ESSENTIALS OF GENETICS-MODIFIED ACCESS
9th Edition
ISBN: 9780134190006
Author: KLUG
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Textbook Question
Chapter 4, Problem 26PDQ
Five human matings numbered 1–5 are shown in the following table. Included are both maternal and paternal
| Parental Phenotypes | Offspring | ||||
| (1) | A, M |
|
A, N | (a) | A, N |
| (2) | B, M |
|
B, M | (b) | O, N |
| (3) | O, N |
|
B, N | (c) | O, MN |
| (4) | AB, M |
|
O, N | (d) | B, M |
| (5) | AB, MN |
|
AB, MN | (e) | B, MN |
Each mating resulted in one of the five offspring shown to the right (a–e). Match each offspring with one correct set of parents, using each parental set only once. Is there more than one set of correct answers?
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Five human matings numbered 1–5 are shown in the followingtable. Included are both maternal and paternal phenotypes forABO and MN blood-group antigen status.
Parental Phenotypes Offspring(1) A, M * A, N (a) A, N(2) B, M * B, M (b) O, N(3) O, N * B, N (c) O, MN(4) AB, M * O, N (d) B, M(5) AB, MN * AB, MN (e) B, MN
Each mating resulted in one of the five offspring shown to theright (a–e). Match each offspring with one correct set of parents,using each parental set only once. Is there more than one set ofcorrect answers?
Five human matings numbered 1–5 are shown in the following table. Included are both maternal and paternal phenotypes for ABO and MN blood-group antigen status.
Each mating resulted in one of the five offspring shown to the right (a–e). Match each offspring with one correct set of parents, using each parental set only once. Is there more than one set of correct answers?
What will be the results of the following crosses, where N is black, n is brown, L is short hair and l is long hair. Make the tables and the corresponding explanation:
1. Crossing a Nnll woman dog with a Nnll dog.
2. Crossing a NnLl woman dog with a NnLl dog
3. Crossing a woman dog nnLl with a dog NNll
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
- In human blood types, what are the genotypes of the following parents? 1. Phenotypes of Parents A x AB Phenotypes of Offspring A 1/2 B 0 AB 1/2 O 0 Genotypes of Parents _________ x ________ 2. Phenotypes of parents. A x AB Phenotypes of offspring  A 1/2 B 1/4 AB 1/4 O 0 Genotypes of the parents _________ x ________ 3. Phenotypes of parents. A x A Phenotypes of offspring A 3/4 B 0 AB 0 O 1/4 Genotypes of the parents _________ x ________ 4. Phenotypes of parents. A x O Phenotypes of offspring A 1/2 B O AB O O 1/2 5. Phenotypes of parents. A x B Phenotypes of offspring A 1/4 B 1/4 AB 1/4 O 1/4 Genotypes of the parents _________ x ________ Genotypes of the parents _________ x ________arrow_forwardTwo parents of the same heterozygous genotype are involved in a tetrahybrid cross. Their genotypes are FfGgHhIi. Assuming independent assortment of these four genes, what are the probabilities that F1 offspring will have the following genotypes? a) ffgghhii b) FfGgHhIi c) FFGGHHII d) FfGGhhIi e) FfGGHHiiarrow_forwardCh. 13-3 Refer to problem 2. Determine the predicted genotype frequencies among the offspring of the following crosses:a. AABB × aaBB b. AaBB × AABb c. AaBb × aabb d. AaBb × AaBbarrow_forward
- As it turned out, one of the tallest Potsdam Guards had an unquenchable attraction to short women. During his tenure as guard, he had numerous clandestine affairs. In each case, children resulted. Subsequently, some of the childrenwho had no way of knowing that they were relatedmarried and had children of their own. Assume that two pairs of genes determine height. The genotype of the 7-foot-tall Potsdam Guard was A9A9B9B9, and the genotype of all of his 5-foot clandestine lovers was AABB. An A9 or B9 allele in the offspring each adds 6 inches to the base height of 5 feet conferred by the AABB genotype. a. What were the genotypes and phenotypes of all the F1 children? b. Diagram the cross between the F1 offspring, and give all possible genotypes and phenotypes of the F2 progenyarrow_forwardUsing the following information perform following crosses using the forked line and predict all possible phenotypes types of the offspring. List the number and description of the phenotypes of the offspring. 1. aa BB Cc Dd Ee X aa bb Cc Dd EE 2. Ff Rr Gg dd Bb X Ff RR Gg Dd Bb 3. Dd gg JJ bb Hh Ee X DD GG Jj Bb Hh Eearrow_forwardThe cross AaBb x aaBB results in which of the following phenotypic ratios? 1:1 1:1:1:1 1:2:1 9:3:3:1arrow_forward
