Question

Where did the 2pq/p^2+2pq come from?

17. (8 points) Below is a pedigree from a family that lives in a population at equilibrium where the frequency of an autosomal recessive genetic allele is 1%. II-1 is affected by the disease. What is the probability that the child (III-1) will have this disease? Show your work and circle your answer. NO = p(dad is Aa) x p(mom is Aa) x 4 (2 points) = 2/3 x (2pq/p^2 + 2pq) x 74 (2 points) = 2/3 x (2*0.01) x 44 = 0.00333=0.333% (4 points)

Answer 1

Ans

Here the Pedigree from of recessive autosomal recessive genetic allele is Aa ( pq)1% = 0.01

In the Pedigree both must be heterozygous allele ( pq) Aa.

गियर ACPI al) समय CP)

During the affected recessive autosomal allele from father and mother allele a , that from generation II - 1 must be aa. Which is dependent on the dominant phenotype A by generation II - 2. From the parents ( generation - I 1 and I - 2 ) Aa.

- Both must be heterozygous Aa affecte alleel

In generation III shows dominant phenotype that is A., So the probability of that generation III-1 is Aa.

Which is the probability that generation II - 2 is heterozygous and passed the a allele to generation III -1

• X( 1/2 ) × ( 1/2 ) = X1/4
• So the conclusion is that generation III- 1 is heterozygous allele with probability of 1/4.
• p ( dad Aa ) × p ( mom Aa) = p (1/2) × p (1/2) = X^{​​​​​1/4}
• So it's dependent on the ancestry of father , that given probability of 1/2 .
• Fom that a allele of dad and mom probability , which presents incorrect . So for the prior probability getting head on a Singal toss is 1/2. Which having two equal possible results A or A ( mom or dad). Which given random AA, Aa, aa, Aa all are equally like . The prior probability of one head is 3/4.
• The prior probability that any combination( Aa = pq) is with one head having two tails is 2/3 ( AA, Aa,aa)
• 2/3 ( 2Aa/ A^2 + 2Aa) X^{​​​​​​1/4} = 2/3 ( 2pq/p^{​​​​​​2} + 2pq)X^{​​​​​​1/4}
• Here Aa = 0.01 = 1%
• 2/3 X ( 2× 0.01) X ^1/4 = 0.00333 = 0.333%
• The probability is 0.333%
• The heterozygous (I-1 and I -2 ) and generation II -2 having dominant phenotype. Which is inherited dominant allele from parents and probability of that chlid wiil inherited recessive allele from the other parent and heterozygous is 1/2.

While 2/3 of unaffected child are heterozygous prior a , having risk is also 2/3. It is knowing of one allele, having lost one degree of freedom.