The pedigree below shows the transmission of albinism in a human family. a. Copy
ID: 191870 • Letter: T
Question
The pedigree below shows the transmission of albinism in a human family.
a. Copy this pedigree and list the genotypes of all individuals.
b. The female on the left in generation I had four children, one of which has albinism. What is the probability that they could have had a total of four children with at least three children having albinism?
c. What is the probability that male on the far right in generation II has no alleles for albinism (is homozygous for the non-albinism allele)?
The pedigree below shows the transmission of albinism in a human family. a. Copy this pedigree and list the genotypes of all individuals. b. The female on the left in generation I had four children, one of which has albinism What is the probability that they could have had a total of four children with at least three children having albinism? c. What is the probability that male on the far right in generation II has no alleles for albinism (is homozygous for the non- albinism allele)?Explanation / Answer
a. List of genotype of generation 1:
From left to right : Aa (mother ), Aa(father) , Aa(mother)
Generation 2. From left to right :
Aa/AA , Aa/AA , aa (diseased), AA/Aa & AA/Aa, AA/Aa , AA/Aa , aa(diseased) , AA/Aa
Where Aa is carrier and AA is normal.
This is a autosomal recessive trait.
b. Probability = No of ways × probability of a particular event
No of ways = nCr = n!/ r!(n - r)!
= 4! / 3! (4-3)!
= 4×3!/ 3!×1!
= 4
Probability of a particular event:
Probability of at least 3 children having albinism among 4 is : 3/4 ×3/4×3/4×3/4
Because probablity of 1st,2nd & 3rd child of being affected
Or chance of affecting 2nd, 3rd &4th child
Or affcted children can be 2nd ,4th,1st or 1st,3rd and 4th child
Probablity = 4 ×(3/4×3/4×3/4×3/4) = 81/64
C.Male on the far right of generation is either homozygous for non albinism gene or heterozygous, means chance is 50%
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