BIO 2133 Lecture Notes - Lecture 5: Lactose Intolerance, Gynaecology, Endocrinology

34 views9 pages

graygiraffe944

3 Feb 2017

School

Department

Course

Professor

For unlimited access to Class Notes, a Class+ subscription is required.

Document Summary

Bio2133 lecture 5: meiosis, sex determination, nondisjunction, and gene linkage. General rule for lof mutations: half the amount of wild type gene product is sufficient to give a wild type phenotype. If you change one amino acid it may change the conformation of the protein. You do(cid:374)"t (cid:374)eed a large a(cid:373)ou(cid:374)t of e(cid:374)z(cid:455)(cid:373)e to get the result. Will have the same genes but different alleles. There are two alleles and each can be template for gene expression. Homozygous mutant, the mutation causes the enzyme to be below the threshold. It is still expressed but it is not viewed. When you look at heterozygotes, will look like mutants or wild type. If the the wild type phenotype is present, the wildtype allele is dominant. Mixed green and yellow, got all green. Mixed green and old green, he got green and yellow peas in 1 3:1 ratio. Dominant: one copy of the allele is enough to have function.

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

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

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.

crimsonimpala955

BIO 2133 Lecture Notes - Lecture 5: Lactose Intolerance, Gynaecology, Endocrinology

Document Summary

Get access

Related textbook solutions

Molecular Cell Biology

Biology: Science for Life with Physiology

Concepts of Biology

Biology

Essentials of Biology

Related Documents

BIO 2133 Lecture Notes - Lecture 3: Intron, Chromosome, Nondisjunction

BIO 2133 Lecture Notes - Lecture 3: Antirrhinum, Missense Mutation, Lactase

BIO 2133 Lecture Notes - Lecture 4: Haploinsufficiency, Start Codon, Lactase

Related Questions