HOMEWORK 3 ? SEX DETERMINATION, SEX-LINKED INHERITANCE, LINKAGE Name _________________________________ Net ID__________________ Lab No______ 1. A female child is determined to be colorblind. Her Mother is not colorblind. Use C for normal vision, c for colorblind. A. Draw a pedigree and give the genotypes and phenotypes of her parents Answer: Parents: Mother Cc X Father cY Daughter: cc B. What is the probability that a bother will be colorblind and why? Answer: Her brother will inherit his X chromosome from the Mother. Thus, he has a 50% chance of being colorblind (C or c). He will inherit the Y chromosome from his father. 2. An X-linked dominant allele causes hypophosphatemia in humans. A man with hypophosphatemia marries a normal woman. What proportion of their sons will have hypophosphatemia and why? Answer: Sons will receive their X-chromosome from the mother who is normal and none of them will have the disease. All of the daughters will have the disease however. 3. Duchenne muscular dystrophy is sex-linked and usually affects only males. Victims of the disease become progressively weaker starting early in life. A. What is the probability that a woman who?s brother has Duchenne muscular dystrophy will have an affected child? Draw a pedigree and show your work. The ?maternal grandmother? had to be a carrier, D/d. The probability that the woman inherited the d allele from her is 1/2. The probability that she passes it to her child is 1/2. The probability that the child is male is 1/2. The total probability of the woman having an affected child is 1/2 ? 1/2 ? 1/2 = 1/8. B. If your mother?s brother had the disease, what is the probability that you have received the allele? Answer: The same pedigree as part (a) applies. The ?maternal grandmother? had to be a carrier, D/d. The probability that your mother received the allele is 1/2. The probability that your mother passed it to you is 1/2. The total probability is 1/2 ? 1/2 = 1/4. C. If your father?s brother had the disease, what is the probability that you have received the allele? Show your work. Answer: Because your father does not have the disease, you cannot inherit the allele from him. Therefore, the probability of inheriting an allele will be based on the chance that your mother is heterozygous. Since she is ?unrelated? to the pedigree, assume that this is zero. 4. Male house cats are either black or yellow. Females are black, yellow or calico. A. If these coat-color phenotypes are governed by a sex-linked gene, how can these observations be explained? This problem involves X inactivation. Let B = black and b = orange. Females Males XB/XB = black XB/Y = black Xb/Xb = yellow Xb/Y = yellow XB/Xb = calico The calico female is heterozygous and shows different coat colors depending on which allele is active in a given area of the skin. B. Using appropriate symbols, determine the phenotypes expected in the progeny of a cross between and yellow female and a black male. P Xb/Xb (yellow) ? XB/Y (black) F1 XB/Xb (calico female) Xb/Y (yellow male) C. Using appropriate symbols, determine the phenotypes expected in the progeny of a cross between and black female and a yellow male. P XB/XB (black) ? Xb/Y (orange) F1 XB/Xb (calico female) XB/Y (black male) D. Half of the females produced by a cross are calico and half are black. Half of the males are yellow and half are black. What colors are the parents? Because the males are black or yellow, the mother had to have been calico. Half the daughters are black, indicating that their father was black. E. Another type of cross produces progeny in the following ratios: ¼ black males, ¼ yellow males, ¼ yellow females, ¼ calico females. What are the phenotypes and genotypes of the parents? Show your work. Males were yellow or black, indicating that the mothers were calico. Yellow females indicate that the father was yellow. 5. In Drosophila, nondisjunction for the X chromosome occurred in the female parent during meiosis II. If mated with a normal male, what sexes will the progeny be? Explain why. Assume that lack of an X chromosome is lethal. Answer: The female genotype is AAXX. Nondisjunction for X in meiosis II will yield two eggs, an egg that is AXX and an egg that is A_. Gametes from an AAXY male can be AX or AY. AX AY AXX AAXXX 3/2=metafemale AAXY ½=male A_ AAX ½=male AAY lethal In corn, the ?minute? gene has two alleles, M = normal height, m = minute; The ?bronze? gene also has two alleles, B = green leaf, b= bronze. A test cross is performed to determine if M and B are linked as follows: MmBm X mmbb. The cross produces the following progeny: 42 normal stature, bronze 40 minute, green leaf 10 minute. bronze leaf 8 normal stature, green leaf Answer questions 6-12 using these data. 6. Why is one parent homozygous recessive? In the homozygous recessive parent, crossing over will occur but will not alter the phenotype of the gametes. All of the gametes will be mb (recessive). This allows crossovers in the other (female) parent to be detected easily by phenotype. (You are essentially examining only crossovers from the female parent.) 