Concept explainers
To review:
The map distance between the genes his-5 and lys-1 using two methods of calculations, where one method considers double crossover while the second does not, and determine which method gave the highest value.
Frequency of single crossover between two genes.
Nonparental ditypes (NPD) expected from the cross. Whether positive interference is happening in this case.
Introduction:
Map distance is used to estimate the distance between two genes present in a chromosome. Its unit is map units (mu). Double crossovers cause the formation of nonparental ditypes. Positive interference is a genetic phenomenon due to which a crossover occurring in one region of chromosome decreases the possibility of a second crossover near it.
Want to see the full answer?
Check out a sample textbook solutionChapter 6 Solutions
Genetics: Analysis and Principles
- The alleles his-5 and lys-1, found in yeast, result in cells that require histidine and lysine for growth, respectively. A cross was made between two haploid yeast strains that are his-5 lys-1 and his+ lys+. 973 tetrads were analyzed, with the following pattern: 7 tetrads with 2 his-5 lys+ spores and 2 his+ lys-1 spores 603 tetrads with 2 his-5 lys-1 spores and 2 his+ lys+ spores 363 tetrads with 1 his-5 lys-1 spore, 1 his-5 lys+ spore, 1 his+ lys-1 spores, and 1 his+ lys+ spore Compute the map distance between these two genes using the method that considered double crossovers and the one that does not. Show your work. Which give the higher value? Why? What is the frequency of single crossovers between these genes? Explain. Based on the frequency of single-crossovers, how many double crossovers would one expect? Is positive interference occurring?arrow_forwardThe alleles his-5 and lys-1, found in baker’s yeast, result in cellsthat require histidine and lysine for growth, respectively. A crosswas made between two haploid yeast strains that are his-5 lys-1and his+ lys+. From the analysis of 818 tetrads, the followingnumbers of tetrads were obtained:2 spores with his-5 lys+ + 2 spores with his+ lys-1 = 42 spores with his-5 lys-1 + 2 spores with his+ lys+ = 5021 spore with his-5 lys-1 + 1 spore with his-5 lys+ + 1 spore withhis+ lys-1 + 1 spore with his+ lys+ = 312A.What is the frequency of single crossovers between these twogenes?B. Based on your answer to part B, how many NPDs are expectedfrom this cross? Explain your answer. Is positive interferenceoccurring?arrow_forwardThe alleles his-5 and lys-1, found in baker’s yeast, result in cellsthat require histidine and lysine for growth, respectively. A crosswas made between two haploid yeast strains that are his-5 lys-1and his+ lys+. From the analysis of 818 tetrads, the followingnumbers of tetrads were obtained: 2 spores with his-5 lys+ + 2 spores with his+ lys-1 = 42 spores with his-5 lys-1 + 2 spores with his+ lys+ = 5021 spore with his-5 lys-1 + 1 spore with his-5 lys+ + 1 spore withhis+ lys-1 + 1 spore with his+ lys+ = 312 Compute the map distance between these two genes using firstthe method of calculation that considers double crossovers andthen the one that does not. Which method gives a higher value?Explain why?arrow_forward
- One yeast strain carries the alleles lys+ and arg+, whereas another strain has lys-3 and arg-2. The two strains were crossed toeach other, and an ascus obtained from this cross has four spores with the following genotypes: lys+ arg+, lys+ arg-2, lys-3arg+, and lys-3 arg 2. This ascus has a. a parental ditype.b. a tetratype.c. a nonparental ditype.d. either a tetratype or a nonparental ditype.arrow_forwardIn the plant Arabidopsis, the loci for pod length (L, long;l, short) and fruit hairs (H, hairy; h, smooth) are linked16 m.u. apart on the same chromosome. The followingcrosses were made:(i) L H/L H × l h/l h → F1(ii) L h/L h × l H/l H → F1If the F1’s from cross i and cross ii are crossed,a. what proportion of the progeny are expected to bel h/l h?b. what proportion of the progeny are expected to beL h/l h?arrow_forwardAn individual heterozygous for a reciprocal translocation possesses the following chromosomes: A B • C D E F G A B • C D V W X R S • T U E F G R S • T U V W X Q. Give the products that result from alternate, adjacent-1, and adjacent-2 segregation.arrow_forward
- The following is a linkage map of chromosome 5 for three genes in tomato: (see image) The cross between the triple heterozygote (Lf J W/ lf j w) and a triple homozygous recessive produced 500 progeny. Assume that there is no interference in the Lf-W region. Give the expected number of individuals for each of the following progeny types and show complete solutions.a. with crossover in the Lf-J and J-W regionsb. with crossover in the Lf-J regionc. with crossover in the J-W regiond. without crossover in the Lf-W regionarrow_forwardIn autotetraploid Chinese primrose (Primula sinensis L.), the gene controlling stigma color is very near the centromere of the chromosome carrying it. The allele G for green stigma is dominant to g for red stigmas. A homozygous green autotetraploid strain is crossed with a homozygous red autotetraploid strain. Each of the F1 GGgg plants would obtain 12 gametes which are 2GG, 8Gg, and 2g. How were these obtained?arrow_forwardThe following results are derived from crosses between Neurospora strain xy and strain ++: Tetrad Class 3 4 ху x+ x+ xy ++ ++ ++ +y +y ху +y 25 ++ 3 124 4 (i) Name the ascus type of each class from 1 to 4 as P, NP or T. (ii) Are genes x and y linked? Explain your answer. (iii) If they are linked, determine the map distance between the two genes. If they are unlinked, provide all the information you can about why you draw this conclusion.arrow_forward
- In a cross between a white-eyed female (ww) and a red-eyed male (w+Y), nearly all the progeny were either red-eyed females (w+w) or white-eyed males (wY). However, about 1 in every 2000 F1 flies had an "exceptional phenotype" and was either a white-eyed female or red-eyed male. How did Bridges explain this unexpected result? A) Crossing over B) Incomplete cytokinesis C) Incorrect synapsis D) Nondisjunction E) Pseudoautosomal regionarrow_forwardThe expected ratio of phenotypes among the progeny of a test cross is 1:1:1:1. Out of 200 total resulting progeny, 48 occur in one of the four phenotypic classes. Given this information, which of the following must also be true? a)At least one additional cell must also contain a count of 48. b)The progeny of this cross do not conform to a 1:1:1:1ratio. c)The value of observed - expected for this cell = -2. d)Since 48 is so close to the expected value, there is no need to calculate chi square before drawing a conclusion about the ratio.arrow_forwardAn ear of corn has a total of 381 grains, including 216 Purple & Smooth, 79 Purple & Shrunken, 65 Yellow & Smooth, and 21 Yellow & Shrunken. This ear of corn was produced by a dihybrid cross (PpSs x PpSs) involving two pairs of heterozygous genes resulting in a theoretical (expected) ratio of 9:3:3:1. Use a chi squarearrow_forward
- Biology (MindTap Course List)BiologyISBN:9781337392938Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. BergPublisher:Cengage Learning