7.012: Introductory Biology-Fall 2004 Instructors: Professor Eric Lander, Professor Robert A Weinberg, Dr. Claudette Gardel Solutions to 7.012 Problem set z Question 1 A female geneticist has difficulty characterizing flies by their eye color because she herself is red-green colorblind(an X-linked recessive phenotype). She has one sister(colorblind) and one brother(not colorblind) a)Sketch a pedigree that includes the geneticist, her sister, her brother, and their parents: indicate genotypes (clearly define your genotype symbols) cb is colorblind allele (recessive phenotype 1 father Xcb/Y 3 geneticist Xcb/Xck 2 mother Xcb/X 4 sister Xcb/ Xcb 5 brother X/Y *With random mating-it's as though there are a pool of gametes from which to choose. So imagine a pool of X chromosomes, a mixture of Xs and XCDs To make a male, you pick one randomly, therefore the chance of XCD/Y Females you pick 2 chromosomes randomly p2 =frequency of:+/+ 2pq=frequ b)If 8%o of males in a human population are red-green colorblind, what are the frequencies of the wild- type and colorblindness alleles? Let p=fraction X Let q= fraction Xco out of total pool of X chromosomes. To make a male, you take/ X chromome probability Xco/r=q ana X+/Y=p therefore q=0.08 and p=0.92 c)What %o of females in this population should be red-green colorblind? probability= XCD/XCb= g2=(0.8)2=0.64% d) What %o of females in this population should be carriers for red-green colorblindness? 14.7 or XCo/X= 2pq =14.7901 Solutions to 7.012 Problem Set 7 Question 1 A female geneticist has difficulty characterizing flies by their eye color because she herself is red-green colorblind (an X-linked recessive phenotype). She has one sister (colorblind) and one brother (not colorblind). a) Sketch a pedigree that includes the geneticist, her sister, her brother, and their parents; indicate genotypes (clearly define your genotype symbols). *With random mating-it's as though there are a pool of gametes from which to choose. So imagine a pool of X chromosomes, a mixture of Xs and Xcbs. To make a male, you pick one randomly, therefore the chance of Xcb/Y = q. Females you pick 2 chromosomes randomly. p2 = frequency of: +/+ 2pq = frequency of cb/+; b) If 8% of males in a human population are red-green colorblind, what are the frequencies of the wild￾type and colorblindness alleles? Let p= fraction X+ Let q= fraction Xcb out of total pool of X chromosomes. To make a male, you take 1 X chromome: ~ probability Xcb/Y = q and X+/Y = p therefore q = 0.08 and p = 0.92 c) What % of females in this population should be red-green colorblind? ~ probability= Xcb/Xcb = q2 = (0.8)2 = 0.64% d) What % of females in this population should be carriers for red-green colorblindness? 14.7% or Xcb/X = 2pq =14.7% 1 2 3 4 5 1 father Xcb/Y 2 mother Xcb/X 3 geneticist Xcb/Xcb 4 sister Xcb/Xcb 5 brother X/Y cb is colorblind allele (recessive phenotype) MIT Biology Department 7.012: Introductory Biology - Fall 2004 Instructors: Professor Eric Lander, Professor Robert A. Weinberg, Dr. Claudette Gardel