Question

I need c and d

13. Genetics. Hereditary characteristics are determined by pairs of genes. A gene pair for a particular characteristic is transmitted from parents to offspring by choosing one gene at random from the mother's pair, and, independently, one at random from the father's Each gene may have several forms, or alleles. For example, human beings have an allele (B) for brown eyes, and an allele (b) for blue eyes. A person with allele pair BB has brown eyes, and a person with allele pair bb has blue eyes. A person with allele pair Bb or bB will have brown eyes-the allele B is called dominant and b recessive. So to have blue eyes, one must have the allele pair bb. The alleles don't "mix" or "blend" a) A brown-eyed (BB) woman and a blue-eyed man plan to have a child. Can the child have blue eyes? chance that the child has brown eyes. Find the chance that the child has brown eyes. b) A brown-eyed (Bb) woman and a blue-eyed man plan to have a child. Find the c) A brown-eyed (Bb) woman and a brown-eyed (Bb) man plan to have a child. A brown-eyed woman has brown-eyed parents, both Bb. She and a blue-eyed man have a child. Given that the child has brown eyes, what is the chance that the woman carries the allele b? b

Answer 1

c)

P(child has Brown eyes)=1-P(child has blue eyes)=1-P(both parents given b allele)=1-(1/2)*(1/2)=1-1/4

=3/4

d)P(mother has BB allele)=P(both parent gives B allele)=(1/2)*(1/2)=1/4

P(mother has Bb allele)=P(one parent given B and other b allele)=2*(1/2)*(1/2)=1/2

P(child has brown eye)=P(mother has BB allele and B allele from mother)+P(mother has Bb allele and child has B allele from mother)=(1/4)*1+(1/2)*(1/2)=1/2

hence P(mother has Bb allele given child has brown eye)

=P(mother has Bb allele and child has B allele from mother)/P(child has brown eye)

=(1/2)*(1/2)/(1/2)=1/2