Ans_SAQ_ProbSet7_chr7,8

Assignment 7 for Biol 311 (2023 Fall, Section 2) Coverage: Part of Chapter 7, Chapter 8; Snustad and Simmons, Principles of Genetics (7th ed) (due 11:59 PM Oct 26, 2023) 1. Multiple Choice (48 points; 3 points each) Check the auto-graded questions. 2. Short-answer questions (33 points) 1) List the three major genetic exchange processes in bacteria. (3 points) Answer: Omitted 2) A homozygous variety of maize with red leaves and normal seeds was crossed with another homozygous variety with green leaves and tassel seeds. The hybrids were then backcrossed to the green, tassel-seeded variety, and the following offspring were obtained: red, normal 124; red, tassel 126; green, normal 125; green, tassel 123. Are the genes for plant color and seed type linked? Explain. (3 points) Hint: no chi-square test is needed. Use the intuitive to assess if they follow independent assortment or there is gene linkage. Answer: No, the genes are not linked. The leaf color and tassel seed traits assort independently because 1:1:1:1 ratio is observed from the backcross. 3) A phenotypically wild-type female fruit fly that was heterozygous for genes controlling body color and wing length was crossed to a homozygous mutant male with black body (allele b ) and vestigial wings (allele vg ). The cross produced the following progeny: gray body, normal wings 126; gray body, vestigial wings 24; black body, normal wings 26; black body, vestigial wings 124. Do these data indicate linkage between the genes for body color and wing length? What is the frequency of recombination? Show your work . (4 points) Hint: Diagram of the cross showing the arrangement of the genetic markers on the chromosomes is given below. Answer: Yes. Recombination frequency = (number of recombinants)/(number of recombinants + number of parental types) 1

= (sum of two small numbers)/( sum of two small numbers + sum of large numbers) = (sum of two small numbers)/( sum of four types) = (24 + 26)/(126 + 1 24 + 24 + 26) = 50/300 = 0.167 . 4) Another phenotypically wild-type female fruit fly heterozygous for the two genes mentioned in the previous problem was crossed to a homozygous black, vestigial male. The cross produced the following progeny: gray body, normal wings 23; gray body, vestigial wings 127; black body, normal wings 124; black body, vestigial wings 26. Do these data indicate linkage? What is the frequency of recombination? Diagram the cross, showing the arrangement of the genetic markers on the chromosomes. (4 points) Hint: Diagram of the cross showing the arrangement of the genetic markers on the chromosomes is given below. Answer: Yes. Recombination frequency = (number of recombinants)/(number of recombinants + number of parental types) = (sum of two small numbers)/( sum of two small numbers + sum of large numbers) = (sum of two small numbers)/( sum of four types) = (23 + 26)/( 23 + 26 + 127 + 124) = 49/300 = 0.163 . 4) For the pedigree shown below, nail-patella syndrome was found to be linked to blood type B. This is a dominant autosomal inheritance. The individual II.1 has a mutation and introduced the nail-patella syndrome into this family. (14 points) 2

recombination frequency between the nail-patella syndrome and blood type B only. Researchers incorporate many case studies and obtain the most accurate recombination frequency and thus genetic distance between the two loci.] 6) Summarize the features of three parasexual processes in bacteria. Use the diagram below to help fill the blanks (yes or no) in the table below. (5 points; deduction of 0.5 points for each blank) Cell contact required? Sensitive to DNase? Bacteriophage needed? Transformation Conjugation Transduction Answer omitted. Bonus questions (8 points) . Determine gene order and genetic distance using 3-point testcross (8 points) In maize, three recessive genes ( z , xt , and cm ) are linked on chromosome 3. A homozygous plant for the recessives is crossed with a wild-type plant. The F 1 is crossed to get an F 2 generation with the following results: z xt cm : 44 z + cm : 460 + xt cm : 39 z xt +: 164 + xt +: 474 z + +: 35 + + cm : 158 + + +: 40 1) What is the order of the three genes on chromosome 3? (2 points) 2) What is the map distance between z and cm ? Show your work. (3 points) 3) What is the map distance between xt and cm ? Show your work. (3 points) Hint: In a three-factor cross, the total number of non-crossover, single, and double crossover 4

chromosomes is determined and the frequency of single and double crossover chromosomes is added and then converted to a percentage which then provides the map distance in cM between the genes. Please refer to the lecture slides and/or come to the instructor for more guidelines to solve this question. Answer: 1) z cm xt (alternatively, xt cm z) (Note when the genotypes are given, the gene order is not necessarily preserved like how it is presented; it could be any order.) Determine the parental genotypes, i.e. the two genotypes with the largest number of individuals: z + cm, and + xt +. Determine the double crossover (CO) genotypes, i.e., the two genotypes with the smallest number of individuals: + xt cm and z + +. The order is determined based on the two genotypes with the smallest number of individuals: + xt cm and z + +. These two genotypes must be the outcome of double crossover events. Only when cm is in the middle will the double crossover generate + cm xt and z + + events from the two parental genotypes z cm + and ++ xt. 2) Determine the genotypes with single crossover between z and cm: + + cm and z xt +. Recombination frequency between z and cm = (single CO + double CO)/total = (158+164 + 35+39)/( 474+460+158+164 + 44+40+35+39) =396/1414 = 0.28. Answer: 28 cM. 3) Determine the genotypes with single crossover between xt and cm: + + + and z xt cm. Recombination frequency between xt and cm = (single CO + double CO)/total = (44+40 + 35+39)/(474+460+158+164 + 44+40+35+39) = 158/1414 = 0.11. Answer: 11 cM. 5