Introduction To Genetic Analysis
12th Edition
ISBN: 9781319114787
Author: Anthony J.F. Griffiths, John Doebley, Catherine Peichel, David A. Wassarman
Publisher: W. H. Freeman
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Chapter 3, Problem 32P
Summary Introduction
To determine: The
Introduction: Females possessing one X-linked recessive mutation are considered carriers. They will generally not manifest clinical symptoms of the disorder. However, differences in X chromosome inactivation can lead to varying degrees of clinical expression in carrier females since some cells will express one X allele, and some will express the other.
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IN 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.
In beetles, an X-linked gene determines body size, with normal size (M)completely dominant to miniature body size (m).
Body color is determined by an autosomal gene with two alleles, where B is incompletely dominant to b such that BB beetles are black, Bb beetles are brown and bb beetles are yellow.
Male beetles are heterogametic (XY).
The following cross is performed:
brown, miniature sized body female X brown, normal sized body male
Based on this information, which of the following statements is FALSE?
Select 3 correct answer(s)
Question 3 options:
A)
1/8 of the female progeny will have yellow miniature bodies.
B)
All of the male progeny will have miniature bodies.
C)
1/4 of the total progeny will be black.
D)
1/4 of the male progeny will have yellow miniature bodies.
E)
All of the female progeny will have miniature bodies.…
In humans, the ABO blood type is under the control of
autosomal multiple alleles. Color blindness is a recessive
X-linked trait. If two parents who are both type A and have
normal vision produce a son who is color blind and is type
O, what is the probability that their next child will be a
female who has normal vision and is type O?
Chapter 3 Solutions
Introduction To Genetic Analysis
Ch. 3 - Prob. 1PCh. 3 - Prob. 2PCh. 3 - Prob. 3PCh. 3 - Prob. 4PCh. 3 - Prob. 5PCh. 3 - Prob. 6PCh. 3 - Prob. 7PCh. 3 - Prob. 8PCh. 3 - Prob. 9PCh. 3 - Prob. 10P
Ch. 3 - Prob. 11PCh. 3 - Prob. 12PCh. 3 - Prob. 13PCh. 3 - Prob. 14PCh. 3 - Prob. 15PCh. 3 - Prob. 16PCh. 3 - Prob. 17PCh. 3 - Prob. 18PCh. 3 - Prob. 19PCh. 3 - Prob. 20PCh. 3 - Prob. 21PCh. 3 - Prob. 22PCh. 3 - Prob. 23PCh. 3 - Prob. 24PCh. 3 - Prob. 25PCh. 3 - Prob. 26PCh. 3 - Prob. 27PCh. 3 - Prob. 28PCh. 3 - Prob. 29PCh. 3 - Prob. 30PCh. 3 - Prob. 31PCh. 3 - Prob. 32PCh. 3 - Prob. 33PCh. 3 - Prob. 34PCh. 3 - Prob. 35PCh. 3 - Prob. 36PCh. 3 - Prob. 37PCh. 3 - Prob. 38PCh. 3 - Prob. 39PCh. 3 - Prob. 40PCh. 3 - Prob. 41PCh. 3 - Prob. 42PCh. 3 - Prob. 43PCh. 3 - Prob. 43.1PCh. 3 - Prob. 43.2PCh. 3 - Prob. 43.3PCh. 3 - Prob. 43.4PCh. 3 - Prob. 43.5PCh. 3 - Prob. 43.6PCh. 3 - Prob. 43.7PCh. 3 - Prob. 43.8PCh. 3 - Prob. 43.9PCh. 3 - Prob. 43.10PCh. 3 - Prob. 43.11PCh. 3 - Prob. 43.12PCh. 3 - Prob. 43.13PCh. 3 - Prob. 43.14PCh. 3 - Prob. 43.15PCh. 3 - Prob. 44PCh. 3 - Prob. 45PCh. 3 - Prob. 46PCh. 3 - Prob. 47PCh. 3 - Prob. 48PCh. 3 - Prob. 49PCh. 3 - Prob. 50PCh. 3 - Prob. 51PCh. 3 - Prob. 52PCh. 3 - Prob. 53PCh. 3 - Prob. 54PCh. 3 - Prob. 55PCh. 3 - Prob. 56PCh. 3 - Prob. 57PCh. 3 - Prob. 58PCh. 3 - Prob. 59PCh. 3 - Prob. 61PCh. 3 - Prob. 62PCh. 3 - Prob. 63PCh. 3 - Prob. 64PCh. 3 - Prob. 65PCh. 3 - Prob. 66PCh. 3 - Prob. 67PCh. 3 - Prob. 70PCh. 3 - Prob. 1GSCh. 3 - Prob. 2GSCh. 3 - Prob. 3GS
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- Human sex chromosomes are XX for females and XY for males. a. With respect to an X-linked gene, how many different types of gametes can a male produce? b. If a female is homozygous for an X-linked allele, how many different types of gametes can she produce with respect to this allele? c. If a female is heterozygous for an X-linked allele, how many different types of gametes can she produce with respect to this allele?arrow_forwardIn goats, development of the beard is due to a recessive gene. The following cross involving true-breeding goats was made and car- ried to the F2 generation: P;: bearded female x beardless male F1: all bearded males and beardless females ´1/8 beardless males | 3/8 bearded males 3/8 beardless females 1/8 bearded females F; × F, Offer an explanation for the inheritance and expression of this trait, diagramming the cross. Propose one or more