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Ch. 5 - Chromosome Mapping in Eukaryotes
All textbooksKlug 12th EditionCh. 5 - Chromosome Mapping in EukaryotesProblem 33
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Chapter 5, Problem 33

The gene controlling the Xg blood group alleles (Xg⁺ and Xg⁻) and the gene controlling a newly described form of inherited recessive muscle weakness called episodic muscle weakness (EMWX) (Ryan et al., 1999) are closely linked on the X chromosome in humans at position Xp22.3 (the tip of the short arm). A male with EMWX who is Xg⁻ marries a woman who is Xg⁺ and they have eight daughters and one son, all of whom are normal for muscle function, the male being Xg⁺ and all the daughters being heterozygous at both the EMWX and Xg loci. Following is a table that lists three of the daughters with the phenotypes of their husbands and children. For each of the offspring, indicate whether or not a crossover was required to produce the phenotypes that are given.

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Hello everyone and welcome to today's video. So the genes responsible for controlling the blood group systems are located on the autism. However, the X. G. Blood group a leo's are located in blank. In order to solve this problem, We simply need to look at our answer choices and first of all identify which ones are going to be autism's. Remember that only the 23rd pair of chromosomes are going to be these what we call the sex chromosomes. Everything else are going to be autism's. So chromosome five and 21 are artisans and we're going to cancel them out because we're told that these X. G. Blood group A leos are not located in the autism's. And then we have left the X. And the Y chromosomes, which are the actual sex chromosomes in chromosome 23. Now because we have the eggs G blood group a leo's, this should already give us a hint that they're going to be part of the X. Chromosome and that is the fact. So we're going to cancel out the Y chromosome and select the X chromosome because the X. G. Blood group a leo's are located in the X chromosome, which is answer choice B. I really hope this video helped you and I hope to see you on the next one
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Textbook Question
Drosophila melanogaster has one pair of sex chromosomes (XX or XY) and three pairs of autosomes, referred to as chromosomes II, III, and IV. A genetics student discovered a male fly with very short (sh) legs. Using this male, the student was able to establish a pure breeding stock of this mutant and found that it was recessive. She then incorporated the mutant into a stock containing the recessive gene black (b, body color located on chromosome II) and the recessive gene pink (p, eye color located on chromosome III). A female from the homozygous black, pink, short stock was then mated to a wild-type male. The F₁ males of this cross were all wild type and were then backcrossed to the homozygous b, p, sh females. The F₂ results appeared as shown in the following table. No other phenotypes were observed. Wild Pink* Black, Black, Pink, Short* Short Females 63 58 55 69 Males 59 65 51 60 *Other trait or traits are wild type. Based on these results, the student was able to assign short to a linkage group (a chromosome). Which one was it? Include your step-by-step reasoning.
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Textbook Question
Drosophila melanogaster has one pair of sex chromosomes (XX or XY) and three pairs of autosomes, referred to as chromosomes II, III, and IV. A genetics student discovered a male fly with very short (sh) legs. Using this male, the student was able to establish a pure breeding stock of this mutant and found that it was recessive. She then incorporated the mutant into a stock containing the recessive gene black (b, body color located on chromosome II) and the recessive gene pink (p, eye color located on chromosome III). A female from the homozygous black, pink, short stock was then mated to a wild-type male. The F₁ males of this cross were all wild type and were then backcrossed to the homozygous b, p, sh females. The F₂ results appeared as shown in the following table. No other phenotypes were observed. Wild Pink* Black, Black, Pink, Short* Short Females 63 58 55 69 Males 59 65 51 60 *Other trait or traits are wild type. The student repeated the experiment, making the reciprocal cross, F₁ females backcrossed to homozygous b, p, sh males. She observed that 85 percent of the offspring fell into the given classes, but that 15 percent of the offspring were equally divided among b + p, b + +, + sh p, and + sh + phenotypic males and females. How can these results be explained, and what information can be derived from the data?
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Textbook Question
In Drosophila, a female fly is heterozygous for three mutations, Bar eyes (B), miniature wings (m), and ebony body (e). Note that Bar is a dominant mutation. The fly is crossed to a male with normal eyes, miniature wings, and ebony body. The results of the cross are as follows. 111 miniature 101 Bar, ebony 29 wild type 31 Bar, miniature, ebony 117 Bar 35 ebony 26 Bar, miniature 115 miniature, ebony Interpret the results of this cross. If you conclude that linkage is involved between any of the genes, determine the map distance(s) between them.
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Textbook Question
How would the results vary in cross (a) of Problem 32 if genes A and B were linked with no crossing over between them? How would the results of cross (a) vary if genes A and B were linked and 20 map units (mu) apart?
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Textbook Question
Because of the relatively high frequency of meiotic errors that lead to developmental abnormalities in humans, many research efforts have focused on identifying correlations between error frequency and chromosome morphology and behavior. Tease et al. (2002) studied human fetal oocytes of chromosomes 21, 18, and 13 using an immunocytological approach that allowed a direct estimate of the frequency and position of meiotic recombination. Below is a summary of information [modified from Tease et al. (2002)] that compares recombination frequency with the frequency of trisomy for chromosomes 21, 18, and 13. (Note: You may want to read appropriate portions of Chapter 8 for descriptions of these trisomic conditions.) Trisomic Mean Recombination Live-born Frequency Frequency Chromosome 21 1.23 1/700 Chromosome 18 2.36 1/3000–1/8000 Chromosome 13 2.50 1/5000–1/19,000 What conclusions can be drawn from these data in terms of recombination and nondisjunction frequencies? How might recombination frequencies influence trisomic frequencies?
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Textbook Question
Because of the relatively high frequency of meiotic errors that lead to developmental abnormalities in humans, many research efforts have focused on identifying correlations between error frequency and chromosome morphology and behavior. Tease et al. (2002) studied human fetal oocytes of chromosomes 21, 18, and 13 using an immunocytological approach that allowed a direct estimate of the frequency and position of meiotic recombination. Below is a summary of information [modified from Tease et al. (2002)] that compares recombination frequency with the frequency of trisomy for chromosomes 21, 18, and 13. (Note: You may want to read appropriate portions of Chapter 8 for descriptions of these trisomic conditions.) Trisomic Mean Recombination Live-born Frequency Frequency Chromosome 21 1.23 1/700 Chromosome 18 2.36 1/3000–1/8000 Chromosome 13 2.50 1/5000–1/19,000 Other studies indicate that the number of crossovers per oocyte is somewhat constant, and it has been suggested that positive chromosomal interference acts to spread out a limited number of crossovers among as many chromosomes as possible. Considering information in part (a), speculate on the selective advantage positive chromosomal interference might confer.
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