ndividuals III-6 and III-7 have two little monstersand are expecting a third one. Their two childrenhave two eyes. What is the c
1 answer:
Answer:
The correct answer is - 75% or 3/4.
Explanation:
In the given question, there is a cross between III-6 and III-7 that produced two double eyes monsters and expected one more. As given in the image that the A or single eye is dominant over the a or double eye then both parents must be heterozygous as they produced double eye offspring (aa) and they do have only one eye.
then chances of single eye offspring would be -
cross: Aa and Aa
gametes: A, a and A, a
Punnett square:
A a
A AA Aa
a Aa aa
so there are 75% chances to have a single eye and 25% chances for a double eye.
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Answer:
a) 1/16, b) 4/16, c) 1/16, d) 1/16, e) 4/16, f) 4/16, g) 0 + alleles, h) 1/256, i) 1 + alleles, j) 4/256, k) 2 + alleles, l) 16/256, m) 3 + alleles, n) 24/256, o) 4 + alleles), p) 36/256, q) 5 + alleles, r) 24/256, s) 6 + alleles, t) 16/256, u) 7 + alleles, v) 4/256, w) 8 + alleles, x) 1/256.
You will find the uncompleted Punnett square (missing in the problem statement) and the complete modified Punnett square in the attached files
Explanation:
Due to technical problems, you will find the complete explanation in the attached files
