Suppose two parents, a father with the genotype AaBbCcDdee and a mother with the genotype aabbCcDDEe, want to have children. Assume each locus follows Mendelian inheritance patterns for dominance. What proportion of the offspring will have each of the specified characteristics?

Answer :

Answer:

A. same genotype as the father (AaBbCcDdee) = (½ x ½ x ½ x ½ x ½) = 0.03

B. phenotypically resemble the father (A_B_C_D_ee) = (½ x ½ x ¾ x 1 x ½) = 0.09

C. same genotype as the mother (aabbCcDDEe) = (½ x ½ x ½ x ½ x ½) = 0.03

D. phenotypically resemble the mother (aabbC_D_E_) = (½ x ½ x ¾ x 1 x ½) = 0.09

E. phenotypically resemble neither parent (simply subtract the probability of phenotypically resembling each parent from 1) = 1 – (0.09375 + 0.09375) = 0.81

Explanation:

The computation of the total probability would simply be the product of individual chances for each particular gene.

This can be better gotten with the punnette sqaure

The study of genes and inheritance is called genetics.

The correct answer is 0.81

What are genes?

  • The structural and functional unit of the DNA is genes and able to code the sequence of the DNA.

The answer is as follows:-

  • Same genotype as the father (AaBbCcDdee) =[tex]\frac{1}{2} *\frac{1}{2} *\frac{1}{2} *\frac{1}{2} *\frac{1}{2} [/tex] = 0.03
  • Phenotypically resemble the father (A_B_C_D_ee) = [tex]\frac{1}{2}*\frac{1}{2}*1*\frac{3}{4}*\frac{1}{2} = 0.09[/tex]
  • Same genotype as the mother (aabbCcDDEe) = [tex]\frac{1}{2} *\frac{1}{2} *\frac{1}{2} *\frac{1}{2} *\frac{1}{2} [/tex]= 0.03phenotypically resemble the mother (aabbC_D_E_) =[tex]\frac{1}{2}*\frac{1}{2}* \frac{3}{4} *1*\frac{1}{2} = 0.09[/tex]

Phenotypically resemble neither parent (simply subtract the probability of phenotypically resembling each parent from 1)[tex] = 1 - (0.09375 + 0.09375) = 0.81[/tex]

Hence, the correct answer is 0.81

For more information about the genes, refer to the link:-

https://brainly.com/question/264225

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