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Question 1. You identify 5 members of amulti-gene family that are expressed in the same tissues. All 5genes appear to be equally expressed in all tissues. You also havediscovered a null allele for one of thegenes.

a. Why might an organism need 5 genes all expressing the sameprotein in a single cell or tissue?

b. What do you suppose the phenotype of an animal homozygous forthis null allele might be? Explain your answer in terms of geneexpression and levels needed to observe a phenotype.

c. What do you suppose the phenotype of an animal homozygous for amissense mutation in the same gene might be? The missense mutationhas been shown to cause the protein to form large aggregates

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Bunny Greenfelder
Bunny GreenfelderLv2
28 Sep 2019

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Related questions

Question 2. Splicing is complicated. There isconstitutive splicing, and then there is regulated alternativesplicing, and these molecular events can occur in the same ordifferent tissues.

a. What is constitutive splicing?

b. What is alternative splicing?

c. Draw a gene with three exons and twointrons. In this example assume exons 1,2,3 are expressed in onetissue and exons 1,3 are expressed in another. Draw the alternativesplicing, and the resulting RNA from this gene.

d. Draw a model illustrating how sterichindrance can promote alternative splicing with in a gene.

e. An individual is heterozygous, one copy ofthe gene has a mutation in exon 1 and the other copy of the gen ehas a mutation exon 3. Draw their genes, with mutations. What RNAmolecules can be expressed from these alleles? Will thisheterozygote have a mutant or a wild - type phenotype? Explain.Assume the 1/3 variant and the 1/2/3 variant are required indifferent tissues

Question 3 . You haveidentified a 4 exon gene expressed in the nervous system of adiploid animal. Different splice variants are expressed in thenervous tissue at the same time. T he organism under study has agene, unc - 250 , that controls movement and has three differentrecessive mutant alleles: allele s a 1 , a2 , and a3 . All threemutations are nonsens e stop mutations. When a diploid organism ishomozygous for allele a 1 animals die. When homozygous for eitherallele a2 , or allele a3 , animals are very uncoordinated in theirmovement (unc). Alleles a2 and a3 fail to complement allele a1 ,that is to say a1 / a2 and a 1 / a3 heteroallelic animals are veryuncoordinated in their movement (unc). However, a2 / a3 animals arewild type, ie they complement. Using alternative splicingconstruct a logical model for these observations and show whereeach of the mutations would be located in the gene.

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

These all relate to exceptions to the inheritance patterns encountered by Mendel.รขย€ย‹

Why do multiple and lethal alleles often result in modifications of the classic Mendelian monohybrid and dihybrid ratios?

Select the four correct statements.

-When an essential gene is mutated, it can result in a lethal phenotype. There are no classic Mendelian monohybrid and dihybrid ratios.
-In the case of codominance, heterozygotes produce gene products from both alleles of a gene. Classic Mendelian monohybrid and dihybrid ratios are modified by codominance.
-In the case of incomplete dominance, the phenotype of the heterozygote is distinct from and often intermediate to the phenotypes of homozygous individuals. Classic Mendelian monohybrid and dihybrid ratios are modified by incomplete dominance.
-Genes exist in a large number of allelic versions and a diploid organism has two homologous gene loci that may be occupied by different alleles of the same gene. This can result in many different phenotypes for traits, which may not follow typical Mendelian ratios.
-When an essential gene is mutated, it can result in a lethal phenotype. This results in a modification of classic Mendelian ratios.
-The phenotype of the heterozygous genotype is distinct from and often intermediate to the phenotypes of the homozygous genotypes. The joint expression of both alleles in a heterozygote is called codominance. There are no classic Mendelian monohybrid and dihybrid ratios.
-Genes exist in a large number of allelic versions, but in a diploid organism, only one allele of the gene can occupy one homologous gene loci. Classic Mendelian inheritance cannot explain this phenomenon.
-Each gene produces a unique gene product. The effect of one allele in a heterozygote completely masks the effect of the other. Classic Mendelian genetics cannot explain this phenomenon.

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