Study Guide for Campbell Biology
11th Edition
ISBN: 9780134443775
Author: Lisa A. Urry, Michael L. Cain, Steven A. Wasserman, Peter V. Minorsky, Jane B. Reece, Martha R. Taylor, Michael A. Pollock
Publisher: PEARSON
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Chapter 15, Problem 3GP
Summary Introduction
To describe: Whether the genes are linked or unlinked.
Introduction: Genetic inheritance is the process by which the hereditary information is passed on to the offspring from the parents. The progeny receives a combination of hereditary material from two parents (the mother and the father). A gene contains two alleles. An allele is known as a variant form of the gene. Generally, the dominant allele masks the expression of the recessive allele.
Summary Introduction
To calculate: How many map units are they apart.
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Students have asked these similar questions
In Drosophila fruit flies, the genes for warped wings (dwp), rumpled bristles (rmp), and pallid wings (pld) are linked. A trihybrid female for all three allleles is crossed with homozygous recessive male for all three alleles and the offspring obtained showed the following phenotypes:
3 pld rmp dwp
428 pld rmp +
427 + + dwp
48 + rmp +
47 pld + dwp
23 pld + +
2 + + +
22 + rmp dwp
What is the order and map distance between these three alleles?
In Drosophila,, the curled mutation (cu, chromosome 3, position 50.0) results in wings that curl up,
while ebony (e, chromosome 3, position 70.7) results in a dark body. True breeding, wild type females
are mated with true breeding males with curled wings and ebony bodies.
Considering Drosophila notation, which of the following correctly diagrams the P1 cross?
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In Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn and ct+ is crossed to a sn+ct male. The F1 flies are interbred. The F2 males are distributed as follows:
genotype
number
sn ct
15
sn ct+
34
sn+ ct
33
sn+ct+
18
What is the map distance between sn and ct?
Chapter 15 Solutions
Study Guide for Campbell Biology
Ch. 15 - Complete the following summary of Morgans crosses...Ch. 15 - Two normal color-sighted individuals have two...Ch. 15 - In a testcross between a heterozygote tall,...Ch. 15 - With unlinked genes, an equal number of parental...Ch. 15 - The following recombination frequencies have been...Ch. 15 - a. What is the difference between an organism with...Ch. 15 - Prob. 7IQCh. 15 - Prob. 8IQCh. 15 - Mendels law of independent assortment applies to...Ch. 15 - You have found a new mutant phenotype in fruit...
Ch. 15 - Prob. 3SYKCh. 15 - Prob. 4SYKCh. 15 - Thomas Hunt Morgan firmly established the location...Ch. 15 - Prob. 2TYKCh. 15 - Sex-linked traits a. are coded for by genes...Ch. 15 - Prob. 4TYKCh. 15 - Prob. 5TYKCh. 15 - Prob. 6TYKCh. 15 - Prob. 7TYKCh. 15 - Prob. 8TYKCh. 15 - Prob. 9TYKCh. 15 - Prob. 10TYKCh. 15 - Consider three genes on the X chromosome: A, B,...Ch. 15 - Prob. 12TYKCh. 15 - Genomic imprinting a. explains cases in which the...Ch. 15 - Prob. 14TYKCh. 15 - Prob. 15TYKCh. 15 - Suppose that alleles for an X-linked character for...Ch. 15 - Some girls who fail to undergo puberty are found...Ch. 15 - Prob. 18TYKCh. 15 - The genetic event that results in Turner syndrome...Ch. 15 - Prob. 20TYKCh. 15 - Prob. 1GPCh. 15 - Prob. 2GPCh. 15 - Prob. 3GPCh. 15 - Prob. 4GPCh. 15 - Prob. 5GPCh. 15 - Red-green color blindness is caused by a...
