🏷️ LIMITED TIME OFFER: GET 20% OFF GRADE+ YEARLY SUBSCRIPTION →
Log in
HomeClass Notes1,200,000US530,000

BISC207 Lecture Notes - Lecture 12: Knockout Mouse, Srb, Metabolic Pathway

18 views3 pages
redcoyote43
9 Oct 2018
School
University of Delaware
Department
Biological Sciences
Course
BISC207
Professor
Olabisi Oyenike

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

BISC207 Full Course Notes
25
BISC207 Full Course Notes
Verified Note
25 documents

Document Summary

Each gene contains the info to make an enzyme. Srb and horowitz tested the hypothesis by studying a three-step metabolic pathway that produces arginine. A knockout mouse is a lab mouse in which researchers have inactivated or knocked out an existing gene by replacing it or disrupting it with an artificial piece of dna. The promoter is a base-pair sequence that specifies where transcription. The coding sequence is a a base-pair sequence that includes coding. The terminator is a sequence that specifies the end of the mrna information for the polypeptide chain specified by the gene. A gene is a sequence of bases in a dna molecule that carries the information necessary for producing a functional product. Transcription is the process of using a dna template to make a complementary rna (making a copy of rna) Translation is the process of using the information in mrna to synthesize proteins.

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

BISC207 Lecture Notes - Lecture 19: Pyrimidine, Polyadenylation, Phosphodiester Bond
violetgrasshopper2
BISC207 Chapter Notes - Chapter 17: Uracil, Adenine, Pyrimidine
violetgrasshopper2
BISC207 Chapter Notes - Chapter 17: Aminoacyl-Trna, Start Codon, Enzyme
harlequinpig55

Related Questions

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




Multiple choice questions, I need the CORRECT ANSWERS, 3500points

1.The promoter is in region_____________.

a.1

b.2

c.3

d.4

e.5

2.The stop codon is identified in region __________

a.4

b.5

c.6

d.7

e.8

3.A mutation is region 8 would be expected to:

a.disrupt transcription

b.disrupt ribosome

c.disrupt DNA polymerase binding

d.disrupt splicing

e.none of the above

4.Which is responsible for the transcription of structural genesin eukaryotes?

a.RNA polymerase I

b.RNA polymerase II

c.RNA polymerase III

d.DNA polymerase

E.RNA Synthesis

5.The eukaryotic TATA box is directly bound by a basaltranscription factor known as:

a.TBF

b.TBP

c.TAF

D.TPP

E.TAT

6.mOST TRANSCRIPTIONAL ACTIVATOR proteins in eukarypotesfunction in the cell as:

a.monomers

b.dimers

c.trimers

d.tetramers

e.pentamers

7.The basal level of transcription of eukaryotic genes is quitehigh in vitro systems. This is because:

a.activation are present

b.formation of chromatin dos not occur in vitro

c.basal transcription factors are present

d.a and c

e.none of the above

8.Transcriptional repression by methylation of DNA involves:

a.methylation of adenine (A) bases of the sequence GATC

b.methylation of cytosine (C) bases of th sequence GATC

c.mthylation of cytosin (C) bases in CpG nuclotides

d.Methylation of guanosine (G) bases in CpG nucleotides

e.acetylation of hisone proteins

9.A eukaryotic DNA sequence that can increase the level oftranscription is called:

a. a transcription factor

b.a transcritpional activator

c.TATA box

d.an enhancer

e.a co-activator

10.Which is generally true of eukaryotic transcriptionalregulation?

a.activators and reprssors bind to the promoter while basalfactors bind to the enhances

b.basal factors bind to the promoter, while activators bind tothe enhancers

c.activators bind to the promoter while basal factors andrprssors bind to th enhancers

d.repressors bind to the promoter whil basal factors andactivators bind to the enhancers

e.activators and basal factors bind to the promtoer whilereprssors bind to the enhancers

11.You wish to sequence the promoters region of the mouse generab.To do this, you need to construct a library and then search forthe gene sequence.You have a DNA probe based on the sequence of theprotein coding region.Which of th follwoing strategies would youfollow in order to carry this out?