- Codominance ABO blood typing is an example of codominance, where A and B are both dominant to O. A woman who is AB type has a child with a man who is O type. Use I allele formatting to designate each genotypes: IA IA, IA i, IB IB, IB i, IA IB, ii A) List the genotype of each parent. B) Complete a Punnett square. List all the genotypes for the predicted offspring of this mating (using the format of genotypes listed above) C) What is the probability that any of their offspring will be O type? D) What is the probability that any of their offspring will be AB type?arrow_forwardThe genetic identity of the female parent is RrGg and the genetic identity of the male parent is Rrgg. They produce 320 offspring together from a single mating: 57 red-eyed females with grey bodies, 61 red-eyed females with yellow bodies, 22 brown-eyed females with grey bodies, 20 brown-eyed females with yellow bodies.59 red-eyed males with grey bodies, 63 red-eyed males with yellow bodies, 20 brown-eyedmales with grey bodies, 18 brown-eyed males with yellow bodies. Show the simultaneous transmission of the two genes involved to give rise to the progeny given with the use of genetic diagrams and summaries as required. (Hint: you are only requiredto show the simultaneous transmission from the P to F1 generations)arrow_forwardIn a diploid plant species, an F1 with the genotype Mm Rr Ss is test crossed to a pure breeding recessive plant with the genotype mm rr ss. The offspring genotypes are as follows: Genotype Number Mm Rr Ss 687 Mm Rr ss 5 Mm rr Ss 68 Mm rr ss 196 mm Rr Ss 185 mm Rr ss 72 mm rr Ss 8 mm rr ss 679 Total 1900 1. Why is the recombination frequency for the outside pair of genes not equal to the sum of the recombination between the adjacent gene pairs?arrow_forward
- Organisms have the following genotypes. What types of gametes will these organisms produce, and in what proportions?a. Aabbb. AABbc. AaBb Given the following matings, what are the predicted phenotypic ratios of the offspring? Use a branch diagram to show the phenotypic ratio. a. AABb x Aabbarrow_forwardIf a parent with a B-negative blood type reproduces with a parent with O-negative blood type what is the probability that an offspring could have an A-negative blood type. 0%, 25%, 50%, 100% If a parent with AB-positive blood type reproduces with a parent of O-negative blood type what is the probability of that offspring could have a A-negative blood type. 0%, 25%, 50%, 100%.arrow_forwardTable 1: F1 ebony flies - 0 F1 non-ebony flies - 560 F1 stubble flies - 560 F1 non-stubble flies - 0 The researcher collects and crosses male and female flies from the F1 generation. In the resulting offspring, F2, there are both stubble and ebony flies. Draw a Punnett Square to illustrate the F1 cross for the stubbly phenotype showing the individual gametes of each parent and the combination in the resulting offspring.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 Biology (MindTap Course List)BiologyISBN:9781305112100Author:Cecie Starr, Beverly McMillanPublisher:Cengage Learning
Human Heredity: Principles and Issues (MindTap Co...
Biology
ISBN:9781305251052
Author:Michael Cummings
Publisher:Cengage Learning
Human Biology (MindTap Course List)
Biology
ISBN:9781305112100
Author:Cecie Starr, Beverly McMillan
Publisher:Cengage Learning
How to solve genetics probability problems; Author: Shomu's Biology;https://www.youtube.com/watch?v=R0yjfb1ooUs;License: Standard YouTube License, CC-BY
Beyond Mendelian Genetics: Complex Patterns of Inheritance; Author: Professor Dave Explains;https://www.youtube.com/watch?v=-EmvmBuK-B8;License: Standard YouTube License, CC-BY