HMB265H1 Lecture Notes - Restriction Fragment Length Polymorphism, Silent Mutation, Southern Blot

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20 Jun 2013

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HMB265H1 Full Course Notes

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Alleles are usually found to be identical in most of their sequences; only differ at one or several nucleotides; essentially, all alleles are different versions of the same gene. Mutation creation of a new allele when the nucleotide sequence changes due to a rare chemical. Alleles with mutations are usually recessive, since it takes only 1 copy of a wild-type gene to provide normal function. A, representing the wild-type allele, means one specific sequence of dna; a is a shorthand that can represent any of the possible types of damage that can lead to non-functional recessive alleles. If we wanted to show dna segregation directly, we could sequence the alleles, and the products would have the a sequence, and would have the a sequence. This would also apply to any dna sequence that differed in the inherited chromosomes. Restriction fragment length polymorphism (rflp) another way of tracking/characterizing a segment of dna.

Related Questions

(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

mauvekangaroo793

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

1. Characters that show a continuous range of variation, such as height and eye color, usually are controlled:

a.	by a single gene with two alleles that are codominant.

b.	by many genes with an additive effect.

c.	by epistatic interactions between two genes.

d.	mainly by the environment, with only a small genetic component.

2. In humans, red-green colorblindness is inherited as a sex-linked recessive trait. In order for a woman to be red-green colorblind, which of the following statements must be true.

a.	Her mother must be red-green colorblind.

b.	All of her brothers must be red-green colorblind.

c.	Her father must be red-green colorblind.

d.	All of the above statements must be true if a woman is red-green colorblind.

3. The x-ray crystallography data collected by Rosalind Franklin suggested to Watson and Crick that the:

a.	structure of DNA is a double helix.

b.	two strands of the DNA molecule are joined by hydrogen bonds between the bases.

c.	four bases within DNA pair in a specific way.

d.	two strands of the DNA molecule are joined by covalent bonds between the bases.

4. In the genetic code, _________ one amino acid.

a.	one nucleotide specifies

b.	two nucleotides specify

c.	three nucleotides specify

d.	four nucleotides specify

5. During Meiosis I, a homologous pair of chromosomes may not separate, resulting in daughter cells that have extra chromosomes or are missing chromosomes. This can lead to genetic disorders, including Down Syndrome. This phenomenon is called:

a.	independent assortment.

b.	nondisjunction.

c.	segregation.

d.	crossing over.

6. You are a human geneticist studying the incidence of retinitis pigmentosa in the residents of Tristan de Cunha, a group of small islands in the middle of the southern Atlantic Ocean. The allele for retinitis pigmentosa, which causes a form of blindness, is inherited as an autosomal recessive. You have determined that the frequency of this allele (r) in the population is 0.4 (40%). Using the principles of the Hardy-Weinberg rule, you would estimate the frequency of individuals who are heterozygous for this allele (Rr) in the population to be:

a.	0.16 (16%)

b.	0.24 (24%)

c.	0.36 (36%)

d.	0.48 (48%)

7. Natural selection acts at the level of the:

a.	phenotype.

gene.

c.	population.

d.	nucleotide.

8. You are working with pea plants, trying to recreate the experiments that Mendel performed. You are doing a dihybrid cross with a plant that is heterozygous for both seed shape and seed color, with the genotype RrYy. Which allelic combinations would you expect to find in the gametes produced by this plant?

a.	This plant would produce only RY and ry gametes.

b.	This plant would produce only RrYy gametes.

c.	This plant would produce RY, Ry, rY, and ry gametes.

d.	You cannot determine which gametes this plant can produce without knowing the genotypes of its parents.

9. Biochemist Erwin Chargaff found that in DNA there is a special relationship between the four bases that we now call Chargaff's rule. His observation was that, in an organism's genome the:

a.	percentage of A nucleotides = the percentage of T nucleotides, and the percentage of C nucleotides = the percentage of G nucleotides.

b.	four bases all occur in an equal frequency (25%) within each organism.

c.	percentage of A nucleotides = the percentage of G nucleotides, and the percentage of C nucleotides = the percentage of T nucleotides.

d.	genetic material is composed of proteins, not DNA.

10. During DNA replication:

a.	each strand of the double helix acts as a template for the synthesis of a new strand.

b.	the enzyme DNA polymerase adds nucleotides to the strand being synthesized.

c.	the bases A,C,G and T are required.

d.	All of the above are true of DNA replication.

11. During translation, amino acids are joined by peptide bonds to make polypeptides. The formation of these peptide bonds is catalyzed by:

12. If an allele (R) at a gene with two alleles shows complete dominance, individuals with the genotypes ______ will have the same phenotype.

a.	RR and rr.

RR and Rr

Rr and rr

d.	Each of the three possible genotypes will have a different phenotype.

mauvefox570

HMB265H1 Lecture Notes - Restriction Fragment Length Polymorphism, Silent Mutation, Southern Blot

HMB265H1 Full Course Notes

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