Problem Set 10 Answers

1. a. “Nonparental ditype” is a tetrad type containing two different genotypes, neither of which is the same as the parents

b. The nonparental ditypes resulting from an a+/+b trans heterozygote are ab/ab/++/++ and ab/++/ab/++. (The important thing here is that this is an ordered tetrad)

1 * a +

2 * a +

3 * + b

4 * + b

ab/ab/++/++ requires two crossovers between the a and b regions of the chromatids; these crossovers occur between 1 and 3 & 2 and 4 or between 1 and 4 & 2 and 3. ab/++/ab/++ requires three crossovers: two crossovers as before, plus one additional crossover between a and the centromere.

d. You would expect to see ab/ab/++/++ more frequently than ab/++/ab/++, because it requires one less crossover.

2. a. Parental ditype refers to a tetrad that only has nonrecombinant chromosomes (i.e., parental chromosomes); it contains only two type of chromosomes. Nonparental ditype refers to a tetrad that only has recombinant chromosomes; it contains only two types of chromosomes.

b. If a and b are linked, the frequency of PD is much greater than NPD. PD results from no recombination, and NPD results from two recombination events, arising with a frequency of p²

c. If a and b are unlinked, you would expect the frequency of PD and NPD to be equal. For a;b X +;+ cross, we expect three possibilities (that come up in a 1:2:1 ratio) from independent assortment:

PD T NPD

a;b a;b a;+

a;b a;+ a;+

+;+ +;b +;b

Thus, PD = NPD.

3. a. (i) a and b segregate as parental ditypes. (ii) a and c segregate as tetratypes. (iii) b and c segregate as tetratypes.

b. (a) a segregates according to a second-division (M_II) segregation pattern. (b) b segregates according to a second-division (M_II) segregation pattern. (c) c segregates according to a first-division (M_I) segregation pattern.

1 a b c

2 a b c

3 + + +

4 + + +

The ordered tetrad could have arisen by a crossover involving 2 and 3 between the centromere and a and another crossover involving 1 and 4 between b and c.

d. The largest map distance is between a and b; therefore, you would expect the most frequent recombinations to occur between these two genes. The most common single crossover tetrad will arise from a crossover between chromatids 2 and 3 between a and b, producing abc/a++/+bc/+++.

4. a. Gene conversion is the meiotic process whereby one allele directs another allele to take on its form. Gene conversion is seen in tetrads B, C, E, and F.

b. Any tetrads with alleles undergoing gene conversion contain hybrid DNA in the region following meiosis (tetrads B, C, E, and F).

c. Hybrid DNA was not repaired in tetrads E and F. E displays an aberrant 4:4 ratio, whereas F displays a half-chromatid conversion (5:3 or 3:5 patterns); hybrid DNA repair is only observed in tetrads with a normal 4:4 ratio or a full chromatid conversion (6:2 or 2:6 patterns).

d. In a repair deficient strain, tetrads B, C, and F would not occur. Both half-chromatid and full chromatid conversions require DNA repair machinery.

e. No conversion will be more frequent than conversion, so tetrads A and D will be the most frequent. Furthermore, experiments in yeast show that 6:2 and 2:6 patterns are more common than 5:3 and 3:5 (repair is much more common than nonrepair). Therefore, tetrads B and C will occur more frequently than F. E, which shows no correction of its heteroduplex DNA, will occur least frequently. Thus, one might expect the frequency to be A, D > B, C > F > E.

5. a. True, most mutations will reduce activity of the gene.

b. False, for unordered tetrads, one need look at the number of nonparental ditypes to determine the number of double crossovers.

c. True, because half of the strand crossover events lead to recombination.

d. False, half of all gene conversion events are associated with recombination.

e. False, full conversion events occur when DNA polymerase repairs a base mismatch.

f. False, recombination is much more frequent than gene conversion (however, half of all gene conversions are associated with recombination).

g. False, an ordered tetrad in a M_II segregation pattern indicates recombination between the centromere and the gene of interest.

6. a. Holliday model: single break on one strand of two homologues, exchange linkage, branch migration, resolution, mismatch repair.

Meselson-Radding: one strand breaks, strand invasion, D-loop formation and degradation, formation of the Holliday junction, branch migration, resolution and repair as before. The essential difference is that one strand is broken and it invades the other double helix.

Double Stranded Break Repair: One homologue is subjected to a double stranded break, 5' ends are chewed back, 3' ends invade other helix and copy off it), formation of two Holliday junctions, branch migration, resolution, and repair.

b. Holliday model explains the presence of gene conversion octads, recombination half the time with gene conversion, co-conversion, and polarity of conversion.

Meselson-Radding explains the same as the Holliday model, but also accounts for the unusually high number of 5:3 octads (the unequal strands at the beginning allow for these to be produced).

Double Stranded Break Repair models explains the same as the two models above, but also explains 6:2 octads in yeast and the correction of plasmids with gaps.

7. a. Branch migration

b. Strand invasion

c. DNA repair of both heteroduplex DNAs

d. In the Holliday model, without repair you expect an equal number of a and + chromosomes (albeit in an aberrant order. In Meselson-Radding, without repair you would also observe 5:3 or 3:5 octads.l

8. 3:5 octads arise in the Holliday model when only one of two half conversions has been corrected. In the other model,s 3:5 octads arise because of asymmetric removal of DNA to form the recombination complexes.

9. In the double strand break model, if the outer or inner strands are cut and religated, then no recombination will occur. If one outer and one inner set of strands are cut, then recombination will occur.