🏷️ LIMITED TIME OFFER: GET 20% OFF GRADE+ YEARLY SUBSCRIPTION →
Log in
HomeClass Notes1,200,000CA670,000

BIOL 1090 Lecture Notes - Lecture 7: Cinnabar, Microphthalmia-Associated Transcription Factor, Null Allele

13 views8 pages
School
University of Guelph
Department
Biology
Course
BIOL 1090
Professor
Andrew Bendall

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

Document Summary

Genes may (and usually) have more than 2 alleles. Diferent alleles may afect the phenotype in diferent ways. A single gene may control several traits. Muliple genes may control a single trait. The most common allele is considered the wild type: designated with a superscript + (ex. Any allele found at appreciable frequencies (at least 1%) in the populaion is considered to be a polymorphism. All other alleles are considered as mutants. Recessive mutaions almost always involved a loss of gene funcion. Loss of funcion mutaions are usually recessive because, for most genes, one funcional copy is enough for a normal (wild type) phenotype. Efect of a null mutaion on the protein transcripion factor protein. Dominant mutaions can involve a loss of funcion or a gain of gene funcion. Loss of funcion mutaion for those genes for which one funcional copy is not enough (haploinsufficiency)

Get access

Grade+20% off
$8 USD/m$10 USD/m
Billed $96 USD annually
Grade+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
40 Verified Answers
Class+
$8 USD/m
Billed $96 USD annually
Class+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
30 Verified Answers

Related Documents

BIOL 1090 Lecture Notes - Zygosity, Peptide, Null Allele
erinprairie-dog747
BIOL 1090 Lecture Notes - Lecture 7: Pleiotropy, Null Allele, Haploinsufficiency
blushrabbit835
BIOL 1090 Lecture Notes - Lecture 9: Null Allele, Haploinsufficiency, Wild Type
indigohound801

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

(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



Expression vectors inprokaryotes do not make functional eukaryotic gene products inbacteria very well because

Answer Not sure which one ?!

· the codon sequence for prokaryotes is differentthan the codon sequence in eukaryotes

· there are no disulfide bridges formed in proteinsnormally made in prokaryotes

· prokaryotic expression vectors cannot translateeukarytic sequences

· eukaryotic genes have introns, and prokaryotesdon't

· eukaryotic genes have exons and prokaryotesdon't


I create a knockout mouseusing the agouti/black fur system \. When I cross the agoutioffspring of the originally engineered mouse, I find a ratio of 2agouti mice to 1 black furred mouse. What is the bestexplanation?

Answer not sure which one?!

· The gene knocked out was recessive.

· The gene knocked out was dominant.

· The gene knocked out was a lethalgene.

· The knockput was integrated into a random spot,and did not knockput the original gene.

A restriction enzyme cuts DNAand leaves the following end

xxCTGCA
xxG

Which of the following could be the sequence of the correspondingend of the other end of the cut DNA?

Answer

· xxG
xxCTGCA

· xxC
xxGTGCA

· xxCCGAT
xxG

· xxGGCTA
xxC


An SNP always occurs dueto

Answer

· a one base-pair change.

· a three base-pair change.

· a single restriction site change.

· a single amino acid change.

· None of these.


I PCR out a mutated gene in apatient with Lisenbee chorea (the inability to dance in acoordinated fashion) and I compare it to another patient with thatsame disease phenotype. One subject had a mutation on chromosome 4,and the other subject couldn't rock it because of a mutation onchromosome 5. This is an exampe of

Answer

· pleitropy

· locus heterogeneity

· allelic heterogeneity

· dominant negative mutation

· penetrance

In his experiments, Mendelnoted that when two traits are involved in a genetic cross, theyare inherited independently of each other. Though Mendel didn'tknow about chromosomes, this still holds true (mostly)because

Answer

· genes on the same chromosome are often mixedthrough meiotic crossover during the formation ofgametes.

· variant alleles of the same gene do not interferewith the phenotypic expression of the gentoype in different geneticloci.

· most higher organisms are diploid, making only twoalleles available for every genetic locus, and there is a 50/50chance of geting either allele from a single parent.

· expression of phenotypes can vary from thegenotypes because of encironmantal influences.

· chromosomes often recombine

Anticipation is caused by amutation that increases in expressivity over subsequentgenerations.

