Solutions to 7.03 Exam 3

Solutions Problem to Exam 3 mean : 64 1 a) Since the super repressor (IS), a trans-acting factor, is present there will be no lacZ expressed under any conditions. O c has no effect since it is cis-acting (on Z-). Thus the answeris 1ow low b) Now Oc is in cis with Z+ so this gene is expressed c) Thus Since there is no promoter (P-), a cis acting no lacZ is expressed under any conditions: d) alter The only a cis-acting way to express element: Z constitutively I*P÷O÷Z'Y÷/F constitutively: element, h i g h to drive and have Y under ' I*P*OeZ÷Y" high any transcription. low low normal regulation is to e) The only way to have constitutive expression of both Z and Y with only one change is to have I-d present. It can be on either the chromosome or the F' since I is trans-acting: I'dP÷O*Z'Y÷/F ' I÷P÷O+Z÷Y" o r I÷P÷O÷Z'Y÷/F' I-dp÷o÷z*Y" A few students changed the same element on both the F' and the chromosome. This would work if they both had 0 c or I_. Getting two mutations at once is very unlikely. Problem 2a) i (6pts) The cross of each elongated mutant to the round wild-type yields all round pumpkins. This tells you that both strains carry recessive mutations. When these two strains are crossed to each other the F1 are all elongated. Since both mutations are recessive and the F1 has the mutant phenotype these two mutations must be in the same gene!!! This is in fact a complementation test. The two mutations fail to complement each other and thus are in the same gene. Note that there is only one gene involved here. Always go for the simplest model that fits all the data because we really are not trying to trick you! Pumpkin Wild type Genotype AA Phen ot y pe round Strain Strain alal a2a2 elongated elongated ala2 elongated F1 1 2 (lx2) F2 ii (2pts) ala2, alal, Both strains a2a2 are elongated recessive to wild-type. i iii (2pts) Complementation (See above) (See above) ! Problem 2b) i.(10 pts.) The fact that 3/16 of the progeny in the F2 generation are elongated tells us that there are two mutant genes segregating independently. Therefore, strain 3 must contain two mutations in two genes. Since it is a true-breeding strain, it must be homozygous for these two mutations. If we designate the two genes as A and B, then the parental, FI, and F2 crosses are as follows: Pompkin wild type strain 3 Genotype AABB a a b b Phenotype r ou n d ro u n d F1 AaBb round (3xWT) F2 9 A_B_ 3 aaB_ 3 A_bb 1 aabb round elongated round r o u nd Therefore, strain 3 contains a mutation (designated as "a") that causes elongated pumpkins. However, strain 3 also contains a mutation that either suppresses, or is epistatic to, the "a" mutation and causes round pumpkins. This mutation (designated as "b") only suppresses, or is epistatic, when it is homozygous. That is the reason we see 3/16 of the progeny in the F2 are elongated, since they are homozygous for aa, but are not homozygous for bb. ii. (10 pts) Since in the F2 generation we see a 3:1 ratio, there must be only one gene segregating that is different between strain 3 and strain 1. We know from above that strain 3 carries a mutation that causes elongated pumkins. We also know that strain 1 carries a mutation for elongated pumpkins, but does not carry a suppressor or epistatic mutation. Therefore, we can conclude that strain 3 and strain 1 carry mutations in the same gene which contribute to the elongated pumpkin phenotype. They differ in the suppressor, or epistatic gene: Pumpkin strain 1 strain 3 Gen0typ_ aaBB a a b b Phenotype elongated round F1 aaBb elongated 1 aaBB 2 aaBb 1 aabb elongated elongated r o und F2 (lx3) Some students tried to explain the 3:1 ratio as actually a 12:4 ratio resulting from a dominant mutation. This model is inconsistent with the fact that strain 1, when crossed to wild type, gives all round pumpkins. Therefore, the mutation in strain I must be recessive (see part a,ii.). iii. (3pts.) When strain 3 is crossed to strain 1 all the resulting progeny are elongated. The phenotype of strain 3 is therefor recessive to that of strain 1. iv. (2pts.) As stated suppresses, or is epistatie addition, these results tell us Therefore, the mutations do the correct answers, but no above, the results show that strain 3 contains a mutation that to, the mutation that causes elongated pumpkins. In that strain 1 and strain 3 carry mutations in the same gene. not complement each other. Credit was given for any of credit was given if an incorrect answer was circled. Problem 3: a) Cotransduction frequencies for A-B: Cotransduction frequencies for A-C: The frequency. closer two markers are to each other the higher the co-transduction Thus the A-B interval is the larger of the two. b) Many derived from the The order of the The map derived students misread this question. The question did not ask for the F' Hfr strain. It asked for the F' that integrated to b¢c0m¢ an Hfr strain. genes was determined from the information in part a). from the data in part a was as follows: _+ A+ ¢+ x1 _A A A A 80% +4% = 84 % 80% + 15% = 9 5 % crossover crossover crossover crossover A- or combinations ¢- ¢- event to generate since they result an Hfr that transfers B+ x1 A- A+B+C+ A+B+CA+B-C+ A+B-C- are not possible A+ x2 outside both B and C outside B and in interval 2 outside C and in interval 1 in interval 1 and interval 2 Note that other The crossover C+ x2 t_- 80% 4% 15% 0.6% in A- A+ and B+ early (we selected A+). and C+ late: / Ft \ Bc) c 1 B+ A-t- _- A- A-¢+ Two answers _ were 17+ I J C- l_+ _ chrom. A+ ¢- C- A+ C+ ¢- _ B+ A+ !_+ A- B- IlL _t/ ¢+ A- B- accepted: The frequency that A and D are both packaged into the same phage, given that A is packaged, is greater than the co-transduction frequency of these two markers since this requires both co-packaging and an appropriate recombination event. The frequency that A and D are packaged into the same phage is actually quite if you do not assume that a phage has either marker. Since you select for A+ transductants you might expect that since D is only 10kb away it will be co-transduced a reasonable frequency. With this explanation it is possible to imagine that the cotransduction frequency of A and D might be greater than the frequency with which they are packaged into the same phage. No points were awarded if there was no explanation to justify the answer. low at