BIOL 1090 Lecture Notes - Zygosity, Peptide, Null Allele

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8 Apr 2013

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Genes may (and usually) have more than 2 alleles. Different alleles may affect the phenotype in different ways. A single gene may control several traits. Multiple genes may control a single trait. The most common allele is considered the wild type. Designated with a superscript + (e. g. a+ or c+) Any allele found at appreciable frequencies (at least 1%) in the populations is considered to be a polymorphism. All other alleles are considered as mutants. Recessive mutations almost always involve loss of gene function. Loss of function mutations are usually recessive because, for most genes, one functional copy is enough for a normal (wild type) phenotype. Dominant mutations can involve a loss of function or a gain of gene function. Loss of function mutation for those genes for which one functional copy is not enough (haploinsufficiency) Dominant negative mutation a loss of function mutation that interferes with the the normal function of the wt allele.

Related Questions

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

11. You suspect that two genes interact with one another to determine the product of a pathway you are studying. You cross two pure breeding plantsâone with white flowers and one with pinkâto make F1 individuals that have blue flowers. You cross the F1 generation and observe the following numbers of individuals with different phenotypes: Number of individuals Phenotype 91 Blue 30 Pink 39 White Which pathway best explains these results? (Note that arrows indicate enzymatic reactions carried out by products of genes 1 and 2. Assume that buildup of metabolites will determine flower color.)

Question 12 and 13: Consider the biochemical pathway presented below. This pathway determines pigment production in pea flowers. (Note: gene products are required for each reaction to proceed.)

12. Do you expect that genes A and B will interact with one another?

A. Yes B. No C. Insufficient information to determine

13. What phenotype ratio would you expect in offspring produced by crossing AaBb parents?

A. 9:3:3:1 (blue : green : yellow : white) B. 9:3:4 (blue : green : white) C. 9:7 (blue : white) D. 15:1 (blue : white)

14. How can haplosufficiency explain dominance of an allele that codes for a functional protein over a null (loss of function) allele?

A) a single functional copy of the gene produces sufficient gene product for normal biological function B) a single functional copy of the gene produces insufficient gene product for normal biological function C) the null allele produces a product that interferes with proper function of the normal allele D) the null allele produces a product that does not interfere with proper function of the normal allele E) the null allele rapidly mutates to reacquire normal functionality

15. In a plant species with genes A and B, you have isolated recessive mutant alleles at each locus, which you call a and b. You would like to know whether these two genes are linked. To test this hypothesis, you obtain an individual that is heterozygous at both genes (AB/ab) and perform a test cross. Which results would be most consistent with these loci being tightly linked? (Results are counts of the numbers of offspring from the testcross.)

A. Ab/ab 390 aB/ab 410 AB/ab 350 ab/ab 420

B. Ab/ab 80 aB/ab 73 AB/ab 760 ab/ab 754

C. Ab/ab 700 aB/ab 695 AB/ab 56 ab/ab 59

D. Ab/ab 400 aB/ab 65 AB/ab 50 ab/ab 659

goldclam880

(tautomerization) The most common form of guanine is the ketoform, not the enol form.

True

False

What term describes the allowed degree of base pair matchingduring annealing, varying from lots of base-pair mismatches allowedto allowing only perfect base-pair matching.

stringency

aneuploidy

polyploidy

quantitative PCR

restriction fragment length polymorphism

I have an individual with a disease we will call SSI-itis,brought on by extreme exposure to cannabis. I do a southern andfind the gene affected. Another of his classmates exhibits theexact same symptoms, but that gene in hiw is wt. A southern in thesecond student finds a mutation in another gene. This is an exampleof

pleitropy

allelic heterogeneity

locus heterogeneity

dominant traits

recessive traits

penetrance

If a person carrying a dominant allele does not demonstrate thephenotype strongly, but still differs from wt, thisdemonstrates?

haploinsufficiency

penetrance

expressivity

monogenic traits

polygenic traits

I find three alleles for a gene in a fruit fly. One allelecreates the ability for the fly to eat graphite. Another allelecreates the ability for the fly to eat coal. The third alleleallows the fly to eat diamond. These alleles are all dominant, andwt cannot eat any of these substances. Which statement below is themost true?

Upon mating a bunch of the flies carrying these alleles, Ishould not find a fly that can eat graphite, coal, and diamond.

I would consider the ability to eat these substances an exampleof pleitropy because they are all carbon.

I would expect the ability to eat diamond to be dominant to theability to eat graphite.

These flies are demonstrating locusheterogeneity.

Expression vectors in prokaryotes do not make functionaleukaryotic gene products in bacteria very well because

the codon sequence for prokaryotes is different than the codonsequence in eukaryotes

there are no disulfide bridges formed in proteins normally madein prokaryotes

prokaryotic expression vectors cannot translate eukaryticsequences

eukaryotic genes have introns, and prokaryotes don't

eukaryotic genes have exons and prokaryotes don't

Which of the following is not used in PCR amplification ofDNA?

Double-stranded DNA template

Oligonucleotide primers

DNA polymerase

DNA ligase

ddNTPs

I create a knockout mouse using the agouti/black fur systemdiscussed in the online lecture and in the book. When I cross theagouti offspring of the originally engineered mouse, I find a ratioof 2 agouti mice to 1 black furred mouse. What is the bestexplanation?

The gene knocked out was recessive.

The gene knocked out was dominant.

The gene knocked out was a lethal gene.

The knockput was integrated into a random spot, and did notknockput the original gene.

A loss of function mutation is usually recessive, but canactually be a dominant mutation because of what effect?

haploinsufficiency

dominant negative mutations

allelic heterogeneity

locus heterogeneity

pleitropy

A genetic carrier of a disease can have a form of the diseasethat only shows intermittent symptoms, while the full homozygousrecessiuve individual would show all of the symptoms of thedisease.

True

False

Wild type, or wt, refers to the genotype that is the typicalgenotype found in nature. Now read that again. Once more. Nowanswer.

Answer

True

False

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BIOL 1090 Lecture Notes - Zygosity, Peptide, Null Allele

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