ESSENTIALS OF GENETICS-MODIFIED ACCESS
9th Edition
ISBN: 9780134190006
Author: KLUG
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Concept explainers
Textbook Question
Chapter 4, Problem 8PDQ
The following genotypes of two independently assorting autosomal genes determine coat color in rats:
A–B– (gray); A–bb (yellow); aaB–(black); aabb (cream)
A third gene pair on a separate autosome determines whether any color will be produced. The CC and Cc genotypes allow color according to the expression of the A and B alleles. However, the cc genotype results in albino rats regardless of the A and B alleles present. Determine the F1
Expert Solution & Answer
Trending nowThis is a popular solution!
Students have asked these similar questions
In rats, the following genotypes of two independently assorting autosomal genes determine coat color: A-B- (gray) A-bb (yellow) aaB- (black) aabb (cream) A third gene pair on a separate autosome determines whether or not any color will be produced. The CC and Cc genotypes allow color according to the expression of the A and B alleles. However, the cc genotype results in albino rats regardless of the A and B alleles present. Determine the F1 phenotypic ratio of the following crosses: (a) AAbbCC * aaBBcc (b) AaBBCC * AABbcc (c) AaBbCc * AaBbcc (d) AaBBCc * AaBBCc (e) AABbCc * AABbcc
The following genotypes of two independently assorting autosomal genes determine coat color in rats:A-B- (gray); A-bb (yellow); aaB-(black); aabb (cream)A third gene pair on a separate autosome determines whetherany color will be produced. The CC and Cc genotypes allow coloraccording to the expression of the A and B alleles. However,the cc genotype results in albino rats regardless of the A and Balleles present. Determine the F1 phenotypic ratio of the followingcrosses: (a)AAbbCC * aaBBcc; (b) AaBBcc * AABbcc;(c) AaBbCc * AaBbcc.
In mice, the wild-type coat color, agouti (AA) [left mouse], is dominant to solid-colored, black fur (aa) [middle mouse]. However, a separate gene (C) is necessary for pigment production. A mouse that is homozygous for a recessive c allele at this locus is unable to produce pigment and is albino [right mouse] regardless of the allele present at locus A. Thus, the following genotypes are listed with their associated phenotypes:
~ A/A; C/C or A/A; C/c or A/a; C/C or A/a; C/c ==> "agouti"
~ a/a; C/C or a/a; C/c ==> "black"
~ A/A; c/c or A/a; c/c or a/a; c/c or a/a; c/c ==> "albino"
This is an example of recessive epistasis, in which the recessive c allele "stands upon" the possible genotypes for the A gene (A/A, A/a, or a/a).
If two agouti mice with the A/a; C/c genotype are mated, what is the expected phenotypic ratio in their offspring?
A.9 albino, 4 agouti, 3 black
B. 9 agouti, 4 albino, 3 black
C.9 black, 4 albino, 3 agouti
D.9 agouti, 4 black, 3 albino
Chapter 4 Solutions
ESSENTIALS OF GENETICS-MODIFIED ACCESS
Ch. 4 - CASE STUDY | But he isn't deaf Researching their...Ch. 4 -
CASE STUDY | But he isn’t deaf
Researching...Ch. 4 - CASE STUDY | But he isn't deaf Researching their...Ch. 4 - HOW DO WE KNOW? In this chapter, we focused on...Ch. 4 - Review the Chapter Concepts list on page 53. These...Ch. 4 - In Shorthorn cattle, coat color may be red, white,...Ch. 4 -
4. With regard to the ABO blood types in humans,...Ch. 4 - In foxes, two alleles of a single gene, P and p,...Ch. 4 - Three gene pairs located on separate autosomes...Ch. 4 - As in the plants of Problem 6, color may be red,...
