Log in
Biology
7
answers
0
watching
372
views
5 Jan 2018

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)?

Watch
7
answers
0
watching
372
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
Already have an account? Log in
Share
Already have an account? Log in
Share
Already have an account? Log in
Share
Already have an account? Log in
Share
Already have an account? Log in
Share
Jamar Ferry
Jamar FerryLv2
5 Jan 2018
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
Biology
2nd Edition, 2013
Openstax
ISBN: 9781947172517
Concepts of Biology
OER Edition,
Openstax
ISBN: 9781938168116
Essentials of Biology
5 Edition,
Mader
ISBN: 9781259660269

Related questions

You identify mutations in a separate gene, gene R. Homozygous rr mutant females form a partial vulva because the mutant r allele produces only 15% as much R protein as the WT allele.

You decide to take advantage of this partial loss of function mutation to isolate mutations in other genes that affect vulva development.

Starting with the rr mutant, you screen for mutations that, in combination with rr, result in complete loss of vulva. You isolate a dominant, gain of function, mutation in a different gene (gene E) that in combination with rr result in a complete loss of vulva. You call the dominant allele E*.

Worms of genotype rr E*E do not form vulvas whereas rr single mutants form partial vulvas and E*E single mutants form normal vulvas.

Which of the following terms is the best description of E* in relation to rr?

a. E* is a suppressor mutation

b. E* is an epistatic mutation

c. E* is an enhancer mutation

d. E* is a resistant mutation

e. E* is a dominant mutation

f. both B and E

What does gene E's specific interaction with gene R suggest about what gene E normally does in wild-type worms?

a. Gene E's normal wild-type function is to inhibit vulva formation.

b. Gene E's normal wild-type function is to promote vulva formation.

c. Gene E can’t have any function in vulva formation because we didn’t look at complete loss of function alleles.

Starting with the rr mutant, you screen for mutations that, in combination with rr, result in normal vulvas. You isolate a dominant, gain of function, mutation in a different gene (gene S) that in combination with rr result in a completely normal vulva. You call the dominant allele S*.

Worms of genotype rr S*S form normal vulvas, whereas rr single mutants form partial vulvas and S*S single mutants form normal vulvas.

Which of the following terms is the best description of S* in relation to rr?

a. S* is a suppressor mutation

b. S* is an epistatic mutation

c. S* is an enhancer mutation

d. S* is a resistant mutation

e. S* is a dominant mutation

f. both A and B

What does gene S's specific interaction with gene R suggest about what gene S normally does in wild-type worms?

a. Gene S's normal wild-type function is to inhibit vulva formation.

b. Gene S's normal wild-type function is to promote vulva formation.

c. Gene S can’t have any function in vulva formation because we didn’t look at complete loss of function alleles

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

1. The eye color of wild-type Drosophila flies is red. Differentmutations in a single or in multiple genes lead to flies with whiteeyes. Mutations in the w gene on the X chromosome lead to whiteeyes instead of the normal red. You have isolated both a white-eyedmutation (designated W-1) that gives a dominant phenotype, and awhite-eyed mutation (designated w-2) that gives a recessivephenotype.

(a) A white-eyed male from the W-1 line is crossed to awild-type female. What color eyes will the female progeny from thiscross have?

(b) What color eyes will the male progeny from the cross in part(a) have?

(c) One of the female progeny from the cross in part (a) ismated to a white-eyed male from the w- 2 line. What fraction of thewhite-eyed progeny from this cross will be female?

2. The body color of Drosophila adult flies is regulated bymultiple genes, including the autosomal ebony gene. Ebony body (e)is recessive to the wild type (yellowish) body (e+). A second gene,which is X-linked, controls bristle shape. Normal bristle shape(f+) is dominant to the mutant forked bristles (f). A female withforked bristles that is heterozygous for the body color gene ismated with an ebony body male (his bristles are normal).

(a) What are the genotypes of the parents?

(b) Using a Punnett square, determine the expected proportionsof each possible progeny phenotype. Since there is a sex-linkedgene here, be sure to include consider sex as part of thephenotype.

3. Describe in your own words what a balancer chromosome is andwhy it is important in fly genetics (you will have to do someresearch for this one!)

Related Documents

MCD BIO 138 Lecture Notes - Lecture 6: Maternal Effect, Morphogen, Developmental Biology
jademoose794
BIOS 1700 Lecture Notes - Lecture 40: Wild Type, Oogenesis, Mutation
maroonwildebeest489
MCD BIO 138 Lecture Notes - Lecture 7: Blastoderm, Pattern Formation, Zygote
jademoose794

Weekly leaderboard

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