Final Exam Practice - Massachusetts Institute of Technology

FALL 2005 Final Exam Practice 1

Final Exam PracticeFinal Exam is on Friday, DECEMBER 16th 1:30 – 4:30 pm in

DuPont sit in rows according to your TA’s name.

BRING PICTURE I.D.

Exam Review (new material only) on Wednesday, Dec. 14th

7-9 PM, Auditorium 10-250

Exam Tutorial Thursday, Dec. 15th 2-4 PM, 68-121NOTE that this practice exam IS WAY LONGER THAN THE REAL FINAL

EXAM IN AN EFFORT TO GIVE YOU EXTRA PRACTICE PROBLEMS!!!!

Tutors:Xiaoyu (Sylvia)

[email protected]

Adriana [email protected]

Jennifer [email protected]

Amy [email protected]

Yiyan [email protected]

Christopher [email protected]

Angelin [email protected]

John [email protected]

Albert [email protected]

Yiyan [email protected]

Office Hours. Each office hour listed lasts one hour.Eager helpers Office Hour

Margaret AckermanPalowski

[email protected] Wed 5 pm, Stata Cafe

Ericka Anderson [email protected] Wed 9 pm, 1st fl Student Center

Garrett Frampton [email protected] Tue 4 pm, 26-204

Emily Miller [email protected] Tues 8 pm, 1st fl Student Center

Shlomo Meislin [email protected] Mon 5:30 pm, 26-204

Hung Nguyen [email protected] Wed 4 pm, 68-329

Renuka Sastry [email protected] Wed 6 pm, 56-180

Sebastian Treusch [email protected] Tues 5 pm, Bio Café, Blg 68

Claudette Gardel [email protected] Wed 1:30, 68-120d


Question 1

In the space provided next to each definition or description, clearly write the letter of theappropriate term from the list of terms given on the last page.

____ A short, single-stranded DNA that serves as the necessary starting materialfor the synthesis of the new DNA strand in PCR

____ The synthesis of DNA using DNA as a template

____ The building blocks of DNA and RNA

____ The synthesis of protein using information encoded in mRNA

____ The location in a eukaryotic cell where the electron transport chain occurs

____ The major component of cell membranes

____ The genetic composition of an organism

____ A gene that lies on one of the sex chromosomes

____ An organism without membrane-bound organelles

____ A cell with 1n chromosomes

____ The building blocks of proteins

____ A cell with 2n chromosomes

____ A major source of energy that has the general formula (CH2O)n

____ An enzyme needed for completion of lagging strand synthesis, but notleading strand synthesis

____ The synthesis of RNA using one strand of DNA as a template

____ An observed characteristic of an organism


Question 1, continued

____ A DNA molecule that is distinct from the chromosome; this molecule can beused to move foreign DNA in or out of a cell

____ The DNA from a eukaryote formed by the enzyme reverse transcriptase; thisDNA lacks introns

____ An organism with 2 identical alleles for the same gene

____ A membrane protein involved in signal transduction; activation involvesbinding a GTP molecule

____ An organism with genetic material inside a nucleus

____ An organism with 2 different alleles for the same gene

____ A measure of the affinity of an enzyme for its substrate

____ A gene that lies on any chromosome except the sex chromosomes

____ The membrane that surrounds the cell

____ One of the alternate forms of a gene found at a given locus on achromosome

____ A technique for the rapid production of millions of copies of a particularregion of DNA

____ Proteins with a signal sequence are directed to this cellular organelle


Question 2

The following double-stranded DNA contains sequence of a eukaryotic gene:

b 1 2 3 5'-ATGGCCTTCACACAGGAAACAGCTATGGCCATGAGCACGC 1 ---------+---------+---------+---------+ 40 3'-TACCGGAAGTGTGTCCTTTGTCGATACCGGTACTCGTGCG

ii CAGTCTCGGCATTATCCTATTAAAGGGAACTGAGGTGA-3' 41 ---------+---------+---------+---------+ 80 GTCAGAGCCGTAATAGGATAATTTCCCTTGACTCCACT-5' i

a) Transcription begins at the underlined A/T at base pair 17 (b) and proceeds to the right.What are the first 12 nucleotides of the resulting mRNA? Indicate the 5' and 3' ends of themRNA.

b) The first 7 amino acids of the protein encoded by this gene are: NH3+ -met-ala-met-ser-thr-pro-his-tyr....COO-

i) underline the nucleotides which correspond to the 5' untranslated region of the primaryRNA transcript made from this gene.

ii) draw a box around the intron region in this gene.

c) Consider each of the following three mutations independently.

i) How would the resulting protein change if the underlined G/C base pair at position 22 (1)was deleted from the DNA sequence? Briefly explain.

ii) How would the resulting protein change if the underlined G/C base pair at position 27 (2)was changed to a C/G base pair? Briefly explain.

iii) How would the resulting protein change if the underlined A/T base pair at position 31(3) was deleted from the DNA sequence? Briefly explain.


