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BIOL 112 Lecture Notes - Lecture 8: Penetrance, Solvent, Deoxyribose

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cyanraccoon413
21 Jul 2016
School
McGill University
Department
Biology
Course
BIOL 112
Professor
Jacalyn Vogel

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BIOL 112 Full Course Notes
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BIOL 112 Full Course Notes
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Document Summary

In other words, the genes assort independently, but the ratio is not 9:3:3:1 because of interactions (epistatis) o. Use an organic solvent to extract lipids and proteins (e. g. phenol) Precipitate with ethanol, and you get just dna o now do the assay test. No o but you can actually do something similar, with bacteria (discovered by. Frederick griffith) o can an extract from dead bacterial cells genetically transform living bacterial cells? o s strains and r strains of bacteria, s is deadly, r is harmless (when injected into a mouse) So take the s strain, heat it up so that the bacteria are dead, inject in mouse - harmless. However, if you kill the s strain, mix it with the r strain, and inject it into the mouse, it will die. Essentially you have transformed the r strain into the deadly s strain, by mixing them (something transferred)

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Related Questions

Q.In the non-9:3:3:1 ratio dihybrid cross below, there are 3 classes of phenotypes: white, pink, red which of the following is NOT true?

9 A_B_ white

3 A_bb pink

3 aaB_ red

1 aabb red

a.) phenotype is dependent on the genotype of gene B

b.) This is an example of recessive episasis

c.) aa is epistatic to B_, and bb

d.) phenotype is dependent on the genotype of gene A

Q,When choosing a selectable marker, you...

a.) choose a enhancer for your gene of interest

b.) choose a rescue gene to save your gene of interest

c.) choose a resistant gene that will allow for colony screening

d.) choose a gene that encodes for DNA replication

Q.The following steps detail an experiment resulting in transgenic mice with a knockout gene for fur color. Recall that agouti (brown) mice are A/A;M/M and black fur coats are a/a;M/M . List the steps in their correct chronological order.

1.Inject transformed agouti embryotic cells into an a/a;M/M embryo to produce A/A;M/m and a/a;M/M altered embryo

2.Self-cross hemizygous mice to produce homozygous transgenic mice (A/-;M/m)

3.Create a a targeting vector containing tk+ for ganciclovir sensitivity, neoR for neomycin resistance, and the gene encoding fur color.

4.Observe mice with chimeric phenotype and cross them with black a/a;M/M mice to produce transgenic hemizygous progeny (A/a;M/m)

5.Isolate all transformed embryonic stem cells that show homologous recombination and inject them into a embryonic stem cells of an agouti mouse.

a.) 3,5,4,1,2

b.) 1,5,3,4,2

c.) 3,5,1,4,2

d.) 5,3,2,4,2

If dog 1 has is black with alleles (A_B_) , and dog 2 is brown with (aaB_), and dog 3 is yellow with (_ _bb) alleles, what kind of epistasis is this and what color would dog 4 be if he was (AAbb).

1. Recessive epistasis, Yellow

2. Recessive epistasis, Black

3. Dominant epistasis, Yellow

4. Dominant epistasis, Black

A gene interaction in which pair of recessive genes at one locus prevents expression of dominant allele at another locus called

a. incomplete dominance

b. epistatis

c. complete dominance

d. polygenic inheritance

Problem A. You cross two homozygous parents: RRYY (round yellow) and rryy (wrinkled green) to produce an F1 generation. Round seeds are dominant, yellow seeds are dominant. You cross two F1s to produce the F2 generation. In the F2 generation, you find an individual with round yellow seeds. You do a testcross to find out the genotype of this “mystery” dominant/dominant phenotype. In the offspring of the cross, you find 1?2 have round yellow seeds and 1?2 have wrinkled yellow seeds.
1. Based on these data what was the genotype of the mystery F2?

Problem B. You carry out a dihybrid cross between a red-flowered rose with long pollen grains (A1A1 LL) and a white-flowered rose with round pollen (A2A2 l l). The flower color gene (A) shows incomplete dominance, and long pollen is completely dominant to round pollen. The genes for these traits are on different chromosomes so assort independently
2. What would be the phenotypes of the F1 generation?
3. What fraction of the F2 generation is expected to have red flowers and long pollen? (As always when traits are independent, what rule can you apply to save time?)

Problem D. In the example of epistasis presented in class, the agouti allele in mice (A) codes for striped hairs and is dominant to the allele for all black hairs (a). A loss-of-function allele for the pigment gene (b) produces no pigment and is recessive to the normal, dominant allele (B). The recessive allele at b masks the effect of the A gene. A dihybrid cross of AABB and aabb is carried out.
5. Name all of the genotypes in the F2 generation that can produce an albino (no pigment) phenotype.

Problem E. A cross is made between homozygous fruit flies that differ for eye color (the st gene) and
body color (the e gene). The F1 double heterozygote is shown below using chromosome notation.
st+ e
--------------------------------------
st e+
st = the recessive allele for scarlet eyes; st+ = the dominant wildtype allele for red eyes e = the recessive allele for ebony body; e+ is the dominant wildtype allele for gray body

6. Are the chromosomes shown in the genotype above in cis or trans configuration?
7. Based on this notation, what were the genotypes and phenotypes of the two parents of the F1? (Think carefully here...)
8. With what phenotype would you cross this F1 to measure recombination frequency?

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