Log in
Biology
2
answers
0
watching
203
views

Two genes (A and B) are located 10cM apart on the ‘X’ chromosome. A women is homozygous for recessive mutant alleles for both genes. She has children with a wild-type man who is homozygous for both genes. What proportion of their children are expected to show mutant phenotype for both genes?

Select one:

a. 50%

b. 0%

c. 5%

d. 95%

e. 100%

Watch
2
answers
0
watching
203
views

For unlimited access to Homework Help, a Homework+ subscription is required.

Unlock all answers

Get 1 free homework help answer.
Already have an account? Log in
Share
Jean Keeling
Jean KeelingLv2
28 Sep 2019
Already have an account? Log in
Share

Related textbook solutions

Molecular Cell Biology
8 Edition,
Lodish
ISBN: 9781464183393
Biology: Science for Life with Physiology
6 Edition,
Belk
ISBN: 9780134555430
Concepts of Biology
OER Edition,
Openstax
ISBN: 9781938168116
Biology
2nd Edition, 2013
Openstax
ISBN: 9781947172517
Essentials of Biology
5 Edition,
Mader
ISBN: 9781259660269

Related questions

Sometimes one gene may have the effect of suppressing an abnormal (mutant) allele of another gene, resulting in the normal phenotype. Flies that are homozygous for the mutant gene, suppressor or hairy wing (s/s) will have a normal phenotype if they are also homozygous for the mutant gene, hairy wing (h/h). The following questions refer to the offspring ratios after the mating of individuals heterozygous for the wild type and mutant alleles of BOTH suppressor of hairy wing (s+/s) and hairy wing (h+/h) genes. The genes, s+ and h+ are located on separate chromosomes. Use the symbols sD, sR, hD, and hR to denote the genotype of a fly that is heterozygous for:

A. Dominant (D) mutations in both s+ and h+

B. Recessive (R) mutations in both s+ and h+

i. What phenotypic ratio of offspring would you expect if suppression occurs when you sib-mate flies heterozygous for dominant mutations in both s+ and h+ ?

ii. What phenotypic ratio of offspring would you expect if suppression occurs when you sib-mate flies heterozygous for recessive mutations in both s+ and h+ ?

iii. Draw the punnett squares showing both the genotype and corresponding phenotype for EACH of the 16 outcomes for both scenarios.

1. You mate flies that you know are heterozygous for a loss-of-function bicoid mutation and a balancer chromosome. The balancer chromosome, in this case, has a marker that allows one to detect the presence of the balancer in embryos. However, embryos without balancers seem normal. Does this imply Bicoid has no effect? Why or why not?

2. you are working with Arabidopsisdopsis, a model for plant development and two different recessive null loss-of-function mutations have been isolated. The mutations are in different genes but both affect the leaf development (specifically the density of the stomata- openings for gas exchange). The homozygous loss of function phenotype for gene A shows twice as many stomata when compared to wild-type leaves, while the homozygous loss of function for gene B shows four times as many stomata compared to wild-type leaves. A double mutant for both gene A and gene B show four times as many stomata. What is the order of action of these two genes and what is their genetic relationship (i.e. activation or inhibition)?

Homozygous recessive loss of function mutations in any one of 4 different genes in C. elegans worms (genes A, B, C, or D) results in female worms that don't form proper egg-laying structures, called vulvas, as shown in Lines 2-5 of the data table below. In vulvaless females, the fertilized eggs hatch and baby worms develop inside the mother, killing her in the process.

genotype phenotype
1 AA BB CC DD (wild type) normal vulva
2 aa BB CC DD No vulva
3 AA bb CC DD no vulva
4 AA BB cc DD no vulva
5 AA BB CC dd no vulva

You would like to figure out the order in which these genes act during the creation of the vulva (you cannot assume it will be A--->B--->C--->D as the gene names are arbitrary). Your friend tells you to perform epistasis analysis by making double mutants between the different homozygous recessive mutants and analyzing the phenotype of the double mutant. For example, she asks you to examine the phenotype of aabbCCDD (double mutant of genes A and B) and compare this to the single mutants to figure out whether gene A acts earlier than gene B, or vice versa. You know this won't work. Why not? Pick the ONE BEST choice:

a. The phenotype of the single mutant is already pretty severe. The double mutant will most likely be dead, therefore epistasis analysis won't be possible.

b. Each of the single mutants (aaBBCCDD) and (AAbbCCDD) has the same mutant phenotype i.e., no vulva. Epistasis analysis between any 2 genes is only possible when the mutant phenotypes for each gene is different.

c. Epistasis analysis involves making triple mutants (such as aabbccDD) in order to learn something about how the genes are ordered.

d. Epistasis analysis can never be carried out with null loss of function mutations. The mutations being analyzed all have to be dominant gain of function alleles.

To address your concern, you first generate overactive alleles of either gene A (denoted as A*) or gene B (denoted as B*). As shown in lines 6-7 of the table below, C. elegans that have one of these overactive alleles produce multiple vulvas.

Genotype Phenotype
6 A*A BB CC DD multiple vulvas
7 AA B*B CC DD multiple vulvas

To find out the order in which these genes act, you combine the overactive alleles with different loss of function alleles and observe the phenotype in double mutants (see Lines 8-11).

Genotype Phenotype
8 A*A bb CC DD multiple vulvas
9 A*A BB cc DD multiple vulvas
10 A*A BB CC dd no vulva
11 AA B*B cc DD multiple vulvas

Based on the data in the table, what is the order in which these 4 genes normally act in wild-type C. elegans in order to produce a wild-type/normal vulva?

a. A----> B----->C----->D ---> Vulva

b. D----> B----->C----->A----> Vulva

c. B----> C----->A----->D----> Vulva

d. C----> B----->A----->D----> Vulva

e. C----> B----->D----->A----> Vulva

f. don't have enough data to make any conclusions

Based on the vulva formation phenotype of the double mutants, which of the following statement(s) accurately describes the genetic interactions between the A* allele and alleles of other genes affecting vulva formation? Pick ALL that apply:

a. The A* allele is epistatic to homozygous recessive loss of function mutations in gene B

b. A homozygous recessive loss of function mutation in gene B is epistatic to the A* allele

c. Homozygous recessive loss of function mutation in gene D is epistatic to the A* allele

d. The A* allele is epistatic to homozygous recessive loss of function mutation in gene D

e. The A* allele enhances the homozygous recessive loss of function mutation in gene D

Weekly leaderboard

Home
Homework Help 3,900,000
Biology 460,000
Start filling in the gaps now
Log in