Can you please explain each part in detail? Especially part C 16. (16 points) Bo
ID: 277056 • Letter: C
Question
Can you please explain each part in detail? Especially part C
16. (16 points) Both hemophilia and red-green color blindness are X-linked recessive traits and complete penetrance. The normal allele, in non-hemophiliacs, is denoted by H and the mutant allele is h. The normal color vision allele is R and the color blindness allele is r. The two genes are separated by 10 map units. In the pedigree below, the male in the first generation is affected by both hemophilia and color-blindness, and the male in the second generation has color- blindness only. Show your work and circle your answer. CHI (both hemophilia and color-blindness) (color-blindness only) IV a. What is the probability that the individual (IV-1) has both hemophilia and color-blindness? P(III-1 is Hr/hR) x P(hr from II1)x P(Y from III-2) (0.1)x (1/2 x 0.1) x (1/2-0.0025 0.25% (4 points) b. What is the probability that IV-1 has normal vision but is affected by hemophilia? (4 points) P(111-1 is Hr/h R) x P(hR from 111-1) x P(Y from 111-2) = (0.1) x (1/2 x 0.9) x (1/2,-0.0225- 2.25% C. (6 points) What is the probability that IV-I is color blind but not affected by hemophilia? P(III-1 is Hr/hR) x P(Hr from III-1) x P(Y from III-2) +P(III-1 is Hr/HR) x P(Hr from III- 1) x P(Y from III-2) (0.1) x (1/2 x 0.9) x(1/2)(0.9) x (1/2) x (1/2) -0.0225 + 0.225-.2475-24.75% d. If IV-1 is a colorblind girl, what is the most likely explanation? (2 points) She must have XO genotype (Turner syndrome). The nondisjunction must happen in 2 (her father) during either meiosis I or II.Explanation / Answer
Answer-
The probability of simulteneous occurance of two independent events is product of their individual probabilities. Probability of hr from Hr/hR will be 1/2 x probability of crossing over between H and R locus.
probabiloty of mutually exclussive event is the sum of individual probabilities of possible events.
(a) Probability . P (hr)
As the genes H and R are 10 cM apart there is 10% probability of crossing over so,
the probability of finding III-1 genotype Hr or hR will be 0.1.
from III-1 the possible genotypes are two and crossing over also needed to get hr so the probability from III-1 will be 1/2 x 0.1. III-2 is unaffected it will contribute Y. P(Y) = 1/2.
so the answer is P (hr) = 0.0025 = 0.25 %.
Answer (b)
normal vision and hemophilia genotype = hR
probability of finding hR from III-1 = 1/2 x (1- probability of crossing over) = 1/2 (1- 0.1) = 1/2 x 0.9
others are same as section a
P (hR) = 0.0225 = 2.25 %
Answer (c)
we need to find probability for finding genotype Hr
P (Hr) = ?
here two possibilities of III-1 genotype must be considered 1) there is cossing over between H and R and there is no crossing over. genotypes Hr/hR and Hr/HR. these are mutually exclussive events , so we need to add their probabilities.
Answer (d)
As the III-2 is normal so it can not contribute colour-blind gene. so the given explaination is correct.