Genetics: Analysis and Principles
6th Edition
ISBN: 9781259616020
Author: Robert J. Brooker Professor Dr.
Publisher: McGraw-Hill Education
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Chapter 6, Problem 20EQ
Let’s suppose that two different X-linked genes exist in mice, designated with the letters N and L. Gene N exists in a dominant, normal allele and in a recessive allele, n, that is lethal. Similarly, gene L exists in a dominant, normal allele and in a recessive allele, l, that is lethal. Heterozygous females are normal, but males that carry either recessive allele are born dead. Explain whether or not it would be possible to map the distance between these two genes by making crosses and analyzing the number of living and dead offspring. You may assume that you have strains of mice in which females are heterozygous for one or both genes.
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Anhidrotic ectodermal dysplasia is an X-linked recessive disorder in humans characterized by small teeth, no sweat glands, and sparse body hair. This trait is usually seen in men, but women who are heterozygous carriers of the trait often have irregular patches of skin with few or no sweat glands (see the illustration below).
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IN DROSOPHILA, AN X-LINKED
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Chapter 6 Solutions
Genetics: Analysis and Principles
Ch. 6.1 - 1. Genetic linkage occurs because
a. genes that...Ch. 6.1 - In the experiment by Bateson and Punnett, which of...Ch. 6.2 - Prob. 1COMQCh. 6.2 - Prob. 2COMQCh. 6.2 - 3. For a chi square analysis involving genes that...Ch. 6.3 - Answer the multiple-choice questions based on the...Ch. 6.3 - Answer the multiple-choice questions based on the...Ch. 6.4 - 1. A tetrad of spores in an ascus is the product...Ch. 6.4 - Prob. 2COMQCh. 6.5 - 1. The process of mitotic recombination involves...
Ch. 6 - 1. What is the difference in meaning between the...Ch. 6 - 2. When a chi square analysis is applied to solve...Ch. 6 - 3. What is mitotic recombination? A heterozygous...Ch. 6 - 4. Mitotic recombination can occasionally produce...Ch. 6 - 5. A crossover has occurred in the bivalent shown...Ch. 6 - A crossover has occurred in the bivalent shown...Ch. 6 - A diploid organism has a total of 14 chromosomes...Ch. 6 - If you try to throw a basketball into a basket,...Ch. 6 - 9. By conducting testcrosses, researchers have...Ch. 6 - In humans, a rare dominant disorder known as...Ch. 6 - 11. When true-breeding mice with brown fur and...Ch. 6 - Though we often think of genes in terms of the...Ch. 6 - 13. If the likelihood of a single crossover in a...Ch. 6 - 14. In most two-factor crosses involving linked...Ch. 6 - Researchers have discovered that some regions of...Ch. 6 - 16. Describe the unique features of ascomycetes...Ch. 6 - Figure 6.1 shows the first experimental results...Ch. 6 - In the experiment of Figure 6.6, Stern followed...Ch. 6 - 3. Explain the rationale behind a testcross. Is it...Ch. 6 - 4. In your own words, explain why a testcross...Ch. 6 - Explain why the percentage of recombinant...Ch. 6 - 6. If two genes are more thanapart, how would you...Ch. 6 - 7. In Morgan’s three-factor crosses of Figure 6.3,...Ch. 6 - Two genes are located on the same chromosome and...Ch. 6 - 9. Two genes, designated A and B, are locatedfrom...Ch. 6 - 10. Two genes in tomatoes areapart; normal fruit...Ch. 6 - In the tomato, three genes are linked on the same...Ch. 6 - A trait in garden peas involves the curling of...Ch. 6 - Prob. 13EQCh. 6 - 14. In the garden pea, several different genes...Ch. 6 - A sex-influenced trait is dominant in males and...Ch. 6 - Three recessive traits in garden pea plants are as...Ch. 6 - In mice, a trait called snubnose is recessive to a...Ch. 6 - 18. In Drosophila, an allele causing vestigial...Ch. 6 - 19. Three autosomal genes are linked along the...Ch. 6 - 20. Let’s suppose that two different X-linked...Ch. 6 - Prob. 21EQCh. 6 - In mice, a dominant allele that causes a short...Ch. 6 - 2. In Chapter 3, we discussed the idea that the X...Ch. 6 - Mendel studied seven traits in pea plants, and the...
