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BIO322: Genetics Douglas J. Burks Department of Biology Wilmington College of Ohio Problem Set 3 Due @ 10:35 AM January27, 2011 Chapter 4: Problems 3, 5, 12, 23, 25, 31, 37, and 41. Chapter 5: Problems 8, 14, 24, and 33. Problem #3. The figure that follows shows the metaphase chromosomes of a male of a particular species. These chromosomes are prepared s they would be for a karyotype, but they have not yet been ordered in pairs of decreasing size. a. b. c. d. e. How many centromeres are shown? 7 How many chromosomes are shown? 7 How many chromatids are shown? 14 How many homologous chromosomes are shown? 3 How many chromosomes on the figure are metacentric? Acrocentric? 2 metacentric and 5 acrocentric f. What is the likely mode of sex determination in this species? What would you predict to be different about the karyotype of female in this species? This appears to be an XO system in which males have a single X and no Y and females have XX. I would predict that females would have two of the small chromosome. 1|Page BIO322 Spring 2011 Problem #5. Indicate which of the cells numbered i – v matches each of following stages of mitosis: a. b. c. d. e. Anaphase iii Prophase i Metaphase iv G2 ii telophase/cytokinesis v Problem #12. The five cells shown in figure a – e below are from the same individual. For each cell, indicate whether it is in mitosis, meiosis I, or meiosis II. What stage of cell division is represented in each case? What is n in this organism? 2|Page BIO322 Spring 2011 a. b. c. d. e. f. is anaphase I of meisosis is metaphase of mitosis is Telophase II of meiosis is anaphase of mitosis is metaphase II of meiosis 1n = 3 Problem #23. A system of sex determination known as haplodiploidy is found in honeybees. Females are diploid, all males (drones) are haploid. Male offspring result from the development of unfertilized eggs. Sperm are produced by mitosis in males and fertilize eggs in the females. Ivory eye is a recessive characteristic in honeybees; wild-type eyes are brown. a. What progeny would result from an ivory-eyed queen and a brown-eyed drone? Give both genotype and phenotype for progeny produced from fertilized and nonfertilized eggs. ♀ ii and ♂ I are P generation. F1 = ♀ iI brown and ♂ i ivory. b. What would result from crossing a daughter from the mating in part a with a browneyed drone? P = ♀ iI and ♂I F1 ♀ iI and II both brown and ♂I and I 50% ivory and 50% brown. Problem #25. Barred feather pattern is a Z-linked dominant trait in chickens. What offspring would you expect from (a) the cross of a barred hen to a non barred rooster Males are WW and females are ZW. The female is Z/WB and the male is Wb/Wb F1 ♂ WBWb, barred ♀ ZWb nonbarred (b) the cross of an FI rooster from part (a) to one of his sisters? ♂ WBWb, barred x ♀ ZWb nonbarred F2 ♂ WBWb barred and WbWb non-barred ♀ ZWb nonbarred and ZWB barred. Problem 31. The pedigree that follows indicates the occurrence of albinism in a group of Hopi Indians, among whom the trait is unusually frequent. Assume that the trait is fully penetrant (all individuals with a genotype that could give rise to albinism will display this condition). a. Is albinism in this population caused by a recessive or a dominant allele? The trait skips generations in several cases. Since the trait is fully penetrant and common this indicates recessive. b. Is the gene sex-linked or autosomal? If the trait is sex linked male II-7 should be deaf as should male as should male III-5. The trait must be autosomal. What are the genotypes of the following individuals? c. individual I-1 dd d. individual 1-8 Dd e. individual 1-9 Dd 3|Page BIO322 Spring 2011 f. individual II-6 Dd g. individual II-8 Dd h. individual III-4 Dd Problem #37. Several different antigens can be detected in blood tests. The following four traits were tested for each individual shown: (IA and IB codominant, i recessive) (Rh + dominant to Rh-) type (M and N codominant) (Xg(a+) dominant to Xg(a-) ABO type Rh type MN Xg(a) type All of these blood type genes are autosomal, except for Xg(a), which is X linked. Mother AB RH- (RH-RH-) MN Xg(a+) Daughter A AB A B O no RH+ (RH+RH-) MN Xg(a-) (a-a-) RH+ M RH- No N Xg(a+) No sex linked Xg(a-) RH+ N Xg(a-) RH- MN Xg(a-) Alleged father 1 Alleged father 2 Alleged father 3 Alleged father 4 a. Which, if any, of the alleged fathers could be the real father? Male 3 b. Would your answer to part a change if the daughter had Turner syndrome (the abnormal phenotype seen in XO individuals)? If so, how? In this case if it was the male who did not contribute an X then both male 1 or 3 could be father. 