Question 3

a) Many patients are coming into the emergency room with a disease caused by an unknownpathogen! A doctor studies this pathogen in order to create a vaccine against it. Shediscovers that the infectious agent is an intracellular bacterium and its cell surface iscoated with human-like proteins. Considering the mechanism of the pathogen, the doctordecides to generate a live-attenuated vaccine instead of a heat-killed vaccine.

i) What are the two advantages of using a live-attenuated vaccine vs. a heat killedvaccine in this case?

ii) What is a disadvantage of using a live-attenuated vaccine?

b) When a rabbit protein is injected into rabbits, no antibodies against this protein aregenerated. If, however, the same rabbit protein is injected into guinea pigs, the guineapigs generate antibodies against the rabbit protein. Briefly (in one or two sentences)explain this observation.

c) The genomes contained in almost all of the somatic cells in an adult human are identical.Name one (diploid) cell type that is an exception to this and specify the process by whichthe genetic variation occurred.

d) Will siblings have the exact same antibody repertoire? What about identical twins?Briefly explain your reasoning.


Question 4a) Below is the pedigree for a family with an autosomal recessive disease, disease X.

?

= unaffected female

= affected female

= unaffected male

= affected male

A

B

C D

i) What is the genotype of individual A at the disease X locus? Use “+” to indicate thewildtype allele and “-“ to indicate the mutant allele.

ii) What is the probability that individual B is a carrier of disease X?

iii) Individuals C and D decide to have a child. What is the probability that the child will havedisease X?

iv) What is the probability that the child of individuals C and D will be a carrier of disease X?

b) The most common mutant allele of the disease X gene is a deletion of three nucleotideswhich eliminates a phenylalanine at amino acid residue 508. Although the mutant X protein ismade, it is not localized to the plasma membrane.

i) Assuming the altered X protein is stable, where might it be found?

ii) Describe another mutation in this gene that could prevent the disease X protein fromlocalizing to the plasma membrane.


Question 4, continued

c) Researchers are currently working on gene therapy for disease X patients. The mostpromising therapy has involved incorporating the disease X gene into an adenovirus. Becauseadenovirus is a double-stranded DNA virus that targets lung epithelial cells, it can be used todeliver the disease X gene to the lung cells of the affected individual.

i) The adenovirus used in these studies is able to produce gp19, a protein that inhibits thedisplay of MHC I molecules on the surface of cells. Why is this a desirable property of thevirus used to deliver the disease X gene?

ii) Using the plasmids and restriction enzymes provided, design a procedure to create a,double-stranded DNA to incorporate into the adenovirus particle. The final product shouldbe linear, contain the majority of the virus genome and have the disease X gene undercontrol of the E1 promoter (PE1). NheI and SpeI create the same sticky ends. All the otherrestriction enzymes create unique cuts.

pBR-Ad2-7

PE1

BamHI

NheIHindIII

SpeI

BamHI

Adenovirusgenome

pCMV-diseaseX

EcoRI

HindIII

SpeI

disease X cDNA

start

stop


Question 5

The figure below shows GDP in the binding pocket of a G protein.

NH

NH

CH N NH

C O

OC

HO

Lys

Arg Asp Glu

Tyr

3+

2 2+

P P

O O

O O O

OH OH

N

N NH

NHN

O

CH 2

O-O-

O-2

O-O-

a) Circle the strongest interaction that exists between:

i) the side chain of Lys and the phosphate group of GDP

van der Waals covalent hydrogen bond ionic

ii) the side chain of Glu and the ribose group of GDP


iii) the side chain of Tyr and the guanine base of GDP


b) You make mutations in the GDP-binding pocket of the G protein and examine their effects onthe binding of GDP. Consider the size and the nature (e.g. charge, polarity, hydrophilicity,hydrophobicity) of the amino acid side chains and and give the most likely reason why eachmutation has the stated effect. Consider each mutation independently.

i) Arg is mutated to a Lys, resulting in a G protein that still binds GDP.

ii) Asp is mutated to a Tyr, resulting in a G protein that cannot bind GDP.


