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Fantom [35]
3 years ago
13

For three autosomal dominant genes, what is the probability that a cross between two heterozygotes (i.e. AaBbCc x AaBbCc) will p

roduce an offspring with the dominant trait for all genes?
Biology
1 answer:
Free_Kalibri [48]3 years ago
5 0

Answer:

27/64 or 0.421

Explanation:

When AaBbCC X AaBbCc :

A_B_C_   =   27

aaB_C_    =   9

A_bbC_    =   9

A_B_cc     =   9

aabbC_     =   3

A_bbcc     =    3

aaB_cc      =   3

aabbcc      =    1

Out of the total 64 offspring, 27 will have at least one dominant allele for each of the three genes hence these 27 offspring will show dominant trait for all genes. So the probability of producing a dominant trait for all genes will be 27/64 or 0.421

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3 years ago
What is the probability that the first offspring from the cross listed below will be heterozygous for all loci?
Tanya [424]

Answer and Explanation:

<u>Cross:</u> aa Bb dd Ee x AA bb Dd Ee

We can calculate the probability of getting heterozygous individuals in the progeny by using the <u>product rule</u>. Assuming that these four genes <u>assort independently</u> (<em>events that occur independently from each other</em>), we can infer that the F1 will have the next genotypic proportions for each gene:

1) aa     x     AA

F1) 4/4=1 Aa

2) Bb    x    bb

F1) ½ bb

     ½ Bb

3) dd    x    Dd

F1) ½ dd

     ½ Dd

4) Ee    x    Ee

F1) ¼ EE

    2/4 = ½ Ee

     ¼ ee

So, to know what the probability is that the F1 of being heterozygous for all loci, we must multiply the respective individual probabilities of getting a heterozygous genotype, like this:  

1 Aa   x   ½ Bb   x    ½ Dd    x    ½ Ee = 1/8 AaBbDdEe

                                 

6 0
3 years ago
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