Chapter 6 problems

From book:
4. 5, 6, 11, 14, 17, 24, 31 and for a challenge, 55, 57, 66

To hand in: (week of Oct 10):

1. In peanuts, a plant may be either a "bunch" plant or a "runner" plant, depending on the rhizome length. Here are the results of various crosses among different varieties. Varieties A and B are true-breeding and both have a "bunch" phenotype, but their F1 produces runners. If two F1’s are crossed, they produce a mix of bunch plants and runner plants, as shown below.

A x A: all bunch
B x B: all bunch
A x B: runner (F1)
F1 x F1: 7 bunch : 9 runner

1. Provide a concise genetic explanation for those results.
2. What cross could you do to test your hypothesis and what would be the predicted outcome of that cross?

2. A, B, and C are three consecutive enzymes in a pathway that ultimately produces a black pigment. Mutant alleles (a, b, and c) at each of those unlinked loci are non-functional and block the pathway, so no pigment is produced.

1. If two triple-heterozygotes are crossed (AaBbCc x AaBbCc), what proportion of progeny will be black?
   (First figure out what the genotypic ratios will be for that tri-hybrid cross, then assign phenotypes to genotypes. Hint: there are 8 genotypic classes to consider: A-B-C-, A-B-cc, A-bbC-, A-bbcc, aaB-C-,aaB-cc, aabbC-, aabbcc).
2. Now assume there is a dominant inhibitor allele at another locus D that inhibits enzyme A. Recessive homozygotes (dd) do not make the inhibitor and have no effect on any of the enzymes.
   Again, two triple-heterozygotes are crossed: AaBbDd x AaBbDd (Assume that both individuals are CC in this cross, so you can ignore the C locus)
   What proportion of the offspring will be black?

3. A series of bacterial mutants were collected that each block steps of a metabolic pathway that produces the essential nutrient X. You know that compounds A-E are all involved in the pathway, but you do not know the order. You collect a set of mutants that cannot grow without supplemental X, so you know they have null mutations somewhere in that pathway.
Finally, you test each of those mutants by adding supplemental A, B, C, D, or E to minimal medium (one at a time) to see which strains are able to grow on which substrate. Here are the results (+ means they can grow; 0 means they cannot).

        substrate
        A B C D E X	
Mutant:
1       + 0 + + + +
2       0 0 + 0 + +
3       + 0 + 0 + +
4       0 0 + 0 0 +
5       0 0 0 0 0 +

1. What is the order of substrates in the pathway?
2. Which step is blocked by which mutant?