BIOL 1090 Lecture Notes - Lecture 6: Pleiotropy, Zygosity, Cinnabar

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orangecamel139

8 Nov 2017

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Outside mendel"s garden, things are not so simple . Genes may and (cid:498)usually(cid:499) have more than 2 alleles: the most common allele is considered the (cid:498)wild type(cid:499) (cid:523)e. g. A+ or c+(cid:524) note: all other alleles are considered (cid:498)mutations(cid:499) considered to be a (cid:498)polymorphism(cid:499) (polymorphic alleles of that gene) Note: view effects in the slides: loss of function mutations are usually recessive because for most genes, one functional copy is enough to confer a normal (wild type) phenotype. Ex: microphthalmia protein (mitf) a transcription factor protein. Ex: effect of a null mutation on the protein mitf. We can see that one allele is responsible for 4 areas of the homozygous protein. The antennapedia mutation in drosophila is a (cid:498)dominant gain of function mutation(cid:499) in a regulatory region of the antp gene. )t"s expression is normally restricted regarding leg formation (segments) of the embryo (thorax) here it is also transcribed in the anterior head segment as shown on the picture.

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 canât have any function in vulva formation because we didnât 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 canât have any function in vulva formation because we didnât look at complete loss of function alleles

Two genes interact to produce various phenotypic ratios among F2 progeny of a dihybrid cross. Design a different pathway explaining each of the F2 ratios below, using hypothetical genes R and T and assuming that the dominant allele at each locus catalyzes a different reaction or performs an action leading to pigment production. The recessive allele at each locus is null (loss-of-function). Begin each pathway with a colorless precursor that produces a white or albino phenotype if it is unmodified. The ratios are for F2 progeny produced by crossing wild-type F1 organisms with the genotype RrTt.

- Its a multiple choice question, this section goes up to Part G, but the next question will not appear after finishing Part D.

Part D. 9/16 red : 7/16 white

Select one of these options:

* At least one copy of both dominant alleles is required for red pigment. The absence of either dominant allele results in white.

* One copy of any dominant allele results in red. The absence of either dominant allele results in white.

* The only way to produce red is to have at least one dominant allele at one specific locus and homozygous recessive alleles at the other locus. The second dominant allele functions as a suppressor; therefore, if both dominant alleles are present, it will suppress the expression of the first.

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

BIOL 1090 Lecture Notes - Lecture 6: Pleiotropy, Zygosity, Cinnabar

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