Ch. 4 -
8. The following genotypes of two independently...Ch. 4 - Given the inheritance pattern of coat color in...Ch. 4 - A husband and wife have normal vision, although...Ch. 4 - In humans, the ABO blood type is under the control...Ch. 4 - In goals, development of the beard is due to a...Ch. 4 -
13. In cats, orange coal color is determined by...Ch. 4 - In Drosophila, an X-linked recessive mutation,...Ch. 4 - Another recessive mutation in Drosophila, ebony...Ch. 4 - While vermilion is X-linked in Drosophila and...Ch. 4 - In pigs, coat color may be sandy, red, or white. A...Ch. 4 - A geneticist from an alien planet that prohibits...Ch. 4 - In another cross, the frog geneticist from Problem...Ch. 4 - In cattle, coats may be solid white, solid black,...Ch. 4 - Consider the following three pedigrees, all...Ch. 4 - Labrador retrievers may be black, brown, or golden...Ch. 4 - Three autosomal recessive mutations in Drosophila,...Ch. 4 -
24. Horses can be cremello (a light cream...Ch. 4 - Pigment in the mouse is produced only when the C...Ch. 4 - Five human matings numbered 1–5 are shown in the...Ch. 4 - Two mothers give birth to sons at the same time at...Ch. 4 - In Dexter and Kerry cattle, animals may be polled...Ch. 4 - What genetic criteria distinguish a case of...Ch. 4 -
30. The specification of the anterior-posterior...Ch. 4 - The maternal-effect mutation bicoid(bcd)is...Ch. 4 -
32. Students taking a genetics exam were...Ch. 4 - In four o'clock plants, many flower colors are...Ch. 4 - Prob. 34PDQ
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- The gene controlling ABO blood type and the gene underlying nail-patella syndrome are said to show linkage. What does that mean in terms of their relative locations in the genome? What does it mean in terms of how the two traits are inherited with respect to each other?arrow_forwardIn a cross in Drosophila involving the X-linked recessive eye mutationwhite and the autosomally linked recessive eye mutation sepia(resulting in a dark eye), predict the F1 and F2 results of crossingtrue-breeding parents of the following phenotypes:(a) white females * sepia males(b) sepia females * white malesNote that white is epistatic to the expression of sepiaarrow_forwardThere are two genetic disorders that result from mutation in imprinted genes: Prader-Willi syndrome, Angelman syndrome. Angelman syndrome results from deletion of UBE3A, which is a gene imprinted such that only the maternal copy is expressed. In the pedigree above, individual I-1 is heterozygous for a deletion of UBE3A and does not have Angelman syndrome. Individual I-2 is homozygous wild type for UBE3A. Which individuals in the pedigree are at risk for exhibiting Angelman syndrome, if any? (Who could potentially have the syndrome, based on what alleles it is possible for them to inherit and express?) Question 8 options: Only I-1 could have been at risk. If he does not have the syndrome, no one in the pedigree could. Only III-1 is at risk I-1, II-2, and III-1 are all at risk Only II-2 is at risk No one in the pedigree is at risk Both II-2 and III-1 are at…arrow_forward
- The pedigree below shows that inheritance of a disease that is caused by a late onset, dominant, autosomal mutation that is rare, but only 50% penetrant. The gene that is mutated in the disease is linked at a distance of 10 cm to a microsatellite marker that has alleles numbered 1, 2, and 3. The marker alleles detected in each individual are indicated below. What is the probability that individual A will develop the disease? Explain using an illustration of this occurs.arrow_forwardIn mice, the trait for high cholesterol is specified by a dominant allele designated HC, whereas the wild-type allele for normal cholesterol levels is designated hc. Black fur is specified by a recessive allele designated bl, whereas the wild-type allele which gives brown fur is designated BL. The genes for both of these traits are 30cM apart on the same autosome. A brown female (#1) with high cholesterol is mated to a black male (#2) with normal cholesterol. The progeny from this cross include a brown male (#3) with high cholesterol and a black female (#4) with normal cholesterol. What is the probability that the black mouse in the progeny of the first cross will also have high cholesterol?arrow_forwardThere are two genetic disorders that result from mutation in imprinted genes: Prader-Willi syndrome and Angelman syndrome. Prader-Willi syndrome results from deletion of region 15q11-q13, which in healthy individuals is a region imprinted such that only the paternal copy is expressed. In the pedigree above, individual I-1 is heterozygous for a deletion of region 15q11-q13 and does not have Prader-Willi syndrome. Individuals I-2 and II-1 are both homozygous wild type for the region. Which individuals in the pedigree might have Prader-Willi syndrome? (Who could potentially have the syndrome, based on what alleles it is possible for them to inherit and express?) Question 9 options: Only II-2 could have Prader-Willi syndrome III-1 could have Prader-Willi syndrome in the presented pedigree; II-2 could only have had it if she were male Both II-2 and III-1 could have Prader-Willi syndrome II-2 could have…arrow_forward
