Introduction to Genetic Analysis
11th Edition
ISBN: 9781464109485
Author: Anthony J.F. Griffiths, Susan R. Wessler, Sean B. Carroll, John Doebley
Publisher: W. H. Freeman
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Chapter 4, Problem 38.3P
Summary Introduction
To determine: The mating type in
Introduction: Mating is the pairing of either opposite-sex or hermaphroditic organisms, usually for sexual reproduction.
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You have already localized the genes to the same chromosome by deletion mapping, and now decide that the best way to accomplish the mapping is to conduct two simultaneous three-point testcross experiments. The genes you are investigating are as follows: N = round leaves, n = notched leaves; H = smooth stems, h = hairy stems; R = purple flowers, r = red flowers; B = grey seeds, b = black seeds; and Y = green pods, y = yellow pods. Earlier experiments you have done already established that gene B is in the middle of this gene cluster, so you design both three-point test crosses to include that gene. Cross #1 is designed as RrHhBb x rrhhbb while cross #2 is NnBbYy x nnbbyy. The results of both crosses are given in the table below.
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A Neurospora colony at the edge of a plate seemed to be sparse (low density) in comparison with the other colonies on the plate. This colony was thought to be a possible mutant, and so it was removed and crossed with a wild type of the opposite mating type. From this cross, 100 ascospore progeny were obtained. None of the colonies from these ascospores was sparse, all appearing to be normal. What is the simplest explanation of this result? How would you test your explanation? (Note: Neurospora is haploid.)
Chapter 4 Solutions
Introduction to Genetic Analysis
Ch. 4 - Prob. 1PCh. 4 - Prob. 5PCh. 4 - Prob. 12PCh. 4 - Prob. 13PCh. 4 - Prob. 14PCh. 4 - Prob. 15PCh. 4 - Prob. 16PCh. 4 - Prob. 17PCh. 4 - Prob. 18PCh. 4 - Prob. 19P
Ch. 4 - Prob. 20PCh. 4 - Prob. 21PCh. 4 - Prob. 21.1PCh. 4 - Prob. 21.2PCh. 4 - Prob. 21.3PCh. 4 - Prob. 21.4PCh. 4 - Prob. 21.5PCh. 4 - Prob. 21.6PCh. 4 - Prob. 21.7PCh. 4 - Prob. 21.8PCh. 4 - Prob. 21.9PCh. 4 - Prob. 21.10PCh. 4 - Prob. 21.11PCh. 4 - Prob. 21.12PCh. 4 - Prob. 21.13PCh. 4 - Prob. 21.14PCh. 4 - Prob. 21.15PCh. 4 - Prob. 21.16PCh. 4 - Prob. 21.17PCh. 4 - Prob. 21.18PCh. 4 - Prob. 21.19PCh. 4 - Prob. 21.20PCh. 4 - Prob. 21.21PCh. 4 - Prob. 21.22PCh. 4 - Prob. 21.23PCh. 4 - Prob. 21.24PCh. 4 - Prob. 21.25PCh. 4 - Prob. 21.26PCh. 4 - Prob. 22PCh. 4 - Prob. 23PCh. 4 - Prob. 24PCh. 4 - Prob. 25PCh. 4 - Prob. 26PCh. 4 - Prob. 27PCh. 4 - Prob. 28PCh. 4 - Prob. 29PCh. 4 - Prob. 30PCh. 4 - Prob. 31PCh. 4 - Prob. 32PCh. 4 - Prob. 33PCh. 4 - Prob. 34PCh. 4 - Prob. 35PCh. 4 - Prob. 36PCh. 4 - Prob. 37PCh. 4 - Prob. 38PCh. 4 - Prob. 38.1PCh. 4 - Prob. 38.2PCh. 4 - Prob. 38.3PCh. 4 - Prob. 38.4PCh. 4 - Prob. 38.5PCh. 4 - Prob. 38.6PCh. 4 - Prob. 38.7PCh. 4 - Prob. 38.8PCh. 4 - Prob. 38.9PCh. 4 - Prob. 38.10PCh. 4 - Prob. 38.11PCh. 4 - Prob. 38.12PCh. 4 - Prob. 38.13PCh. 4 - Prob. 38.14PCh. 4 - Prob. 38.15PCh. 4 - Prob. 38.16PCh. 4 - Prob. 38.17PCh. 4 - Prob. 38.18PCh. 4 - Prob. 38.19PCh. 4 - Prob. 38.20PCh. 4 - Prob. 38.21PCh. 4 - Prob. 38.22PCh. 4 - Prob. 38.23PCh. 4 - Prob. 38.24PCh. 4 - Prob. 39PCh. 4 - Prob. 40PCh. 4 - Prob. 41PCh. 4 - Prob. 42PCh. 4 - Prob. 43PCh. 4 - Prob. 44PCh. 4 - Prob. 45PCh. 4 - Prob. 46PCh. 4 - Prob. 47PCh. 4 - Prob. 48PCh. 4 - Prob. 49PCh. 4 - Prob. 50PCh. 4 - Prob. 51PCh. 4 - Prob. 52PCh. 4 - Prob. 53PCh. 4 - Prob. 54PCh. 4 - Prob. 55PCh. 4 - Prob. 56PCh. 4 - Prob. 57PCh. 4 - Prob. 58PCh. 4 - Prob. 59PCh. 4 - Prob. 60PCh. 4 - Prob. 62PCh. 4 - Prob. 63PCh. 4 - Prob. 64PCh. 4 - Prob. 65PCh. 4 - Prob. 66PCh. 4 - Prob. 67PCh. 4 - Prob. 68PCh. 4 - Prob. 69P
