Genetics 371B, Autumn 1999 Answer key, Problem set 3

1.	(a)	The F1 progeny are fully heterozygous, with the dominant alleles in cis (+++/+++ crossed with qdn/qdn giving +++/qdn). The testcross is +++/qdn x qdn/qdn. The progeny phenotypes and numbers are shown:
	(b)	Here, n is unlinked to q and d. So we can calculate the progeny numbers with respect to q and d; half of each of those classes will additionally be n+, and half of each class will be n. With respect to q and d, the crossover progeny will be + d and q + , and will account for (0.08 x 2000) = 160 progeny. Each of these two classes will be evenly split between n+ and n; likewise, each of the two non-crossover products (2000 - 160 = 1840 total) will also be split evenly between n+ and n: + + + = 1840/4 = 460 + d + = 160/4 = 40 + + n = 1840/4 = 460 + d n = 160/4 = 40 q d + = 1840/4 = 460 q + + = 160/4 = 40 q d n = 1840/4 = 460 q + n = 160/4 = 40
	(c)	Here, we can simply add up the progeny numbers from the two separate crosses and calculate the apparent map distances from those: SCO(q-d) = 64 + 64 + 40 + 40 = 208 SCO(d-n) = 184 + 184 + 460 + 460 = 1288 DCO = 16 + 16 + 40 + 40 = 112 q - d map distance: (208 + 112) / 4000 = 0.08; map distance = 8 cM d - n map distance: (1288 + 112) / 4000 = 0.35; map distance = 35 cM Note that if the assistant knew he was combining two separate maps, he could simply take the mean of the two (d-n) map distances and get the same answer.

2.	(a)	The strategy is to look at the genotype of the offspring and ask if that genotype could be derived in more than one way (i.e., with recombination or without). If so, then the meiosis in the father is uninformative. Note: In all four pedigrees, II-2 received allele18 along with the brachydactyly allele; in all four pedigrees, he transmits the dominant brachydactyly allele. Pedigree (i): Uninformative. II-2 is homozygous for the polymorphic site, so it's not possible to tell whether recombination has occurred. Pedigree (ii): Uninformative. II-2 could have transmitted either allele 18 (without recombination) or allele 23 (with recombination). Pedigree (iii): Informative. II-2 transmitted the brachydactyly allele and allele 18, so no recombination occurred. Pedigree (iv): Informative. II-2 transmitted the brachydactyly allele and allele 18, so no recombination occurred.
	(b)	To be an affected boy with {21,23} at the polymorphic site, the child has to receive b 21 from the mom, B 23 from the dad, and a Y chromosome from the dad (where B = brachydactyly, b = unaffected). p(b 21 from mom) = 0.5 p(B 23 from dad) = 0.06 (because B/b - PS map distance = 12 cM; recombinants B 23 and b 18 together account for 12% of gametes) p(Y from dad) = 0.5 Overall probability = (0.5)(0.06)(0.5) = 0.015

3.	(a)	The size of the fragment is dictated by the number of repeats. At allele {25} there are 25 repeats, so the length of the fragment = 1000 + (25 x 4) = 1100 bp. Likewise, the length of the fragment with allele {50} = 1000 + (50 x 4) = 1200 bp.
	(b)	The two bands in each lane represent the two copies of the locus in each person; clearly, they are all heterozygotes, each having two different-length alleles at the site. Each child is expected to receive one copy of the locus from the mom and one from the dad so the mom should have one band in common with each of her children. The only lane that has one band in common with each of the other six lanes is lane 3 so lane 3 has the mom's DNA. The father's genotype can be deduced from bands present in the children but absent in the mom it must be {47, 76}.

4.	(a)	T will be incorporated across from the alkylated G: Only one of the original alleles was alkylated, so each cell has at least one normal copy so each cell will give at least one copy of the 1200 bp fragment. In addition, the two mutated alleles (the left two in the last row of the diagram above) will fail to be cut by Hind III, giving a fragment of size (1100 + 1200) = 2300 bp in addition to the 1200 bp fragment from the normal allele. (We are assuming that Hind III will cut any site that has all the correct basepairs and will not cut any site that has any incorrect basepair.)
	(b)	There are many ways of destroying a functional Hind III site (any one of the six base pairs could be mutated, and each base could be changed into any of the other three). In contrast, a reverse mutation to restore the site has to be precise -- only the incorrect base pair can be changed, and it has to be changed to exactly the correct base pair. So the probability of a forward mutation is higher than the probability of a reverse mutation.

5.

Remember that if a portion of a Lod score curve is at +3 or higher, it means that we have 95% confidence in favor of linkage at the corresponding recombination (map) distances; if the curve is at -2 or lower, we have 95% confidence against linkage at those distances. (Between +3 and -2, positive values favor linkage and negative values favor non-linkage, but the confidence level is less than 95%.) PS1: Evidence favors linkage to the disease gene at less than 10 cM, with maximum probability of linkage at 5 cM PS2: Evidence favors linkage, with a map distance of at least 10 cM and no more than 20 cM; the most probable distance is ~15 cM. The curve dips below -2, so linkage at 5 cM or less can be excluded. PS3: The curve remains <0, so there is no evidence of linkage; the curve drops below -2 at 15% recombination, so linkage at 15 cM or less can be excluded. PS4: The curve doesn't go below -2 or above +3, so the data are inconclusive. However, the curve stays negative, so it argues against linkage.

.. Color