Question 6The bos/seven receptor is required for differentiation of a particular cell, called R7. It is areceptor tyrosine kinase with the structure below. As a monomer, the protein is inactive.Binding of ligand causes the receptor to dimerize, causing phosphorylation of the intracellulardomain, activating the protein. During processing of the protein, the extracellular domain iscleaved and a disulfide bridge forms between two cysteines, tethering the ligand-bindingdomain to the rest of the protein.

-S-S-

ligand-bindingdomain

intracellular

extracellular

membrane

-S-S- -S-S-

ligand

ACTIVEINACTIVE

PP

a)i) How would receptor activity be affected by changing one of the two cysteines shownabove to an alanine? Explain.

ii) What effect would this mutation have on the differentiation of R7?

b) Name three amino acids that would be likely to be found in the transmembrane domain.What property do those amino acids have in common, and why do they cause thetransmembrane domain to stay in the membrane?

c) Activation of the above receptor causes Ras to exchange GDP for GTP, thereby activating it.This activated Ras can activate a signal transduction cascade, which ultimately results in thetranscription of genes required for R7 differentiation. In different cells in the same animal, Rascan be activated by an activated growth factor receptor. This leads to transcription of genesrequired for cell division.


i) How is it possible for the activation of Ras to lead to transcription of different sets ofgenes?

ii) Given that these cells exist in the same animal, name one component in the pathway thatcould be mutated to give each of the following results (consider each situationindependently). Describe how the mutant component differs from the wild-typecomponent, and whether it is a loss-of-function or gain-of-function mutation.

• You never see differentiation of R7 cells.

• You see uncontrolled cell proliferation.

boss/sev

RasGDP

GTP

transcription of genesfor R7 development

RasGDP

GTP

EGF

receptorEGF

transcription of genesfor cell division


Question 7You are studying a common genetic condition. The mutant allele differs from the wild-typeallele by a single base-pair (bp) substitution. This substitution eliminates a NheI restriction sitethat is present in the wild-type allele. (The mutant allele is not cut by NheI.) A pedigree of afamily exhibiting this condition is shown below:

normal male

affected male

normal female

affected female1 2 3 4

5 6 7 8

You isolate DNA from four individuals in the pedigree. Using PCR techniques, you amplify a1000 bp portion of their DNA that includes the site affected by the mutation. You digest thePCR products with NheI and analyze the resulting DNA fragments on a gel:

a) Based on these data, is this gene located on an autosome or the X-chromosome? Briefly justifyyour reasoning.

b) Based on these data, is the mutant phenotype dominant or recessive to wild-type and why?

c) If individuals 3 and 4 have a daughter, what is the probability that she will be affected?Justify your reasoning.

1000 bp

600 bp

400 bp

5 Individual:NheI NheI NheI NheI

6 7 8


You sequence the region around the NheI site in the wild-type PCR product. You thensequence the corresponding region in the mutant PCR product and discover that not only didthe mutation eliminate the NheI site in the mutant allele but it has created a new PvuIIrestriction site. The recognition sites for the two enzymes are indicated below.

NheI cuts at: 5' GCTAGC 3' PvuII cuts at: 5' CAGCTG 3'3' CGATCG 5' 3' GTCGAC 5'

A portion of one strand of the wild-type DNA sequence is shown below:

5’....GCTAGCTG...3’

d) What is the sequence of this same region in the mutant allele?Indicate the 5' and the 3' ends of the DNA sequence.

e) Individuals 1 and 2 have another child, 9, who is affected by the genetic condition.

1 2

5 6 9

You PCR amplify the 1000 bp region affected by the mutation from individuals 1, 2, and 9,digest the PCR products with NheI or PvuII, and analyze the restriction fragments on a gel:

NheI PvuII9Individual:

NheI PvuII2

NheI PvuII

1000 bp

600 bp

400 bp

1

What event occurred and how does this explain the data shown above?