crosses to test your hypothesis.arrow_forwardIn the common daisy, genes A and B control flower color. Both genes have a dominant allele (A or B) and a recessive allele (a or b). At least one copy of each dominant allele is required for flowers to be colorful instead of white. (Explain and Justify your answers) 21.1) Predict the genotypes and phenotypes of the F1 progeny of a cross between two white-flowered plants, one homozygous AA and the other homozygous BB. A) AA bb, white B) aa BB, white C) Aa Bb, colorful D) Aa Bb, white E) aa bb, colorful 21.2) Predict the phenotypic ratio of the F2 progeny of a cross between two white-flowered plants, one homozygous AA and the other homozygous BB. A) 3 colorful : 1 white B) 9 colorful : 7 white C) 9 white : 7 colorful D) 15 white : 1 colorful E) 15 colorful : 1 white 21.3) The inheritance pattern of daisy flower color provides an example of what type of gene interaction? A) additivity…arrow_forward
- 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.arrow_forwardPigment in mouse fur is only produced when the C allele is present. Individuals of the cc genotype are white. If color is present, it may be determined by the A, a alleles. AA or Aa results in agouti color, while aa results in black coats. (a) What F1 and F2 genotypic and phenotypic ratios are obtained from a cross between AACC and aacc mice? (b) In three crosses between agouti females whose genotypes were unknown and males of the aacc genotype, the following phenotypic ratios were obtained: (1) 8 agouti (2) 9 agouti (3) 4 agouti 8 white 10 black 5 black 10 white What are the genotypes of these female parents?arrow_forwardColour-blindness is the result of an X-linked recessive allele, Xc. The allele for normal eyesight is XC. (a) A woman with normal colour vision whose father was colour-blind marries a colour-blind man. Give the genotypes and phenotypes of their children. What ratio of their children can be expected to be colour-blind? (b) A man with normal colour vision whose father was colour-blind marries a woman carrier of the colour-blind allele. What is the likelihood that their children will be colour-blind? Carriers of the trait?arrow_forward
- The Delacour family is descended from the Veela race, a semi-human, semi-magical humanoid people reminiscent of the Sirens in Greek mythology with special powers. Imagine that being a Veela and having these powers is a X-Linked Dominant trait that is passed on genetically. (Hint: Use the Punnett Square.) What are the possible genotypes for Fleur and Bill’s daughter, Victoire? What are the genotypic frequencies related to the female children (in percentage format)? What is the likelihood (in percentage format) that Victoire would express Veela powers?arrow_forwardIn Drosophila, the white gene located on the X chromosome affects eye color; an autosomal gene, wingless, is on an autosomal chromosome. Use the following allele symbols: Xw+ _ , Xw+Y = wild type red eyes; X-linked dominant allele Xw Xw , XwY = white eyes; X-linked recessive allele Y = Y sex chromosome vg+ = wild type wings; autosomal dominant vg = wingless; autosomal recessive Predict ratios/proportions of genotypes and phenotypes of offspring from the following cross, of a white-eyed male with wild type wings and a wild type red eyed female with wild type wings: indicate sex of offspring along with phenotypes. XwY vg+ vg x Xw+Xw vg+vgarrow_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
- 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_forwardIn a species of bird, feather colour is determined by a single, non-autosomal gene that shows complete penetration. A red-feathered female mates with a blue-feathered male, and all the F1 males are blue-feathered and the F1 females are red-feathered. When the F1 individuals are crossed, all the F2 males are again blue-feathered and the F1 females are again red- feathered. Provide the genotype of the F2 males and females. Clearly indicate which genotype is associated with males and females, respectively. Binomial Expansion 1 a+b 2 a' + 2ab + b? 3 a'+ 3a'b + 3ab + b a + 4a'b + 6a'b + 4ab + b 5 a + Sa*b + 10a'b2 + 100'b + Sab + barrow_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: genotype number sn ct 15 sn ct+ 34 sn+ ct 33 sn+ct+ 18 What is the map distance between sn and ct?arrow_forward
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