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- Males of many diploid species (like us) have X and Y sex chromosomes. They are hemizygous for most X- linked genes. Thus, males express most X-linked alleles, whether they are dominant or recessive in females. In the fruit fly Drosophila, it is common to achieve the equivalent of a test cross of X-linked genes in females by assessing the readily observed phenotypes of their male progeny. Since males do not receive X-linked genes from their father, sires of these crosses can be normal or wild-type flies. In fly genetics, it is conventional to name a gene after the mutant phenotype that enabled its discovery. Your challenge is to establish gene order and map distances between three X-linked genes in Drosophila. Each gene is represented by recessive mutant alleles that express rather distinctive phenotypes relative to their dominant wild-type alternative alleles. Flies expressing fruitless (f) are bisexual, lush (1) have a heightened responses to ethanol, and ken&barbie (kb) lack external…arrow_forwardThe following X-linked recessive traits are found in fruit flies:vermilion eyes are recessive to red eyes, miniature wings are recessiveto long wings, and sable body is recessive to gray body. A cross wasmade between wild-type males with red eyes, long wings, and graybodies and females with vermilion eyes, miniature wings, and sablebodies. The heterozygous female offspring from this cross, whichhad red eyes, long wings, and gray bodies, were then crossed tomales with vermilion eyes, miniature wings, and sable bodies. Thefollowing data were obtained for the F2 generation (including bothmales and females):1320 vermilion eyes, miniature wings, sable body1346 red eyes, long wings, gray body102 vermilion eyes, miniature wings, gray body90 red eyes, long wings, sable body42 vermilion eyes, long wings, gray body48 red eyes, miniature wings, sable body2 vermilion eyes, long wings, sable body1 red eyes, miniature wings, gray bodyWhat information do you know based on the question and your…arrow_forwardThe following X-linked recessive traits are found in fruit flies:vermilion eyes are recessive to red eyes, miniature wings are recessiveto long wings, and sable body is recessive to gray body. A cross wasmade between wild-type males with red eyes, long wings, and graybodies and females with vermilion eyes, miniature wings, and sablebodies. The heterozygous female offspring from this cross, whichhad red eyes, long wings, and gray bodies, were then crossed tomales with vermilion eyes, miniature wings, and sable bodies. Thefollowing data were obtained for the F2 generation (including bothmales and females):1320 vermilion eyes, miniature wings, sable body1346 red eyes, long wings, gray body102 vermilion eyes, miniature wings, gray body90 red eyes, long wings, sable body42 vermilion eyes, long wings, gray body48 red eyes, miniature wings, sable body2 vermilion eyes, long wings, sable body1 red eyes, miniature wings, gray bodyA. Calculate the map distances separating the three genes.B. Is…arrow_forward
- The following X-linked recessive traits are found in fruit flies:vermilion eyes are recessive to red eyes, miniature wings are recessiveto long wings, and sable body is recessive to gray body. A cross wasmade between wild-type males with red eyes, long wings, and graybodies and females with vermilion eyes, miniature wings, and sablebodies. The heterozygous female offspring from this cross, whichhad red eyes, long wings, and gray bodies, were then crossed tomales with vermilion eyes, miniature wings, and sable bodies. Thefollowing data were obtained for the F2 generation (including bothmales and females):1320 vermilion eyes, miniature wings, sable body1346 red eyes, long wings, gray body102 vermilion eyes, miniature wings, gray body90 red eyes, long wings, sable body42 vermilion eyes, long wings, gray body48 red eyes, miniature wings, sable body2 vermilion eyes, long wings, sable body1 red eyes, miniature wings, gray bodyAnalyze data. Make a drawing. Make a calculation.arrow_forwardThe following X-linked recessive traits are found in fruit flies:vermilion eyes are recessive to red eyes, miniature wings are recessiveto long wings, and sable body is recessive to gray body. A cross wasmade between wild-type males with red eyes, long wings, and graybodies and females with vermilion eyes, miniature wings, and sablebodies. The