Question 8

While walking through the sub-basement of the Infinite Corridor late one night you comeupon an enclave of gnomes. You are struck by the color of their beards, which are all blue.(Gnomes are diploid organisms, both male and female gnomes have beards, and you canassume that the gnomes are true-breeding for this trait.) The following week you are busypulling a hack at Harvard when you spy another enclave of gnomes. All of these gnomeshave yellow beards. (Again assume that the gnomes are true-breeding for this trait.) Curious,you collect a few yellow-bearded gnomes from Harvard and bring them back to MIT. Lateryou discover that several of the yellow-bearded gnomes and blue-bearded gnomes havemated. The offspring of these matings are all green-bearded. Below are two possibleexplanations for these results.

a) Possibility 1: Beard color is controlled by a single locus. Give the genotypes in the blanksbelow.

Yellow beards Blue beards

Green beards

X

b) Possibility 2: Beard color is controlled by a pathway of two distinct enzymes encoded bythe A and Q genes.i) Give one genotype in each of the blanks below. Use A and Q to designate the wild-typealleles. Use a and q to designate the loss-of-function alleles.


Green beards

X

ii) When two F1 green-bearded gnomes mate, they produce 64 green-bearded gnomes, 27blue-bearded gnomes, and 22 yellow-bearded gnomes. Given your answer to i) above, drawthe pathway for beard color. Be sure to include at which step each of the genes functions.


Question 9Bob, a sophomore at MIT, failed 8.01 his freshman year. His parents are both physicists, buthe remembers that his great-grandfather also failed physics. Bob constructs the followingfamily pedigree and is convinced that his poor performance in physics is an inherited genetictrait.

Bob

1 2

3 4 56 7

8 9

unaffected maleaffected male

unaffected femaleaffected female

10

a) If Bob's hypothesis is true, what is the most likely mode of inheritance?_____________

b) Individuals marrying into the family are homozygous for the wild-type allele. Complete thetable below. Use G or g to denote the alleles of this gene. Be sure to note any ambiguities.

individual genotype

2

3

4

9

10

c) Bob meets Leah in his remedial physics class. Bob is a hard worker (homozygous for the Hallele). Leah is lazy (homozygous for the h allele). The H locus is linked to a chromosomalmarker, which exists in two forms Ω+ or Ω-. Circle the non-recombinant genotypes of Boband Leah's grandchildren.

BobLeah

?

hhΩ-Ω-

HHΩ-Ω- Ω+Ω+ hh

genotypes

hhΩ+Ω+ hhΩ-Ω- hhΩ+Ω- HHΩ+Ω+ HHΩ-Ω-

HHΩ+Ω- HhΩ+Ω+ HhΩ-Ω- HhΩ+Ω-


Question 10

The virus that causes AIDS (acquired immune deficiency syndrome) is called HIV (humanimmunodeficiency virus). HIV is a retrovirus, and its genome is a single (+) strand RNA.Two copies of this RNA genome are packaged with two copies of the retroviral enzyme,reverse transcriptase, within a protein capsid. This capsid is further packaged into anenvelope derived from the plasma membrane of the host cell in which the virus was formed.The surface of the envelope is covered with an envelope glycoprotein, called gp120.

a) Shown below is a schematic representation of HIV. Identify each component(numbered i - iv) in the figure below, indicate if the component is of viral origin or hostorigin.

i)

ii)

iii)

iv)

v)

ORIGIN?

b) HIV specifically infects the helper T cells of the human immune system. If receptor-ligand mediated viral entry is the means by which HIV enters TH cells, what would be themost likely ligand on HIV?

c) Once infection has taken place, the RNA genome has to be made into double-strandedDNA. This process is mediated by reverse transcriptase. Once a double-stranded DNAcopy of the HIV genome has been made, it is integrated into the host cell genome. Theintegration event is mediated by an enzyme, “integrase”.

i) What are the three steps required to produce double-stranded DNA from the single (+)strand RNA genome?

ii) Given that these three steps are all performed by reverse transcriptase, what arethree enzymatic activities that reverse transcriptase must possess?


In recent years, therapies have been developed to fight AIDS using nucleotide analogs.The drug most widely used to combat AIDS is Azidothymine (AZT). The structure of AZTis very similar to thymidine except that in AZT, the 3'-hydroxyl (OH) group on thedeoxyribose ring has been replaced by an azido (N3) group.

d) Which process of the life cycle of the HIV do you think is inhibited by AZT?

e) What other phase in the HIV life cycle may be more appropriate to disrupt, one thatwon't likely affect host cellular processes? Explain your answer.