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- What is probability, and how is it applied in genetic analysis?arrow_forwardWhat is an interrupted mating experiment? What type of experimentalinformation can be obtained from this type of study? Whyis it necessary to interrupt mating?arrow_forwardYou cross two yeast strains one is an ade auxotroph the other is a pro auxotroph and allow the diploid to sporulate. When you score each spore in the ascus you find the following proportions: 518 PD, 8 NPD, and 225 T. a.) What are the genotypes of each spore in all three types of the tetrads. b) Are these genes linked why or why not? c.) If these genes are unlinked what would you expect the progeny numbers and ratios to be? d.) What is the formula to determine the most accurate distance between these genes? If linked what is the map distance?arrow_forward
- In linkage mapping, how do we measure the distance between genes on a chromosome? options: A) The number of gametes used is proportional to the percent recombination in offspring. B) The number of offspring produced is equal to the percent recombination during crossover. C) The number of chromosomes resulting from crossover is equal to the distance the genes are apart in any one gamete. D) The percentage of recombinant offspring is directly proportional to how far apart the genes are on the chromosome.arrow_forwardHow is the mating type of a yeast cell determined?arrow_forwardChoose which of the following is true: a) Haploid cells have one allele per gene b) Haploid cells have two alleles per genearrow_forward
- A heterozygous diploid yeast Aa Bb went through meiosis. What percentage of the haploid spores will have recombinant combinations of alleles? What if genes A and B are unlinked? Explain What is genes A and B are linked? Explainarrow_forwardIn an electrophoretic gel across which is applied a powerful electrical alternating pulsed field, the DNA of the haploid fungus Neurospora crassa (n = 7) moves slowly but eventually forms seven bands, which represent DNA fractions that are of different sizes and hence have moved at different speeds. These bands are presumed to be the seven chromosomes. How would you show which band corresponds to which chromosome?arrow_forwardUnder what circumstances could nonhomologous endjoining be said to be error prone?arrow_forward
- To understand the genetic basis of locomotion in the diploid nematode Caenorhabditis elegans, recessive mutations were obtained, all making the worm “wiggle” ineffectually instead of moving with its usual smooth gliding motion. These mutations presumably affect the nervous or muscle systems. Twelve homozygous mutants were intercrossed, and the F1 hybrids were examined to see if they wiggled. The results were as follows, where a plus sign means that the F1 hybrid was wild type (gliding) and “w” means that the hybrid wiggled.a. Explain what this experiment was designed to test. b. Use this reasoning to assign genotypes to all 12 mutants. c. Explain why the phenotype of the F1 hybrids between mutants 1 and 2 differed from that of the hybrids between mutants 1 and 5arrow_forwardIn fruit flies red (A) eyes are dominant to apricot (a) eyes, and normal (P) wings and dominant to pointed (p) wings. Based on the information above, what are the possible gametes that could be produced by a fly that is homozygous for red eyes and heterozygous for normal wings? Group of answer choices: A) AA, Pp B) AP, Ap, aP, ap C) AP, Ap D) AAPparrow_forwardA) He might be more likely heterozygous black B He might be more likely homozygous brown C) He must be heterozygous brown. D) He must be homozygous black E He must be homozygous brown 2-11 "Dumpy" is a commonly used mutant phenotype in the nematode wom C. elegans. Two "Dumpy individuals are crossed to each other and this cross produces 210 "Dumpy" and 68 wild-type individuals. If one of the "Dumpy" individuals used in this cros5 was mated with a wild-type, what ratio of "Dumpy" wild-type would we observe in the offspring? A) 0:1 B) 1:0 C) 1:1 D) 1:3 E) 3:1 2-12 If genes assort independently, a testcrossed dihybrid characteristically produces progeny phenotypes in the ratio: A) 1:1 B) 1:1:1:1 c) 1:2:1 D) 3:1 E) 9:3:3:1 2-13 A fish of genotype a/a; B/b is crossed to a fish whose genotype is 4/a; B/b. What proportion of the progeny will be heterozygous for at least one of the genes? (Assume independent assortment.) A) 1/8 B) 1/4 C) 1/2 D) 5/8 E) 3/4 2-14 In hogs, a dominant allele B…arrow_forward
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