In the plant Arabidoposis thaliana, a geneticist is interested in the developmen
ID: 23356 • Letter: I
Question
In the plant Arabidoposis thaliana, a geneticist is interested in the development of trichomes (small projections) on the leaves. A large screen turns up two mutant plants (A and B) that have no trichomes, and these mutants seem to be potentially useful in studying trichome development. (If they are determined by single-gene mutations, then finding the normal and abnormal function of these genes will be instructive.) Each plant was crossed with wild type; in both cases, the next generation (F1) had normal trichomes. When the F1 plants were selfed, the resuling F2's were as follows:F2 from mutant A: 602 normal; 198 no trichomes
F2 from mutant B: 267 normal; 93 no trichomes
a.) what do these results show? Include proposed genotyes of all plants in your answer.
b.) Assume that the genes are located on separate chromosomes. An F1 is produces by crossing the original mutant A with the original mutant B. This F1 is testcrossed: What propotion of testcross progeny will have no trichomes?
Explanation / Answer
The data for both crosses suggest that both A and B mutant plants are
homozygous for a recessive allele. Both F2 crosses give 3:1 normal to
mutant ratios of progeny. For example, let A = normal and a = mutant, then
P --------- A / A´ a / a
F1 A / a
F2
F1 A / a x F1 A / a
A
a
A
A / A
A / a
a
A / a
a / a
1 ------ A / A phenotype: normal
2 -- A / a phenotype: normal
1 -----a / a phenotype: mutant (no trichomes)
The cross is A/A ; b/b´ a/a ; B/B to give the F1 of A/a ; B/b. This is then test
crossed (crossed to a/a ; b/b) to give
1/4 A/a ; B/b (normal)
1/4 A/a ; b/b (no trichomes)
1/4 a/a ; B/b (no trichomes)
1/4 a/a ; b/b (no trichomes)
1 normal : 3 no trichomes
A
a
A
A / A
A / a
a
A / a
a / a
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.