AZT resistant forms of HIV have been isolated. These mutant viruses have mutations inone of the genes of the virus that result in resistance to AZT.

f) In which viral gene do you think are the mutations most likely to be found? Explain.

g) Starting with the virus in the bloodstream, order the following steps in the life cycle ofHIV.

A- Entry of the viral genome into the nucleus.

B- Adsorption of the viral proteins to cell-surface receptors.

C- Degradation of DNA-associated viral RNA.

D- Fusion of the viral membrane with the plasma membrane.

E- DNA-dependent RNA synthesis.

F- Integration of the viral nucleic acid into the host genome.

G- Encapsidation of the viral genome into protein coats.


Question 11To investigate the yeast metabolic pathway for serine biosynthesis, you screen for serineauxotrophs (mutants which are unable to grow without serine supplied in their growthmedium).You isolate four such mutants, which are recessive to the wild-type strain, and you test themfor growth on medium supplemented with several intermediates (A, B and C) known to bepart of the pathway. The results are shown below ("+" represents growth,"-" represents no growth).

Strainminimalmedium

minimal+ A

minimal+ B

minimal+ C

minimal+ serine

wild-type + + + + +

m1 - + + - +

m2 - - - - +

m3 - + - - +

m4 - + + - +

You then mate the haploid m1 strain with the haploid m4 strain to create a diploid yeast straincarrying both the m1 and the m4 mutations. You test the diploid for growth on the sameconditons as above and observe that the diploid exhibits the same growth requirements as them1 or the m4 haploid.a) Are the m1 and m4 mutations in the same gene or in different genes? Briefly explain yourreasoning.

b) Draw the metabolic pathway for the synthesis of serine, consistent with the data givenabove. Include the intermediates (A, B, and C) and serine, and indicate which mutants (m1,m2, m3, m4) are defective at each step in the pathway.

c) You create a haploid strain that has both the m1 and m3 mutations.i) This haploid mutant will grow on media supplemented with which of the followingintermediate(s): A, B and/or C?

ii) When grown on minimal medium this haploid will accumulate which of the followingintermediate(s): A, B and/or C?


Question 12

Shown below is a schematic representation of an antibody made in B cells.

a) On the diagram label the following structures (i) – (v) on the antibody diagram.

i) the heavy chains

ii) the light chains

iii) the variable regions

iv) the antigen binding sites

v) If this were a secreted antibody, indicate the region of the antibody that would interactwith phagocytic cells.

vi) If this were a membrane-bound antibody, indicate the membrane-spanning region ofthe antibody.


b)i) List two ways that combinatorial joining generates antibody diversity.

ii) What are the interactions between the light and heavy chains that make them associatewith each other?

c) Name the cell type of the immune system that would:

i) bind an antigen floating around in the blood.

ii) recognize an antigenic peptide on a MHC Class II protein displayed by a macrophage.

iii) recognize an antigenic peptide on a MHC Class I protein displayed by a skin cell.

iv) nonspecifically engulf and digest a variety of pathogens (disease-causing organisms).

v) secrete large amounts of antibody in response to an infection.

vi) recognize a viral infected body cell and destroy it.

vii) Children immunized against measles virus by injection of viral proteins have long-livedimmunity to measles due to this cell type.

d) Will an individual who does not express MHC Class II proteins on the surfaces ofB cells and macrophages produce a humoral immune response?

e) Will siblings have the exact same antibody repertoire? What about identical twins? Why orwhy not?


Question 13

In 1930, 30 astronaut couples were secretly sent to Titan, where no humans previously resided.However their spacecraft was damaged on landing and they could not return. It took more then 70years to make another spacecraft and reach the newly formed human population on Titan.

The new population was analyzed for some traits and it was compared to the population on earth.Blood samples had been taken of each of the 60 people (astronauts) so the frequencies of these traits inthe Titan colonizers are known, as well as those of current Titan population. See data below.

Trait Earth in 1930 Earth 2004 Astronauts Titan 2004Tongue rolling 0.7 0.7 0.5 0.1O blood type 0.05 0.05 0.1 0.3

a) The frequencies of tongue rolling and blood type traits of the astronauts were different fromthose of the earth population in 1930. Which of the following phenomena best explains this?

a) genetic drift

b) founder effect

c) mutation

d) natural selection

b) Trait frequencies of Titan today are different from those of the astronauts. Given that therewas no selective pressure in the planet for any trait, which of the following is the best reason ofthese changes?

a) genetic drift

b) founder effect

c) migration into the population


Blood samples from all 30 couples were taken prior to their departure from earth. Yousequence a specific region of each woman’s mitochondrial DNA or man’s Y chromosome. Allwere grouped into haplotypes. The frequency of haplotypes were compared to the currentpopulation on Titan.