7. Are the bronze and minute loci located on the same chromosome? How do you know? Bronze and minute are on the same chromosome. If there were not (or were widely separated on the same chromosome), the frequency of offspring would be equal (25% each or 1:1:1:1) 8. What is the configuration of the genes on the chromosomes in the heterozygous parent? (Are both dominant alleles on the same chromosome or different chromosomes?) Show your work The most frequent phenotypic categories represent chromosomes that have not recombined and thus represent the chromosomes present in the heterozygous parent. The most frequent offspring are: 42 normal stature, bronze 40 minute, green leaf For the 42 normal stature, bronze offspring, you know that mb comes from the homozygous recessive parent. Therefore, the heterozygous parent must be Mb to get the observed phenotype. Using the same reasoning, the 40 minute, green leaf offspring must get an mB from the heterozygous parent. The heterozygous parent must be Mb/mB (trans configuration). 9. What percentage of the gametes from the female parent represent recombination events? 18 % of the offspring have chromosomes from the heterozygous parent that have recombined. Therefore, 18% of the gametes from the heterozygous parent represent crossover events. 10. How far apart are the genes on the chromosome? Show your work. 18 map units. 1% crossing over = 1 map unit. 11. A third gene is identified, F = smooth leaf, f = fuzzy leaf. Two-point (two gene) test crosses are performed to map the position of this gene and the following results were obtained. The map distance between bronze and fuzzy was 3.6 units. The map distance between minute and fuzzy was 21. Draw the map arrangement for these three loci. M ___(18)_____________________________B_(3.6)__F 12. If map distances are for M to B and B to F are added, the distance is 21.6. The map distance between M and F, however, was calculated to be 21. Why the difference? When the distance between M and F is determined based on only those two loci, double crossovers that occur are missed and the distance is underestimated. 13. The following pairs of alleles are present in tomatoes: R = red fruit, r = yellow fruit; H = hairy leaves (Pubescent), h = smooth leaves; S = short plants, s = tall plants. A test cross is made between a heterozygote for all three genes and a plant that is homozygous recessive. The offspring from that cross are as follows: Determine the distance between genes R, H and S. Show all work. RHs//rhs = 40 rhS//rhs = 39 rhs//rhs = 4 RHS//rhs = 4 rHS//rhs = 6 Rhs//rhs = 5 RhS//rhs = 2 rHs//rhs = 0 Total progeny 100 In this cross, you are examining only crossovers from the female parent = NCOs (Since the male parent is homozygous, no recombination can be detected.) Chromosomes from the female parent that have not undergone recombination are represented by the largest category of offspring and are RHs and rhS = DCOs Chromosomes from the female parent that have undergone two crossovers (a double crossover event) are the smallest category or rHs and RhS. Since with a double crossover, the allele that is exchanged is the middle allele, the order of the genes can be determined by comparing the NCOs and DCOs RHs vs rHs rhS RhS The ?r? alleles have changed their position with respect to the other genes. R is therefore in the center. The gene order is HRS (using the gene names and ignoring for the time being the actual alleles) or SRH. Rewriting then we get HRs//hrs = 40 hrS//hrs = 39 hrs//hrs = 4 HRS//hrs = 4 HrS//hrs = 6 hRs//hrs = 5 hRS//hrs = 2 Hrs//hrs = 0 Total progeny 100 We can see that the genes are in trans configuration from the NCOs HRs hrS Now the problem is easier to solve. The distance between H and R is represented by HrS and hRs = 6 + 5 + 2 + 0 (DCOs are included because they represent a crossover event also) = 13/100 = 13% The distance between R and S is represented by hrs and HRS = 4 + 4 + 2 + 0 = 8/100 = 8% _H________________13_______________________R__________8_______________S__ PAGE PAGE 1