A.construct a genomic library and screen for the promoter usingPCR

b.construct a cDNA library and screen for the promotoer usingPCR

c.construct a genomic library and screen for the promoter usingPCR

d.Construct a cDNA library and screen for the promoter usingcolony hybridization

e.all of the above

12.if one were to examine th clones in a cDNA library derivedfrom th heart muscle cells and the clones in a cDNA library derivdfrom prancreatic cells, you would find that:

a.the two libraries would hav the same cloned sequences

b.the two libraries would have completely different clonesequences from one another

c.the two libraries would share some common clond sequences, butwould each have some sequences unique to that library, not to befound in the other

d.the sequences from both would also b the same as those foundin the genomic library

e.none of th above

13.In a particular population at Hardy-Weinberg equilibrium, thefrequency of recssive allele b for a particular gne is0.01.Assuming that there are only two alleles for the gene, whatwould b th expected frequency for dominant allele B in th nextgeneration?

a.0.01

b.0.02

c.0.98

d.0.99

e.none of the above

14.White wool in the sheep is controlled by a dominant allele R,and black wool by the recessive allele r.In an isolated popualtionof 6530 sheep, 514 are balck, 4981 are hterozygous, 1035 arhomozygous white. When will this population be expected to reachHardy-Weinberg equilibrium?

a.in the next generation

b.in two generations

c.it already is in equilibrium

d.impossible to tell

e.it will not reach equilibrium

15.A population has th following genotypic disturbutions:

A^M A^M=1787, A^M A^O=2011, A^OA^O=2331.Is this population atH-W equilibrium?

a.yes, with allele frequencies of p=0.45562 and q=0.54437

b.yes, with frequencies of p=0.54437 and q=0.37862

c.no, with allele frequencies of p=0.45562 and q=0.37862

d.no, with xpectd genotypic ratios of p^2=1272, 2pq=3040, andq^2=1826

e.yes, with expctd genotypic frequencies of p^2=1778, 2pq=2015,and q^2=2336

16.The result of the Luria-Delbruck fluctuation exprimentindicatd that:

a. most bacteria are naturally resistant to phage

b.a low level of bacteria become resistant to phage by mutationupon exposure to phage

c.most bacteria become resistant to phage by mutation uponexposure to phag

d.bacteria become resistant to phag by random spontaneousmutation

e.bacteria become resistant to phag by acquiring resistanttraists upon repeated exposure

17.DNA probes are used for:

a.restriction ndonuclease analysis

b.sequencing

c.PCR

d.DNA library analysis

e.all of th above

18.Colony hybridization is used to:

a.isloate a gne of interest froma DNA library

b.analyze the presence or number of copies of a particular DNAsegment in a chromosome

c.determine the size of a gene or sequencing of interest

d.replace the more tediuos method of library screening.

e.all of the above

19.For the colony hybridization procedure, which of thefollowing is true:

a.individual colonies are isolated and analyzed by DNAsequencing

b.the DNA fragments are separated by size prior to transfer ontonitrocellulose membrane

c.the plasmid DNA must be denatured before hybridization

d.the relative positions of DNA fragments on the agarose gel arepreserved after transfer to the nitrocllulose membrane

e.the cloned target sequence is detected by staining withthidium bromide

20.The following strand of DNA is to be used for sequencing3'CGACTGACTCGACGGCCTAGCTATAG.The primer is GCTGACTG.The number ofDNA fragments resulting from the ddCTP tube is expectd to be:

a.7

b.5

c.8

d.4

e.6

21.Which of the following is not true of heterozygosity for agene deletion?

a.it is often detrimental if there is a gene dosage effect.

b.as long as one wildtype copy of any deleted gene is presnt onthe homologous chromosome, no phenotypic consequnces willresult.

