The diagram below is a pedigree of three generations showing the occurrence of t
ID: 146191 • Letter: T
Question
The diagram below is a pedigree of three generations showing the occurrence of two genetically-transmitted
diseases. Phenotypes of affected individuals are shown. Circles represent females, and squares represent males.
Assume that the diseases are rare, unlinked, and that individuals not blood related to I-a sad I-b (“in-laws") do not have either disease-producing allele.
2. The genotype of Individual l-b for Disease 2 is:
a. homozygous recessive.
b. heterozygous.
c. homozygous dominant.
d. spontaneous mutant
e. Not enough information to determine
3. lf Individual III - r marries a normal man with a homozygous wild-type genotype. what is the probability that their son would have Disease 2?
a. 0
b. 1/4
c. 1/2
d. 3/4
4) If Individual III-v marries a non-disease carrying male, what is the probability that their son would have both diseases?
a-0
b-1/16
c-1/8
d-1/4
5) What is the probability that offspring produced by a mating between Individuals III-j and III-m, which would be afflicted by either Disease 1 or 2?
a-1/4
b-1/2
c-3/4
d-1
6) In a certain randomly-mating population, the frequency of the allele causing Disease 1 is 0.1. what would be the frequency of affected individuals in this population?
a--0.01
b--0.1
c--0.19
d--0.9
Please answer all if you want a high rating, if you can't answer all, leave it so somebody else can answer, do not just answer one or two.
Explanation / Answer
Answer
2) b. heterozygous.
Explanation :genotype of Individual l-b for Disease 2 must be heterozyous because I-b do not shows symptoms of disease(he's not diseased) but her offspring acquired the disease 2. For this She must have atleast one allele of disease
3 a.0
Explanation : as individual II-h has both diseased1 and 2. And from above explanation disease 2 is recessive disorder.
Individual II-h must be homozygous recessive for disease 2. So if II-i which is wife of II-h is homozygous dominant then probability of carrier in their children would be 1/2.
Hence III-r is carrier with a probability of 1/2 if she marries a homozygous dominant male then probability of carrier in their children would be 1/4 and probability of having disease would zero. Because for disease 2 both must contribute one diseased allele to the progeny
4)c-1/8
Explanation : Individual III-v is having disease 1 and her sibling III-y have both the disease. Hence chance that she might be carrier for disease 2 is 1/2.
Therefore if she marries normal without any diseased allele then chance of his offspring with disease 1 would be 1/2 as disease 1 is dominant disorder
And probability of disease 2 would be 1/2*1/2=1/4
Hence probability of having both will be 1/4*1/2=1/8
5) c-3/4
6)a. 0.01
Explanation : 0.1 person means 1/10 probability. Hence all individual in the population have 1/10 probability of having 1 allelea Hence if the two mates then probability of disease in children would be 1/10*1/10=1/100=0.01