18. Brown eyes (B) are dominant to blue eyes (b). The presence of freckles (F) is dominant to the absence of freckles (f). What
is the probability that two people who are heterozygous for both traits will have a child that has blue eyes and no freckles?
1 answer:
Answer:
For bbff we have only 6.3% probability
Step-by-step explanation:
If the parents are heterozygous for both traits, them they are represented by:
BbFf × BbFf
Parent 1: BbFf
Parent 2: BbFf
We have to find the percentage of occurence of bb × ff, which is a child that has blue eyes and no freckles, with no dominant factor.
By distributing the possibilities in a Punnett square, <em>vide</em> picture. We have the following possibilities:
Genotype Count Percent
bBfF 4 25
BBfF 2 12.5
bBFF 2 12.5
bBff 2 12.5
bbfF 2 12.5
BBFF 1 6.3
BBff 1 6.3
bbFF 1 6.3
bbff 1 6.3
For bbff we have only 6.3% probability
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Answer:
5.43×10^14 or
Step-by-step explanation:
Scientific notation, the number must always be less than 10, in this case 5.43. The exponent represents how much times I moved the decimal point to the left.