Genetic Analysis: An Integrated Approach (3rd Edition)
3rd Edition
ISBN: 9780134605173
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 4, Problem 16P
Summary Introduction
To analyze:
The expected
Introduction:
In cattle, short stature and short limbs are caused because of autosomal recessive mutation called Dexter. Cattles with a heterozygous genotype show short stature and limbs; cattle embryos with homozygous recessive genotype either spontaneously abort or are stillborn. This means that recessive homozygosity is lethal, and the individuals do not survive.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Hemophilia is caused by an X-linked recessive mutation in humans. If a
man whose paternal uncle (father's brother) was a hemophiliac marries a
woman whose brother is also a hemophiliac, what is the probability that
their first child will have hemophilia? (Assume that no other cases of
hemophilia exist in the pedigree.)
1/3
0
1/8
0 1/4
1/2
A mutant sex-linked trait called “notched” (N) is deadly in Drosophila when homozygous in females. Males who have a single N allele will also die. The heterozygous condition (Nn) causes small notches on the wing. The normal condition in both male and females is represented by the allele n.
a) Indicate the phenotypes of the F1 generation from the following cross: XNXn x XnY
b) Explain why dead females are never found in the F1 generation no matter which parents are crossed.
c) Explain why the mating of female XNXn and a male XNy is unlikely.
Cystic fibrosis in humans is caused by mutations in a single gene and is inherited as an autosomal
(non-sex chromosome) recessive trait. An unaffected couple has two children. The first child has
cystic fibrosis, and the second child is unaffected. What is the probability that the second child is a
carrier (heterozygous) for the mutation that causes the disease?
2/3
1
1/2
3/4
Chapter 4 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
Ch. 4 - 1. Define and distinguish incomplete penetrance...Ch. 4 -
2. Define and distinguish epistasis and...Ch. 4 - When working on barley plants, two researchers...Ch. 4 - Fifteen bacterial colonies growing on a complete...Ch. 4 - 5. In a type of parakeet known as a “budgie,”...Ch. 4 - 6. The and blood groups are given below for four...Ch. 4 - The wild-type color of horned beetles is black,...Ch. 4 - 8. Two genes interact to produce various...Ch. 4 - Prob. 9PCh. 4 - 10. In rats, gene produces black coat color if the...
Ch. 4 - 11. In the rats identified in Problem, a third...Ch. 4 - Using the information provided in Problems 10 and...Ch. 4 - 13. Total cholesterol in blood is reported as the...Ch. 4 - 14. Flower color in snapdragons results from the...Ch. 4 - 5. A plant line with reduced fertility comes to...Ch. 4 - Prob. 16PCh. 4 - The coat color in mink is controlled by two...Ch. 4 - Prob. 18PCh. 4 - 19. Feather color in parakeets is produced by the...Ch. 4 - Brachydactyly type D is a human autosomal dominant...Ch. 4 - 21. A male and a female mouse are each from...Ch. 4 - Xerodermapigmentosum (XP) is an autosomal...Ch. 4 - 23. Three strains of green-seeded lentil plants...Ch. 4 - Blue flower color is produced in a species of...Ch. 4 - 25. The following crosses are performed between...Ch. 4 - Two pure-breeding strains of summer squash...Ch. 4 - Marfan syndrome is an autosomal dominant disorder...Ch. 4 - 28. Yeast are single-celled eukaryotic organisms...Ch. 4 - Prob. 29PCh. 4 - Dr. Ara B. Dopsis and Dr. C. Ellie Gans are...Ch. 4 - Human ABO blood type is determined by three...Ch. 4 - In rabbits, albinism is an autosomal recessive...Ch. 4 - Dr. O. Sophila, a close friend of Dr. Ara B....Ch. 4 - In a breed of domestic cattle, horns can appear on...Ch. 4 - Prob. 35PCh. 4 - Prob. 36PCh. 4 - 37. Epistatic gene interaction results in a...Ch. 4 - 38. Draw a pedigree containing two parents and...
