a mother has one allele for color blindness and one allele for normal vision. what is the probability that her gamete will have
2 answers:
Answer: The correct answer is -
The probability that gamete of female will have the allele for color blindness is 1/2 that is half.
Explanation:
Color blindness is a X linked recessive disorder. It means that a female can only have color blindness if the gene for color blindness is carried by both the X chromosomes (that is homozygous recessive genotype) .
As per the given information in the question, mother contains one allele for color blindness and one for normal vision
This means that mother is a carrier of colorblindness.
Therefore, two different gametes produced by her are- X and .
Thus, probability of gamete having allele for colorblindness is 1/2.
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Answer:
1.875 is the expected number of children the couple will have.
Explanation:
<em>The probability of an event </em><em>X</em><em> is denoted by </em><em>P(X)</em><em>.</em>
Let assume that X is the number of children to whom the couple will give birth. The couple can have at most <em>four children</em> with the following probabilities.
- The couple will have one child if the first is boy, thus:
For one child P(X)=1/2=0.50
- The couple will have two children if the first is girl and the second is a boy, thus:
For two children P(X)=0.5×0.5=0.25
- The couple will have three children if the first two are girls and the third is a boy, thus:
For three children P(X)= 0.5×0.5×0.5=0.125
- The couple will have four children if the first three are girls and the fourth is a boy, or if all the four are girls. thus:
For four children P(X) =(0.5×0.5×0.5×0.5)+(0.5×0.5×0.5×0.5)=0.125
<u>Now calculating the expected number of children the couple will have:</u>
NOTE: Formula is given in the attached pic
Expected number of children = E(X)= (1×0.50)+(2×0.25)+(3×0.125)+(4×0.125)
E(X)= 1.875