- In Drosophila melanogaster body color is controlled by one gene while wing shape is controlled by a second gene. Gray body color is dominant to black body color, and normal wings are dominant to vestigial wings. Flies homozygous for gray body color and vestiial wing are crossed with flies homozygous for black body color and normal wings Compare the possible F2generation genotypes and phenotypes and proportions if these two traits are autosomally linked in comparison to the non-linked How does your answer change if one of the original parents is homozygous for gray body color and normal wings while the other has black body color and vestigial wingsarrow_forwardIn rats, the following genotypes of two independently assorting autosomal genes determinecoat color:A–B– (gray)A–bb (yellow)aaB– (black)aabb (cream)A third gene pair on a separate autosome determines whether or not any color will be produced.The CC and Cc genotypes allow color according to the expression of the A and B alleles.However, the cc genotype results in albino rats regardless of the A and B alleles present.Determine the F1 phenotypic ratio of the following crosses: (1) AaBBCc × AaBBCc (2) AABbCc × AABbcc (3) AaBbCc × AaBbccarrow_forwardSepia eyes, spineless bristles, and striped body are three recessive mutations in Drosophila found on chromosome 3. A genetics student crosses a fly homozygous for the alleles encoding sepia eyes, spineless bristles, and striped body with a fly homozygous for the wild-type alleles—encoding red eyes, normal bristles, and solid body. The female progeny are then test-crossed with males that have sepia eyes, spineless bristles, and striped body. Assume that the interference between these genes is 0.2 and that 400 progeny flies are produced by the testcross. Based on the map distances provided in Figure , predict the phenotypes and proportions of the progeny resulting from the test cross.arrow_forward
- Take the example of B-thalassemia, an autosomal recessive genetic disease that particularly affects people from around the Mediterranean. This disease is associated with an anomaly of hemoglobin, a protein essential for the transport of oxygen, which is composed of four chains: two alpha (a) and two beta (B). In case of B-thalassemia, the ẞ chains are produced in insufficient or no quantity in an individual homozygous recessive resulting in insufficient production of overall hemoglobin leading to anemia and other physiological challenges. The gene that controls the synthesis of the ẞ chains is located on chromosome 11. Here is part of the coding portion of this gene (which controls a total of 146 amino acids and of which you only see the portion 36 to 41) and one of the targeted mutations: 1. Give the sequence of amino acids from the template and mutated strands. 2. What type of point mutation is it? 3. Using the principles of the theory of evolution, explain briefly and generally why…arrow_forwardA standard three-point mapping is conducted for recessive mutations in autosomal genes purple eye (pr), curved wing ( c) and black body (b). Their wild type alleles are also used for genetic mapping. An F1 Drosophila female heterozygous for purple eye (pr), curved wing (c) and black (b) is crossed to a triply homozygous mutant male. The observed numbers and phenotypes of the offspring are as follows: 360 pr c b 380 pr+ c+ b+ 104 pr c+ b 96 pr+ c b+ 30 pr c b+ 20 pr+ c+ b 6 pr c+ b+ 4 pr+ c b PROVIDE THE FOLLOWING: A) State the order of genes on this chromosome. B) Calculate map distances between the gene pairs: pr-c, pr-b, c-b. Show calculations, state the number of map units and which gene pairs they refer to.arrow_forwardIn Drosophila, three autosomal genes have the following map:20 m.u. 10 m.u.a b ca. Provide the data, in terms of the expected numberof flies in the following phenotypic classes, whena+ b+ c+/ a b c females are crossed to a b c / a b c males. Assume 1000 flies are counted and that nointerference exists in this region.a+ b+ c+a b ca+ b ca b+ c+a+ b+ ca b c+a+ b c+a b+ cb. If the cross was reversed, such that a+ b+ c+ / a b cmales are crossed to a b c / a b c females, howmany flies would you expect in the same phenotypic classes?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning
Human Heredity: Principles and Issues (MindTap Co...
Biology
ISBN:9781305251052
Author:Michael Cummings
Publisher:Cengage Learning
Mitochondrial mutations; Author: Useful Genetics;https://www.youtube.com/watch?v=GvgXe-3RJeU;License: CC-BY