Answer

· True

· False

Question 44

I have a genotype that shouldproduce a specific phenotype, but some of the individuals with thegenotype do not demonstrate any evidence of the phenotype. I wouldconsider this an example of

Answer

· allelic heterogeneity.

· locus heterogeneity.

· penetrance.

· expressitivity.

· sheer luck.

Question

Which of the following doesnot occur during the PCR reactions?

Answer

· Changes in the reaction temperature by athermocycler

· Denaturation at high temperatures

· Synthesis of oligonucleotide primers

· Extension of primers by DNA polymerase

· Annealing of oligonucleotide primers

Question

DNA markers, or variantnon-coding regions of DNA, are DNA polymorphisms that are usefulfor genetic mapping.

Answer

· True

· False

A mosaic is an organismwith

Answer

· multiple genotypes within one organism

· multiple alleles within one genotype

· more than one color of fur

· transgenes added to the zygote beforedevelopement

· a wt phenotype but a mutated genotype

Question

Genotype causesphenotype.

Answer

· No, gentoype just influences phenotype.

· Yes, genotype is the DNA sequence that createsphenotype.

Question

A true genetic chimera can becreated by

Answer

· mutating a gene early on in the development of anorganism resulting in different alleles being present in theadult

· multiple fertilized eggs or zygotes fusing to formone embryo

· adding a transgene to the genome of an organismduring fetal development only

· adding cells of a different species to an adultorganism

· adding a transgene to the genome of an animal atany stage in development

Question

The ABO blood group can bestbe explained by the concept of _______.

Answer

· dominant traits

· recessive traits

· allelic heterogeneity

· locus heterogeneity

· vampirism

If a loss of functionmutation creates a dominant phenotype, it may be becauseof

Answer

· haploinsufficiency

· penetrance

· expressivity

· allelic heterogeneity

· locus heterogeneity

Please select the best matchfor each.

Answer

· PCR: Technique used to generatecopies of a DNA fragment

· Southern blot : Screeningprocedure used to assay DNA fragments

· RFLPs : DNA fragments generatedby restriction enzymes that vary in length betweenindividuals

· Northern blot : Screeningprocedure used to assay mRNA fragments

· Electrophoresis Movement ofmacromolecules in an electrical field

Answer

A.

Screening procedure used toassay DNA fragments

B.

Movement ofmacromolecules in an electrical field

C.

Screening procedure used toassay mRNA fragments

D.

DNA fragments generated byrestriction enzymes that vary in length betweenindividuals

E.

Technique used to generatecopies of a DNA fragment

A gene mutates, and theprotein produced has a novel way of interacting with the cell, andcreates a new phenotype because of this new functionality. This iscalled

Answer

· pleitropy

· locus heterogeneity

· allelic heterogeneity

· dominant negative mutation

· gain of function dominant mutation

Question 62

Mutations in the somaticchromosomes may be inherited by the next generation.

Answer

· True

· False


A degenerate PCR primer withmany variant sequences must be used to make multiple copies of agene

Answer

· if only the protein sequence of the gene productis known to construct the primers

· if the DNA you are using is cDNA to constructthe primers

· if the DNA you are using is genomic DNA to becopied

· if the DNA you are trying to copy iscDNA

· if the vector is prokaryotic and the transformedcell is eukarytotic

Question

Please select the best matchfor each term.

Answer

· Restriction enzyme


· PCR C

· Plasmid

· YAC A

· Dideoxynucleotide

Answer

A.

Endonuclease isolated frombacteria

B.

Eukaryotic cloningvector

C.

Method for amplification ofDNA sequences

D.

Small circular fragment ofDNA

E.

Molecule used in DNAsequencing

Question

A couple goes on MauryPovitch, and the results are in: you are not the father. But noother man impregnated the female (granted, unlikely for a MauryPovitch guest, but work with me here) and he must be the father.DNA analysis claims otherwise, though the child definitely wasmom's (poor thing). What may be going on here?

Answer

· the child is parthenogenic because the motheractually impregnated herself like a shark, and the child's DNA isall mom's

· the child had a mutation that changed the genethat is used to trace paternity

· the child is a mosaic because he is actually a setof twins fused early during fetal development, and therefore camefrom 2 eggs and 2 sperm

· the dad may have germ-line mosaicism, meaning thatthe genotype of his sperm is different from his somaticgenotype

· mitochondrial DNA only comes from mom, so there isno way to tell whobthe baby's father is