Woman # Haplotype % of polymorphismsin old population

% of polymorphismsin new population

1,29,18,10,13 A 17 217,3,8,20 B 13.3 102,5,24,21 C 13.3 530,27,11,25 D 13.3 304,19,9,16 E 13.3 1328,6,12,22 F 13.3 2515,14,26,23 G 13.3 37 H 3.2 11


Man # Haplotype % of polymorphismsin old population


1, 8,18 A 10 217,3,8,29 B 13.3 112,4,24,15 C 13.3 812,9,11,23 D 13.3 225,19,27,16 E 13.3 40

14,6,30,22F F 13.3 321,28,26 G 10 4

7,25,10,13H H 13.3 10c) Females having which haplotype (A-H) mated most successfully? Circle one.A B C D E F G H

d) Males having which haplotype (A-H) mated most successfully? Circle one.

A B C D E F G H

e) The small subunit ribosomal RNA gene is often used to infer evolutionary relationships between distantlyrelated organisms because… (Circle one.)

i) It evolves slowly.ii) It evolves rapidly.iii) It is maternally inherited.iv) It is a Y-linked gene.v) It is found in eukaryotes, prokaryotes and viruses.

f) While studying the relationships between isolated populations of giant sea turtles (labeled A-F), you usePCR to obtain the following DNA sequences.

(‘-‘ indicates an identical nucleotide)A: CCGTATCGAAACGTAB: -G-A-CG--------C: -G--C-----T-T--D: -G-A-C--C------E: -G---A----T-T--F: -G-----C--T---C

Which phylogenetic tree is consistent with the sequence data? Circle one.

i) ii) iii) iv)B C F B

E E E EA A A A

F F C C

C B B F

D D D D


SolutionsQuestion 1

AA A short, single-stranded DNA that serves as the necessary starting material for the synthesis ofthe new DNA strand in PCR

DD The synthesis of DNA using DNA as a templateS The building blocks of DNA and RNA

H H The synthesis of protein using information encoded in mRNAQ The location in a eukaryotic cell where the electron transport chain occursW The major component of cell membranesL The genetic composition of an organism

FF A gene that lies on one of the sex chromosomesB B An organism without membrane-bound organellesM A cell with 1n chromosomesB The building blocks of proteinsG A cell with 2n chromosomesD A major source of energy that has the general formula (CH2O)n

T An enzyme needed for completion of lagging strand synthesis, but not leading strand synthesis.G G The synthesis of RNA using one strand of DNA as a templateU An observed characteristic of an organismY A DNA molecule that is distinct from the chromosome; this molecule can be used to move foreign

DNA in or out of a cellE The DNA from a eukaryote formed by the enzyme reverse transcriptase; this DNA lacks intronsO An organism with 2 identical alleles for the same geneK A membrane protein involved in signal transduction; activation involves binding a GTP moleculeJ An organism with genetic material inside a nucleus

N An organism with 2 different alleles for the same geneP A measure of the affinity of an enzyme for its substrateC A gene that lies on any chromosome except the sex chromosomesX The membrane that surrounds the cellA One of the alternate forms of a gene found at a given locus on a chromosomeZ A technique for the rapid production of millions of copies of a particular region of DNAH Proteins with a leader peptide are directed to this cellular organelle


Question 2

a) 5' AAACAGCUAUGG 3'

b)c) Consider each of the following three mutations independently.

i) The mutation is before the start codon so does not change the protein sequence.

ii) The start codon (at nucleotides 25 - 27) would be changed and protein synthesis would nowstart at the next start codon (position 31 - 33). The protein would be shorter by two aminoacids.

iii) This frameshift mutation will result in the protein being terminated prematurely because anew stop codon was created. As a result of this deletion, the new sequence of the peptide wouldbe : H3N+-methionine-alanine-COO-

d) Puromycin functions by entering the A site of the ribosome. Here, because puromycin isstructurally similar to the aminoacyl-tRNA, it can participate in formation of a peptide bond with thenascent polypeptide chain. Puromycin causes peptide release from the ribosome because there is no t-RNA anticodon to link the mRNA to the peptide chain.