4|Page BIO322 Spring 2011 Problem #41. The pedigree at the bottom of the page shows the inheritance of various types of cancer in a particular family. Molecular analysis (described in subsequent chapters) indicate that with one exception, the cancers occurring in the patients in this pedigree are associated with a rare mutation in a gene called BRCA2. Which individual is the exceptional cancer patient whose disease is not associated with a BRCA2 mutation? I would say individual III-5 since he exhibits a cancer but is not part of the lineages carrying the BRCA2 allele. If we knew the cause of death of others they too might be exceptions. Is the BRCA2 mutation dominant or recessive to the normal BRCA2 allele in terms of its cancer-causing effects? If you assume that individuals who are deceased may have exhibited some effects and that expressitivity and penetrance are not complete then I would say dominant. Is the BRCA2 gene likely to reside on the X chromosome, the Y chromosome, or an autosome? How definitive is your assignment of the chromosome carrying BRCA2? As a hypothesis, I would say sex linked. More females than males show the trait which is expected and males who show the trait have daughters with the trait. The pedigree is also consistent with autosomal. I would need to see several more pedigrees. If autosomal then the trait is sex-limited to some extent.`` Is the penetrance of the cancer phenotype complete or incomplete? Individuals II-1 and II-2 indicate that penetrance is incomplete Is the expressivity of the cancer phenotype unvarying or variable? Appears variable Are any of the cancer phenotypes associated with the BRCA2 mutation sex-limited or sex-influenced? Males can’t have ovarian cancer and there is no mention of testicular cancer so this is sex-limited. The breast cancer penetrances appears to be sex influenced. One could study estrogen and other sex hormone levels in males who exhibit breast cancer. How can you explain the absence of individuals diagnosed with cancer in generations I and II? Low penetrance and/or incomplete records of cause of death. Length of life different or not good records. 5|Page BIO322 Spring 2011 Problem #8. In Drosophilia, males from a true-breeding stock with raspberry-colored eyes were mated to females from a true-breeding stock with sable-colored bodies. In the F1 generation, all the females had wild-type eye and body color, while all the males had wild-type eye color but sable-colored bodies. When F1 males and females are mated, the F2 generation was composed of 216 females with wild type eyes and bodies, 223 females with wild-type eyes and sable-bodies, 191 males with wild-type eyes and sable-bodies, 188 males with raspberry eyes and wild-type bodies, 23 males with wild-type eyes and bodies, and 27 males with raspberry eyes and sable bodies. Explain these results by diagraming crosses, and calculate any relevant map distances. P = ♂ + raspberry and ♀ sable +. That the males have the mothers phenotype and females are wild type for both traits suggest sex linkage P = ♂ + ras/Y and ♀ sab sab ++ F1 is ♂ sab+/Y and ♀ sab +/ + ras. Notice the linkage Let’s look at males from F2 since they inherit there X genes from the mother and only have one copy so that their phenotype reflects their genotype directly. ♂ sab + + ras ++ Sab ras 191 188 23 27 429 Recombinants/ total 0.117 m.u. m.u P = Female Male F1 = Female Males F2 = Females = males parental types Recombinant types = RF = Rf = RF x 100 = 11.7 ++, ss r+, Y r+, s+ +s, Y Sex linked ?+, +s = 223 ?+,ss = 188 + s , Y = 191 r+, Y =188 ++, Y = 23 Rs, Y = 27 Problem # 14. In corn, the allele A allows the deposition of anthocynanin (blue) pigment in the kernels (seeds), while aa plants have yellow kernels. At a second gene, W produces smooth kernels, while ww kernels are wrinkled. A plant with blue smooth kernels was crossed to a plant with yellow wrinkled kernels. The progeny consisted of 1447 blue smooth, 169 blue wrinkled, 186 yellow smooth, and 1510 yellow wrinkled. 