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4 Jan 2018

In a complementation test involving 4 monogenic recessive mutants with the same phenotype, a, b, c, and d, only these two crosses: a X b and c X d produce the F1 progeny that do not resemble wild type, while all other crosses do. What conclusion is correct?
A. These mutants represent only one gene
B. These mutants represent 2 genes
C. These mutants represent 3 genes
D these mutants represent 4 genes
E. The information is not sufficient to draw a conclusion.

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Collen Von
Collen VonLv2
5 Jan 2018
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Related questions

18) In your research on gene regulation in E. coli, you produce 2 mutant strains with the same phenotype: they can grow on lactose but not on glucose. To investigate these 2 mutants further, you perform a complementation test using them and discover that the diploid combination strain can grow on glucose or lactose. From this, you can deduce that:

a.)Both mutants have a mutation in the same gene

b.)The 2 mutants have mutations in different genes
c.)Glucose and lactose metabolic pathways in E. coli utilize at least one shared gene in common
d.)At least 2 different genes are required for glucose metabolism in E. coli
e.) b) and d)

19. In your research on the lac operon, you produce a unique regulatory mutatant strain of E. Coli where the lac l gene's repressor protein binding site for the operator is normal (wild-type) but its lactose binding site no longer binds lactose. The expected phenotype of this mutant will be:

a. inability to grow on lactose only medium

b. ability to grow on lactose only medium

c. blue colony color when grown on glycerol plus x-gal medium

d. blue colony oclor when grown on lactose plus x-gal medium

e. b & d

20.

20) In your research on the lac operon, you create a diploid E.coli strain with the following genotype (+) = wild type gene, (-) = mutant gene.

I- /O+ /Z- /Y+

I+ /Oc /Z+ /Y+
Can this diploid mutant grow on medium containing lactose as the only carbon source?

a) Yes

b) No

c) Maybe, it depends on if the Oc mutation is dominant or recessive

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

Drosophila diploid 2n=8 tolerate trisomy and monosomy of their smolest chromosome 4 so all trisomic and monosomic for this chromosome are viable. However, nullisomic and tetrasomics for chromosome 4 are letal and this type of progeny never born.

Consider the segment polarity gene cubitus interruptus (ci), that affets cuticle characteristics (mutant has naked cuticle) and also alters the wing vein pattern. Consider that ci recessive mutations results in vein interraption in the wing while the ci+ wild type allows the normal vein pattern in the wing. Deduce the phenotypic proportions in the progeny of the following crosses involving trisomcs. What are the expected phenotypes and genotypes of the progeny from these different fly crosses?

Assume tat the presence ofone wild type ci+ allele is sufficient to confer the wild type wing vein pattern phenotype.

write the diagramm of the cross, genotype, phenotype

for F1 - genotypic and phenotypic ratio - # wild type (split as # diploid # trisomic or # ?) : # mutant (split as # diploid # trisomic or # ?)

1) ci+/ci+/ci x ci+/ci

2) ci/ci/ci x ci+/ci/ci

3) ci+/ci/ci x ci+/ci

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