Question 3

i) What are the two advantages of using a live-attenuated vaccine vs. a heat killed vaccine in this case?

It’ll mimic the disease by invading cells, thus it will illicit both a humoral and cellular response.

Surface proteins will not be denatured by heat.

ii) What is a disadvantage of using a live-attenuated vaccine?Could acquire virulence factors, Need a “cold chain” (expensive refrigeration), it may make people sick.

b) The rabbit protein is recognized as foreign (non-self) by the guinea pig.

c) B cells, by gene rearrangement of Ab genes (VDJ rearrangement). Also, T cells (byrearrangement of T cell receptor genes).

d) Neither siblings nor identical twins will produce the same antibodies, because the DNArearrangement process that produces the antibody repertoire is a random event in each B cell.Question 4

a) i) + / -

ii) 2/3. B is not affected, therefore must be either +/+ or +/-.+ cf

+ +/+ +/cfcf +/cf cf/cf

5'- ATGGCCTTCACACAGGAAACAGCTATGGCCATGAGCACGC ---------+---------+---------+---------+ 3'- TACCGGAAGTGTGTCCTTTGTCGATACCGGTACTCGTGCG CAGTCTCGGCATTATCCTATTAAAGGGAACTGAGGTGA -3' ---------+---------+---------+-------- GTCAGAGCCGTAATAGGATAATTTCCCTTGACTCCACT -5'


iii) Since both parents are carriers (+/cf), the probability of an affected child is 1/4.

iv) Again, both parents are carriers, the probability of having a child who is a carrier is 1/2.

b) i) Since the signal sequence would be unaffected by the mutation, the protein could be found ineither the Golgi, the ER, or some membrane vesicle.ii) Deletion or mutation of the signal sequence would create a protein which would not betranslated into the ER.

c) i) Since gp19 prevents MHC I display, virus-infected cells which are expressing the wild-type disease

X gene would not be attacked by the cellular immune system.ii) 1) Digest pBR-Ad2-7 with NheI and HindIII.

2) Digest pCMV-disease X with SpeI and HindIII. Isolate the CFTR cDNA fragment.3) Ligate the products of steps 1 and 2.

4) Cut the resulting plasmid with BamHI to obtain a linear fragment with disease X gene at the PE1promoter.Question 5

b) i)

ii)

iii)




b) i) Arg and Lys are both positively charged, thus the ionic interaction with the phosphate groupis preserved. The side chains of both amino acids are also of similar size.ii) Tyr is much larger than Asp. Although Tyr can form a hydrogen bond, GDP will no longerfit into the binding pocket. The Tyr side chain is also much more hydrophobic than the Asp sidechain.

Question 6a) i) This would eliminate the disulfide bridge tethering the ligand-binding domain to the rest of the

protein. The receptor would be inactive.

ii) This would prevent the differentiation of the R7 cell type.

b) Leucine, alanine, isoleucine, valine, phenylalanine, glycine, tryptophan are all hydrophobic aminoacids. The hydrophobic effect causes these amino acids to cluster away from water and stay in theinterior of the plasma membrane.c) i) Ras can activate several different proteins, each of which leads to a different signal

transduction cascade. Different cells express different genes, and the specific protein that agiven cell expresses will determine the outcome of Ras activation.

ii) • You never see diferentiation of R7 cells.A loss-of function mutation in the boss/sev receptor would prevent signaling throughRas and activation of the differentiation pathway.

• You see uncontrolled cell proliferation.A gain-of function mutation in the EGF receptor such that it signals to Ras in theabsence of growth factor would allow uncontrolled cell division.

a)


Question 7a) An autosome, because individual 6, a male, has 2 alleles.b) The mutant phenotype is recessive, because individuals 5 and 6 each have one copy of the mutantallele, m, and are both phenotypically normal.c) 1/4. Since individuals 3 and 4 already have an affected child, then they must both be heterozygotes.d) 5’...GCCAGCTG...3’e) A mutation occurred which led to the production of a new mutant allele, m*. This mutant allele has arecessive phenotype and its PCR product is cut by neither NheI nor PvuII. Individual 9 has thegenotype m/m*.Question 8Below are two possible explanations for these results.

a) Possibility 1: Beard color is controlled by a single locus. Give the genotypes in the blanks below.