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
- You have a Drosophila line that is homozygous for autosomal recessive alleles a, b, and c, linked in that order. You cross females of this line with males homozygous for the corresponding wild-type alleles. You then cross the F1 heterozygous males with their heterozygous sisters. You obtain the following F2 phenotypes (where letters denote recessive phenotypes and pluses denote wild-type phenotypes): 1364 + + +, 365 a b c, 87 a b +, 84 + + c, 47 a + +, 44 + b c, 5 a + c, and 4 + b +.a. What is the recombinant frequency between a and b? Between b and c? (Remember, there is no crossing over in Drosophila males.)b. What is the coefficient of coincidence?arrow_forwardA geneticist crosses two yellow mice with straight hair and obtains the following progeny: 1/2 yellow straight, 1/6 yellow fuzzy, 1/4 gray straight and 1/12 gray fuzzy What additional cross might be carried out to determine if your explanation is correct? Note: recessive lethal epistatisarrow_forwardIn goats, a beard is produced by an autosomal allele that is dominant in males and recessive in females. We’ll use the symbol Bb for the beard allele and B+ for the beardless allele. Another independently assorting autosomal allele that produces a black coat (W) is dominant over the allele for white coat (w). Give the phenotypes and their expected proportions for the following crosses. a. B+ Bb Ww male × B+ Bb Ww female b. B+ Bb Ww male × B+ Bb ww female c. B+ B+ Ww male × Bb Bb Ww female d. B+ Bb Ww male × Bb Bb ww femalearrow_forward
- In some plants, a true-breeding, red-flowered variety gives all pink flowers when crossed with a white-flowered strain: RR (red) X rr (white) = Rr (pink). Flower position in this plant may either be axial (dominant) or terminal (recessive). What will be the phenotypic ratios of the F1 generation resulting from the following cross: axial-red (true-breeding) X terminal-white? What will the phenotypic ratios of the F2 be? SHOW ALL WORKarrow_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_forwardYou have a Drosophila line that is homozygous for autosomal recessive alleles a, b, and c, linked in that order. Youcross females of this line with males homozygous for thecorresponding wild-type alleles. You then cross the F1 heterozygous males with their heterozygous sisters. You obtain the following F2 phenotypes (where letters denoterecessive phenotypes and pluses denote wild-type phenotypes): 1364 + + +, 365 a b c, 87 a b +, 84 + + c,47 a + +, 44 + b c, 5 a + c, and 4 + b +.a. What is the recombinant frequency between a andb? Between b and c? (Remember, there is no crossingover in Drosophila males.)b. What is the coefficient of coincidence?arrow_forward
- Hemophilia is caused by an X-linked recessive mutation in humans. If a woman whose paternal uncle (father's brother) was a hemophiliac marries a man whose brother is also a hemophiliac, what is the probability of their first child having hemophilia (Note: assume that there are no other cases of hemophilia in the pedigree and no hidden carriers)? Select one: a. 1/2 b. 1/8 c. 0 d. 1/4 e. 1 Oarrow_forwardTwo unlinked genes, A and B, encode very similar enzymes involved in the production of purple flower pigments in a particular plant species. Recessive, non-functional alleles a and b are available. If the A and B genes are functionally redundant, so a single functional copy of either gene is sufficient for purple pigmentation, what proportion of the progeny of a cross between AaBb double heterozygotes is expected to have white flowers? 1/8 O 1/4 O 1/16 O 1/32arrow_forwardA wild-type male and a wild-type female Drosophila with red eyes and full wings are crossed. Their progeny are shown below: Males Females 3/8 Full wing, Red eye ¾ Full wing, Red eye 3/8 Miniature wing, Red eye ¼ Full wing, Purple eye 1/8 Full wing, Purple eye 1/8 Miniature wing, Purple eye 1. What is/are the genotype(s) of females with purple eye? Of males with purple eye and miniature wing? Draw both out with appropriate symbolsarrow_forward
- A certain species of morning glories produces flowers that are blue, red, or purple. Two pure-breeding purple lines are crossed and produce F1 progeny that all make blue flowers. The F1 are allowed to self and produce 320 F2 progeny with the following distribution: 185 blue, 115 purple, and 20 red. Which the following is NOT consistent with this information? A) Red-flowering plants are homozygous recessive for both genes. B) The pure-breeding parental parents are homozygous recessive for mutations in two different genes. C) Dominant gene interaction appears to result in a 9:6:1 ratio. D) Blue-flowering plants are either A_bb or aaB_. E) Analysis of the F1 and F2 progeny phenotypes suggests epistasis.arrow_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_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_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Human Biology (MindTap Course List)BiologyISBN:9781305112100Author:Cecie Starr, Beverly McMillanPublisher: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