heterozygous female offspring from this cross, whichhad red eyes, long wings, and gray bodies, were then crossed tomales with vermilion eyes, miniature wings, and sable bodies. Thefollowing data were obtained for the F2 generation (including bothmales and females):1320 vermilion eyes, miniature wings, sable body1346 red eyes, long wings, gray body102 vermilion eyes, miniature wings, gray body90 red eyes, long wings, sable body42 vermilion eyes, long wings, gray body48 red eyes, miniature wings, sable body2 vermilion eyes, long wings, sable body1 red eyes, miniature wings, gray bodyWhat topic in genetics does this question address?arrow_forward: In Drosophila, yellow body is due to an X-linked gene that is recessive to the gene forgray body.(a) A homozygous gray female is crossed with a yellow male. The F1 are intercrossed toproduce F2. Give the genotypes and phenotypes, along with the expected proportions, of theF1 and F2 progeny.(b) A yellow female is crossed with a gray male. The F1 are intercrossed to produce the F2.Give the genotypes and phenotypes, along with the expected proportions, of the F1 and F2progeny.(c) A yellow female is crossed with a gray male. The F1 females are backcrossed with graymales. Give the genotypes and phenotypes, along with the expected proportions, of the F2progeny.(d) If the F2 flies in part b mate randomly, what are the expected phenotypic proportions offlies in the F3??arrow_forward
- In humans, the genes for red-green color blindness (R = normal, r = color blind) and hemophilia A (H = normal, h = hemophilia) are both X-linked and only 3 map units apart. Suppose a woman has four sons, and two are color blind but have normal blood clotting and two have hemophilia but normal color vision. What is the probable genotype of the woman? HR/hr Hr/hr hr/hR Hr/hR HR/Hrarrow_forwardNormal vision (XA) in humans is dominant to color blindness (Xa) and is X-linked. A man with normal vision, whose father was colorblind, marries a colorblind woman. What are the chances that a son will be colorblind? What are the chances that a daughter will be colorblind? The determiner for brown eyes (B) is dominant to blue eyes (b) and is not X-linked. A colorblind man with brown eyes, whose mother was blue-eyed, marries a blue-eyed woman having normal vision, whose father was colorblind. Show the expected phenotypes ratio of their children involving eye color, color blindness, and sex.arrow_forwardIn Drosophila, the X-linked recessive mutation vermilion (v) causes bright red eyes, in contrast to the brick-red eyes of wild type. A separate autosomal recessive mutation, suppressor of vermilion (su-v), causes flies homozygous or hemizygous for v to have wildtype eyes. In the absence of vermilion alleles, su-v has no effect on eye color. Determine the F1 and F2 phenotypic ratios from a cross between a female with wild-type alleles at the vermilion locus, but who is homozygous for su-v, with a vermilion male who has wildtype alleles at the su-v locusarrow_forward
- Miniature wings in Drosophila result from an X-linked allele (w) that is recessive to the allele for long wings (+). In a cross of a long winged male with a long winged female, the following offspring were obtained: 100 long winged males, 106 miniature winged males, and 480 long winged females. Given this result, the female must be a carrier of the miniature wing allele. True Falsearrow_forwardMiniature wings (Xm) in Drosophila result from an X-linked allele that is recessive to the allele for long wings (X*). Give the genotypes of the parents in the following cross: Male parent Female parent Male offspring Female offspring Long Miniature 750 miniature 761 long O male: X* / X* and female X™ /x+ O male: X*/Xt and female Xm /xm O male: X*/ Y and female Xm /xm O male: Xm/ Y and female Xm /xmarrow_forwardColor blindness in humans is controlled by an X-linked completely recessive allele (Xc), while breast cancer is controlled by an autosomal completely dominant allele, B. A color blind male, who is a heterozygote carrier for breast cancer has three children/n with a normal eyed female (whose mother was color blind), who is homozygote recessive for the breast cancer allele. What is the probability that out of three children, 2 will be color blind males, and not show breast cancer, and one will be a color blind female, who shows breast cancer?arrow_forward
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