Answers to Problem Set 2 |
1a. Complementation Test - This test is used to determine if mutations are in te same or different genes. If a cross between 2 mutants produces wild type progeny the mutations are in different genes. If the mutant phenotype remains then you can assume they are in the same gene.
1b. Minium is 4, Maximum is 6 - Notice that mutations b and C show dominant inheritance (except when crossed to each other, in which case there is incomplete dominance). Because B&C are dominant to all other alleles you cannot determine from a complementation test whether they are separate genes or dominant alleles of one of the genes described by te other complementation groups. Complementation Groups: (A,G,H) (D,I) (E,J) (F) - B & C Either part of one of the previous groups, 2 alleles of one new gene group or 2 different genes not described by any of the previous groups.
1c. - You can map the mutations to determine the exact number of genes involved. If B&C mutations map to different locations than the other mutations, they are probably in different genes.
2.
| Minimal media supplemented with: | ||||||||
| Mutant for enzyme | Minimal media | A | B | C | D | E | F | G |
| 1 | - | - | + | - | - | + | + | + |
| 5 | - | - | - | - | - | - | + | + |
| 2&3 | - | - | - | - | - | + | + | + |
| 4&6 | - | - | - | - | - | - | - | + |
| 1&3 | - | - | - | - | - | + | + | + |
Because all of the mutants have the G-phenotype (they cannot synthesize G) both pathways must function to create E, one of the G precursors. If there are mutations in both pathways, providing an intermediate in one of the pathways will not enable the organism to grow.
3. - 1=e, 2=b3=d, 4=g, 5=a, 6=f, 7=h, 8=c - Step a (label human DNA probe fragment) can be any of the steps 1-5 as long as it is done before f (add probe to filter)
i. - the mutation could be in a non-coding region (either between genes or in an intron). The mutation could change the base but not the amino acid it encoded (such as a mutation in the 3rd codon). The mutation may change the amino acid but not affect the structure or function of the protein.
4a. -
i) 6kb from allele A and 20kb from allele B
ii) A band greater than 14kb from both alleles A&B
iii) 6kb from allele A and 14 kb from allele B
4b. -
i) 14 kb from allele A and 20 kb from allele B
ii) 12 kb from both alleles A & B
iii) 6 kb from both alleles A & B
4c. - ai, aiii, and bi - All three give different size fragments from the different alleles.
5. - for diagram, please see handout
a) RFLP 1&3 appear to be linked because there are few recombinations present.
b) RFLP 1: allele 1 inherited from II-1 (who inherited it from I-2) is linked to the disease. Notice that individual II-2 is homozygous for allele one but does not have the disease. Therefore, if allele 1 is inherited from the mother and not from the father, it is not linked to the disease and the offspring should not show the disease phenotype. RFLP3: allele 2 is linked to the disease.
c) Divide the number of recombinants by the total number of offspring (samples) to get the recombination frequency. A smaller recombination frequency means that the marker is closer to the disease gene.
| RFLP 3, allele 2 is closer | RF = 1/12 = 0.08 = 8 m.u. |
| RFLP 1, allele 1 is farther | RF = 2/12 = 0.16 = 16 m.u. |
6) C + M = D in all cases and B = K in all cases (= does not have to mean exactly equal) therefore we can say that D pairs with either C or M and B pairs with K.