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- Miniature 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_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 231 long, 250 miniature Long Long 560 long O male: Xm/Y and female X* /X* male: X* / Y and female Xm /x O male: X* /Y and female X* /X* male: Xm/Y and female Xm /x+arrow_forwardMiniature 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_forward
- A female from true breeding line of Drosophila with white eyes is crossed with a male from a true breeding line with brown eyes. All of the offspring have wild type brick red eyes. Which of the following explanations is most likely? A) There are many alleles for the single gene for eye color. Wild type brick red eyes result only when the fly is heterozygous. B) The alleles for brown, white, and brick red eyes are alleles for a single locus. The allele for brown eye color is dominant to the allele for brick red eye color and to the allele for white eyes. C) There is more than one gene for eye color. The brown mutation and the white mutation occur in separate genes and are both recessive to the wild type alleles. The offspring are heterozygous for both genes, so they are phenotypically wild type. D) None of the above. It is not possible for a cross between a white-eyed and a brown-eyed fly to produce wild type offspring.arrow_forwardIn Drosophila, the brown mutation (bw, chromosome 2, position 104.5) results in brown eyes, while miniature (min, chromosome X, position 36.1) results in wings that are 2/3 the length of wild type. True breeding, wild type females are mated with true breeding males with brown eyes and miniature wings. Using Drosophila notation, diagram the P1 and F1 crosses. P1 F1 Fill in the chart with phenotypic ratios that would be expected in the F2 generation. Use the space provided to show your work. Phenotype Females Males Overall (♀and ♂) =1 =1 =1arrow_forwardIn Drosophila,, the curled mutation (cu, chromosome 3, position 50.0) results in wings that curl up, while ebony (e, chromosome 3, position 70.7) results in a dark body. True breeding, wild type females are mated with true breeding males with curled wings and ebony bodies. Considering Drosophila notation, which of the following correctly diagrams the F1 cross? X X 3+ cu e + X X e + + + + + cu e + O + ■ 3+ X X X X Y Y + + ■ cu cu cu ' + ■ cu ■ ' + e + e e e e e + cu +arrow_forward
- Multiple crosses were made between true-breeding lines of black and yellow Labrador retrievers. All the F1 progeny were yellow. When these progeny were intercrossed, they produced an F2 consisting of 121 yellow, 9 black and 30 chocolate. What epistatic ratio and what kind of epistasis is approximated in the F2? Propose a biochemical pathway for coat color in Labrador retrievers based on the type of epistasis. Correlate each genotype with the phenotype that would occur in your pathway. Also show the frequency of each genotype. A-B- A-bb aaB- aabbarrow_forwardIn Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn+ and ct+ is crossed to a sn ct male. The F1 flies are interbred. The F2 males are distributed as follows sn ct 36 sn ct+ 13 sn+ ct 12 sn+ ct+ 39 What is the map distance between sn and ct?arrow_forwardIn Drosophila, the sepia mutation (se, chromosome 3, position 26) results in dark brown eyes, while cinnabar (cn, chromosome 2, position 57.5) results in bright orange-red eyes. True breeding, wild type females are mated with true breeding males homozygous recessive for both traits. Using Drosophila notation, diagram the P1 and F1 crosses. P1 F1 Fill in the chart with phenotypic ratios that would be expected in the F2 generation. Use the space provided to show your work. Phenotype Females Males Overall (♀and ♂) =1 =1arrow_forward
- Part 2 Spotty and Cutie have puppies and 50% of the puppies are unaffected for both traits (not blind and not BHFD). What is Spotty's genotype? bbDD bbdd BBDd BbDd BbDD BBddarrow_forwardWild-type mice have brown fur and short tails. Loss of function of a particular gene produces white fur, while loss of function of another gene produces long tails, and loss of function at a third locus produces agitated behavior. Each of these loss of function alleles is recessive. If a wild-type mouse is crossed with a triple mutant, and their F1 progeny is test-crossed, the following recombination frequencies are observed among their progeny. Produce a genetic map for these loci. Brown, short tailed, normal: 955 White, short tailed, normal: 16 Brown, short tailed, agitated: 0 White, short tailed, agitated: 36 Brown, long tailed, normal: White, long tailed, normal: Brown, long tailed, agitated: 46 0 14 White, long tailed, agitated: 933arrow_forwardIN DROSOPHILA, AN X-LINKED RECESSIVE MUTATION, Xm CAUSES MINIATURE WINGS. LIST THE F₂ PHENOTYPIC RATIOS IF: A MINIATURE-WINGED FEMALE IS CROSSED WITH A NORMAL MALE AND A MINIATURE-WINGED MALE IS ● ● CROSSED WITH A NORMAL FEMALE. WHAT WOULD THE PHENOTYPIC RATIO FROM (A) BE IF THE MINIATURE- WINGED GENE WERE AUTOSOMAL? ASSUME IN ALL CASES THAT THE P1 INDIVIDUALS ARE TRUE-BREEDING.arrow_forward
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