6|Page BIO322 Spring 2011 a. Perform a Chi Square analysis of data. Predict a 1:1:1:1. Total sample is 3312 so you expect 828 of each type. X2 = [(1447 – 828)2/828] + [(169 – 828)2/828] + [(186– 828)2/828] + [(1510– 828)2/828] 2 X = 462 + 524 + 497 + 561 = 2041 actual 1447 169 186 1510 predicted 828 828 828 828 2 X = [(= a -p)2/p] 462.8 524.5 497.8 561.7 2046.8 p= 0.0000 Null hypothesis (genes are not linked) is rejected. Genes are linked. b. Are the a and w loci linked? If so how far apart are they? RF = (169 + 186)/ 3312 = 0.107 m.u. = 0.107 x 10 = 10.7 m.u. c. What was the genotype of the blue smooth parent? Include the chromosome arrangement of alleles. Since all four phenotypes are present then AaWw x aaww Aw/aW X aw/aw d. If a plant grown from a blue wrinkled progeny seed is crossed to a plant grown from a yellow smooth F1 seed, what kinds of kernels would be expected and in what proportions? A- w/ a-w x a-W/a-w All four possible phenotypes in approximately 1:1:1:1 since opposite parents are homozygous recessive for opposite traits. Problem # 24. In the tubular flowers of foxgloves, wild-type coloration is red while a mutation called white produces white flowers. Another mutation, called peloria, causes the flowers at the apex of the stem to be huge. Yet another mutation called dwarf, affects the stem length. You cross a white-flowered plant (otherwise phenotypically wild type) to a plant that is dwarf and peloria but has wild type red flower color. All of the F1 plants are tall with white normal-sized flowers. You cross an F1 plant back to the dwarf and peloria parent, and you see 543 progeny shown in the chart. (only mutants traits are noted.) dwarf, peloria white dwarf, peloria, white wild type dwarf, white peloria dwarf peloria, white 7|Page BIO322 dp+ ++w dpw +++ d+w +p+ d ++ +pw Spring 2011 172 162 56 48 51 43 6 5 a. Which alleles are dominant? In the F1 you observe white (dominant ) tall (dominant) and normal flower size normal (dominant) b. What were the genotypes of the parents in the original cross? dp+/++w x ddwwrr c. Perform a Chi Square analysis to determine if linkage is present Actual Predicted (a-p)2/p 172 67.875 159.735 162 67.875 130.5269 56 67.875 2.077578 48 67.875 5.819751 51 67.875 4.195442 43 67.875 9.116252 6 67.875 56.40539 5 67.875 58.24332 2 X 426.1197 P less than 0.05 genes are linked d. Draw a map showing the linkage relationships of these three loci? Parental linkage is and dco are d++ dp+ ++w +pw d to p d+ and +p dwarf, peloria white dwarf, peloria, white wild type dwarf, white peloria dwarf peloria, white dp+ ++w dpw +++ d+w +p+ d ++ +pw 19.33 d—p 21.17 p – w 8|Page BIO322 Spring 2011 172 162 56 48 51 43 6 5 e. Is there interference? If so, calculate the coefficient of coincidence and the interference values. Yes the expected number of dco is 22 and the observed is 11. 11/22 = .5 for co and interference is 50% Problem #33. Neurospora of genotype a + c are crossed with Neurospora of genotype + b +. The following tetrads are obtained (note the genotype of the four spore pairs in an ascus are listed, rather than listing all eight spores): a+c a+c +b+ +b+ 137 abc abc +++ +++ 141 ++c a+c +b+ ab+ 26 +bc abc +++ a++ 25 ab+ ab+ ++c ++c 2 a+c abc +++ +b+ 3 a. In how many cells has meiosis occurred to yield these data? 334 b. Give the best genetic map to explain these results. Indicate all relevant genetic distances, both between genes and between each gene and its respective centromere. First designate the type of asci present. Remember that this will be done for each pair of genes (PD (P), NPD (n), and T . When PD = NPD then the genes are not linked. A – b PD = 137 and NPD = 141 + 2 = 143 PD = NPD therefore not linked. B – c PD = 137 + 26 + 2 = 165 and NPD = 141 + 25 = 166 PD = NPD therefore not linked. A – c PD = 137 + 141 = 278 and NPD = 3 PD > NPD therefore linked. Mu = [(PD + 1/2T)/total ] x 100 [2 + 1.2(26 +25)/334] x 100 = 8.2 cM = second division segregation * ½ / total For a + and a + = [(26 + 25)x .5]334 = 0.076 = RF m.u = 7.6 For b + and b + = (3 x 0.5)/334 = 0.004 Rf = 0.4 m.u. For + c and + c = 0 on top of centromere c. Diagram a meiosis that could give rise to one of the three tetrads in the class at the far right in the list. 9|Page BIO322 Spring 2011