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PuaAnela808 PuaAnela808
wrote...
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9 years ago
I was able to figure out most of the questions for my genetics assignment except for 2 questions, can someone please help? Thank you!

It has been reported that the purple flower color in one species of the Japanese morning glory may be caused by dominant alleles at either one of two separate gene pairs, e.g., A-bb or aaB-. When dominant alleles are present at both gene pairs (A-B-), the color of the flower is blue, and when both are homozygous recessive (aabb), the color is scarlet. If AAbb was crossed with aaBB the flower color would be blue. If these blue flowers (F1) were crossed,
what would be the phenotype and genotype of the F2? If the F1 were crossed to either of the parents, what
would be the phenotype and genotype of the F2?


A woman has migraine headaches (homozygous condition), is left-handed, and carries the trait for Tay-Sachs disease. She marries a man who has no headaches, has no trait for Tay-Sachs disease, and is right-handed (however, his mother is left-handed). What are the possible phenotypes of their children?


Present   Absent   M?   mm
Right handed   Left handed   L?   ll
Absent   Present   T?   tt
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wrote...
Staff Member
9 years ago
I'll show you the work...
AaBb x AaBb Leftwards Arrow- parents
...
(AB, aB, Ab, ab) x (AB, aB, Ab, ab) Leftwards Arrow- gametes
...
A_B_ (blue) = 9 Leftwards Arrow--F2 genotypes
A_bb (purple) = 3
aaB_ (purple) = 3
aabb (scarlet) = 1

Blue: purple: scarlet = 9:6:1 Leftwards Arrow- F2 phenotype ratio

Backcross (F1 crossed to either of the parents) =
AaBb (F1) x AAbb
...
(AB, aB, Ab, ab) x Ab
...
AABb (Blue) : AaBb (Blue) : AAbb (Purple) : Aabb (Purple) = 1:1:1:1
Blue : Purple = 1:1

AaBb (F1) x aaBB
...
(AB, aB, Ab, ab) x aB
...
AaBB (blue) : aaBB (purple) : AaBb (blue) : aaBb (purple) = 1:1:1:1
Blue: Purple = 1:1
Source  https://answers.yahoo.com/question/index?qid=20141207132301AA90N7v
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