosome, what approximate recombination frequency would you detect between the Hs and D genes in flies that are:
	(a)	heterozygous for the paracentric inversion?
	(b)	homozygous for the paracentric inversion?
	(c)	heterozygous for the deletion?
	Show how you arrived at your answers.

2.	Dosage compensation in human females results in the genetic silencing of one of the X chromosomes -- but XO females (who have one active X chromosome) show distinct phenotypes (Turner syndrome) not shown by XX females. Suggest two explanations for this difference.

3.	As discussed in class, the Rb (retinoblastoma) protein binds to the transcription factor E2F and prevents it from promoting transcription of genes needed for entry into S phase. Phosphorylation of specific amino acids of the Rb protein causes it to dissociate from E2F, allowing transcription activation by E2F.
	(a)	BRIEFLY discuss (stating your reasons) whether you expect the following mutations to give a dominant or a recessive phenotype: (i) a mutation in Rb such that it can no longer be phosphorylated (ii) a mutation in Rb such that it can no longer bind to E2F (iii) a mutation in Rb such that it remains bound to E2F even when phosphorylated (iv) a mutation in E2F such that it can no longer bind to Rb protein (v) a mutation in E2F such that it cannot be released from Rb (vi) a promoter mutation in the E2F gene, preventing transcription of that allele
	(b)	Which one(s) of the six mutations above do you think would be capable of promoting progression toward cancer? Why?

4.	Wildtype slugs are slimy, olive in color, and are unable to sense when someone is about to step on them. A slug geneticist mutagenizes a colony of slugs and identifies mutants that are black (b) instead of olive (+), and dry (d) instead of slimy (+). In addition, by applying a suitable selection, he discovers a Jedi-like mutant (J) that can sense a descending foot and slide out of the way. The three genes are linked; J is in the middle.
	(a)	He X-irradiates fully heterozygous young slugs that are olive, slimy and Jedi. Some of these slugs develop lone skin patches that are black instead of olive, while an equal number of black patches are located next to patches that are dry instead of slimy. Which of these two loci is closer to the centromere? Diagram the recombination event that gave rise to each of these two outcomes.
	(b)	If he starts with bJ+/++d heterozygotes and induces mitotic recombination, which of the following cells would be most likely to be JJ? -- Explain briefly. (i) cells in a lone black patch in the skin of the slug (ii) cells in a black patch that is next to a dry patch (iii) cells in a dry patch that is next to a black patch
	(c)	In one rare instance, he finds a black patch that is next to a dry patch -- but some clever DNA testing reveals that cells in both of these patches are J/+ heterozygotes. Explain how these patches must have arisen.

.. Color