Solutions For An Introduction to Genetic Analysis Chapter 6 Problem 35P

2. The wild type phenotype is blue in color.

3. Anything that differs from the phenotype of the wild type strain is known as the variant.

4. Two variants in the form of pink and white are observed in this problem.

5. “In nature” means, that the variants that did not appear in the laboratory stock, but would be found growing in the wild.

6. Probably the variants should have appeared as a small patch or a single plant within a larger patch of wild type.

7. To check the outcome of each cross, seeds would be used.

8. If there is no sex linkage appears then there will not be any changes in the offspring of both the blue x white and white x blue. Similar results would be observed as the trait studied appears to be autosomal.

9. The first two rows showing a ratio of 3:1 in F₂ suggesting segregation of one gene, whereas, the F₂ generations in the third cross showing a ratio of 9:4:3 suggests that two genes are segregating.

10. Blue is dominant to both white and pink.

11. Complementation is the development of wild-type progeny from two mutant strains. The only thing that needs to be considered is that the mutation should have happened in two different genes and not on the same one.

12. The development of blue color necessitates the work of two enzymes that are individually defective in the pink or white strains. The F₁ progeny of this cross is blue since the non-mutated gene from either of the strains would be complemented aiding it to produce the two functional enzymes.

13. Blue color from a pink X white cross arises through complementation.

14. Ratio observed = 3:1; 9:4:3

15. Yes. The monohybrid ratios are observed in the first two crosses.

16. A modified 9:3:3:1 ratio is observed in the third cross.

17. The monohybrid ratio indicates that there is only one gene segregating while in the case of a dihybrid ratio indicates that there are two genes segregating.

18. 1:2:1, 2:1, 9:7, 9:4:3, 9:6:1, 12:3:1, 15:1

19. Yes. There is a modified dihybrid ratio in the third cross.

20. An interaction of two or more genes is the fact that is indicated by a modified Mendelian ratio.

21. Phenomenon of recessive epistasis is indicated by the modified dihybrid ratio.

22. The below diagram represents the chromosomes in the meiosis when blue colored plants are crossed with the white colored plants.

23. (a) A = wild-type; a = white; B = wild type and b = pink

Cross 1

Parent

Blue X White

A/A; B/B X a/a; B/B

F ₁

A/a; B/B all blue

F₂

3 A/-; B/B blue

1 a/a; B/B white

Cross 2

Parent

Blue X White

A/A; B/B X a/a; B/B

F ₁

A/a; B/B all blue

F₂

3 A/A; B/- blue

1 A/A; b/b white

Cross 3:

Parent

Pink X white

A/A; b/b X a/a; B/B

F ₁ :

A/a; B/b all blue

F ₂ :

9 A/-; B/- blue

4 a/a; B/- (3): a/a; b/b (1) white

3 A/-; b/b pink

When allele ‘a’ is homozygous the expression of B or b would be blocked. White phenotype is epistatic to the pigmented ones. The product of gene A develops an intermediate which would be modified by the gene B. The plant a/a does not develop the intermediate product; hence, the phenotype is the same regardless of the ability of the development of functional B product.

(b) The cross between certain blue F₂ plant and a certain white F₂ plant is as follows:

F ₂ :

Blue X White

F ₃ :

3/8 blue

1/8 pink

4/8 white

Begin by writing as much of each genotype as can be assumed

F ₂ :

A/-; B/- X a/a; -/-

F ₃ :

3/8 A/-; B/-

1/8 A/-; b/b

4/8 a/a; -/-

It should be noted that both a/a and b/b are formed in the F₃ progeny. For the formation of both these genotypes to appear the parents should have at least one a and one b.

F ₂ :

A/a; B/b X a/a; -/b

F₃:

3/8 A/a; B/b

1/8 A/a; b/b

4/8 a/a; b/-

The only remaining question to be answered is whether the white parent was homozygous, b/b or heterozygous B/b. In case if the white parent is a homozygous recessive then the cross would be a test cross with the blue parent and the following progeny ratio would be observed.

1 blue: 1 pink: 2 white,

1 A/a; B/b: 1 A/a; b/b: 1 a/a; B/b: 1 a/a; b/b

Since this ratio was not observed the white parent is not a homozygous recessive parent and should be a heterozygous one. This makes the F2 cross A/a; B/b X a/a; B/b.