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ki77a [65]
3 years ago
11

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?

Mathematics
1 answer:
MrRissso [65]3 years ago
6 0

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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See attachment below

Step-by-step explanation:

For the first option, we are given m = - 2 / 3, b = 3. This is likely in the point - slope form y = mx + b, where m = slope and b = y - intercept. Thus, the equation of the line in point - slope form given this information, should be y = - 2 / 3x + 3. It seems as if none of the following equations match this form, so let us interchange the equation a bit,

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_______________________________________________________

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_______________________________________________________

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Take a look at the attachment below for further help;

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