c.it could be detected by Southern blot analysis

d.it could be detected by PCR analysis

e.none of the above

22.Robertsonian translocation arise from:

a.breaks at or near the chromosome of two acrocentricchromosomes

b.breaks at or near th centromeres of two metacentricchromosomes

c.breaks at or near th centromeres of one acrocentric and onemetacentric chromosome

d.breaks in a large and a small chromosomes

e.transposable elements

13. The arms of a human and the wing of a bat are _________________ structures, composed of the same bones that have been modified for different functions.

a. analogous
b. vestigial
c. homologous
d. convergent

14. Xeroderma pigmentosum is an autosomal recessive genetic disorder. People who are homozygous for the disease-causing allele have skin that is damaged by exposure to ultraviolet radiation; this can lead to skin cancer at an early age. The mutation that causes this disorder is in a gene that functions:

a. to repair DNA by excising (removing) nucleotides damaged by UV light.
b. in lysosomes, to break down lipids that otherwise accumulate in the nervous system.
c. to produce the UV blocking pigment melanin in the skin cells
d. in red blood cells to carry oxygen to the skin.

15. The Amish are a religious sect descended from a small group of colonists who came to the United States over 200 years ago. Their religious beliefs have kept them isolated from society. The Amish have a much higher than average incidence of both polydactyly (extra fingers and toes) and dwarfism in their communities. This is most likely explained by:

a. inheritance of acquired characters.
b. the founder effect.
c. disruptive selection.
d. natural selection.

16. In the structure of DNA that Watson and Crick proposed, the sides of the "ladder" of the DNA molecule are always the same distance apart, making the molecule a constant width. This occurs because when the bases hydrogen bond to form the rungs of the "ladder", a:

a. purine always pairs with a pyrimidine.
b. purine always pairs with a purine.
c. pyrimidine always pairs with a pyrmidine.
d. The structure of the double helix is not determined by how the bases pair.

17. The RNA transcript of DNA that travels to the cytoplasm, carrying the instructions to make a protein, is called:

a. rRNA.
b. tRNA.
c. mRNA.
d. RNA polymerase.

18. When DNA is replicated, the error rate is approximately one error for every 10,000 nucleotides copied. However, that error rate is reduced to only 1 error for every 1 billion nucleotides. The DNA sequence is "corrected" by:

a. repeating S phase to see if the error rate is lower the second time around.
b. enzymes that proofread the DNA and repair errors.
c. messenger RNA during the process of transcription.
d. Any errors made during DNA replication cannot be corrected. If errors are detected after S phase, the cell is destroyed.

19. Mendel's principle of segregation says that:

a. when gametes are formed, each gamete receives only one allele for a particular gene.
b. some genes are dominant to others.
c. a testcross must be used to determine the genotype of an organism with a dominant phenotype.
d. all of the above

20. Tay-Sachs disease is lethal before reproductive age, but the allele persists in Louisiana French Canadians because:

a. it is a dominant allele.
b. new mutations causing this disease are common in that population.
c. it is not expressed in the phenotype of heterozygous individuals.
d. the disease is contagious and can be transmitted by nonheritable means.

21. Physical features that are similar in organisms which are not closely related to each other are most likely a result of ____________. A good example of this is the similarities seen between some marsupial and placental mammals.

a. natural selection.
b. adaptation to similar environments
c. convergent evolution
d. All of the above phrases can be used to correctly fill this blank.

22. In humans, which of the following sex chromosome compliments could be found in males?

a. XY
b. XX
c. Y

23. The proteins that are associated with eukaryotic chromosomes:

a. help to package the DNA within the nucleus.
b. are involved in gene expression, through binding with the DNA.
c. may be replicating or transcribing the DNA.
d. All of these are true of proteins that are associated with eukaryotic chromosomes.

24. A sequence of DNA nucleotides that contains the information to produce a single protein is a(n):

a. codon.
b. gene.
c. polypeptide.
d. anticodon.