Green beards

X

B By b

B By y B Bb b

b) Possibility 2: Beard color is controlled by a pathway of two distinct enzymes encoded bythe A and Q genes.i) Give one genotype in each of the blanks below. Use A and Q to designate the wild- type alleles. Use a and q todesignate the loss-of-function alleles.

B)

A)


Green beards

XAAqqaaQQ

AaQq


Green beards

XAAqq aaQQ

AaQq

ii) Given your answer to i) above, draw the pathway for beard color. Be sure to include at which step each of the genes functions.

gene or enzyme A gene or enzyme QIf A) blue --------------> yellow--------------> green

gene or enzyme Q gene or enzyme AIf B) blue --------------> yellow--------------> green


Question 9a) Autosomal Recessive

b) individual genotype

2 gg

3 Gg

4 Gg

9 Gg

10 GG or Ggc)

genotypes:

hhΩ+Ω+ hhΩ-Ω-

hhΩ+Ω- HHΩ+Ω+

HHΩ-Ω- HHΩ+Ω-

HhΩ+Ω+ HhΩ-Ω-

HhΩ+Ω-

Question 10

a)

i)

ii)

iii)

iv)

v)

glycoprotein

lipid

protein

RNA

Reverse Transcriptase

Host

Viral

Viral

Viral

Viral

b) The gp120 protein on the surface of the HIV envelope binds to the CD4 receptor, and this ligand-receptor binding event is the first step in infection.

c) i)1. Synthesis of single (-) strand of DNA using the single (+) strand RNA genome as the template.2. Degradation of the single (+) strand RNA template.3. Synthesis of the complementary (+) strand of DNA using the single (-) strand of DNA as the template toform double stranded DNA.

ii) 1. RNA-directed DNA polymerase activity.


2. RNase activity.3. DNA-directed DNA polymerase activity.

d) While AZT will inhibit DNA synthesis mediated by reverse transcriptase, it will also inhibit nucleic acidsynthesis in other cells of the body, mediated by DNA polymerases and thus kill cells which are dividing. Allbodily cells that are actively replicating (e.g.: red blood cell precursors, skin cells, etc.) can be affected bythese nucleotide analogs. However, HIV reverse transcriptase will use AZT and thymidine interchangeablywhile host cell DNA polymerase prefers thymidine over AZT. Therefore, AZT may be used atconcentrations that inhibit HIV replication but are not yet toxic to host cell polymerases.

e) It would be effective to target the integration of the double-stranded DNA into the host cell genome.Not only is the integrase enzyme a retroviral-specific enzyme, but developing therapies to disrupt thisenzyme's activity won't likely affect other cellular properties, because we do not possess any enzymes whichhave this type of activity.Additionally, viral protease inhibitors will affect the processing of the viral polyprotein into active proteins.The host proteins are not synthesized as polyproteins.

f) The mutations are most likely to be in the viral gene coding for reverse transcriptase (the pol gene)because mutations in this gene may directly cause the mutated reverse transcriptase not to prefer AZTduring replication and thus become resistant to AZT.

g) Order of steps: B D C A F E G

Question 11a) The m1 and m4 mutations are in the same gene. This is a complementation test. The diploid strainhas the same growth phenotype as the haploid single mutants; the two mutations fail to complement (failto produce the wild-type phenotype) in the double heterozygote. The m1 and m4 mutations must bothinactivate the same gene (which codes for an enzyme essential for serine biosynthesis) so that the diploiddouble mutant has two mutant alleles of the same gene.

b) ABC serinem1, m4 m3 m2

c) i) This haploid mutant will grow on media supplemented with intermediate A.ii) Intermediate C will accumulate when this haploid mutant is grown on minimal medium.


Question 12 Shown below is a schematic representation of an antibody made in B cells.

a)

Light chainLight chain

Heavy chain

Variable regionVariable region

Antigen binding site Antigen binding site

Membrane-spanning region of a membrane-bound antibody

Region of secreted antibody that interacts with phagocytes

Heavy chain

Membrane-spanning region of a membrane-bound antibody

b)i) List two ways that combinatorial joining generates antibody diversity.

1) Association of different light chains and different heavy chains2) Recombination of different V (D) and J segments on the immunoglobulin genes.3) Imprecise joining of nucleotides at the joining sites.

ii) What are the interactions between the light and heavy chains that make them associatewith each other?Disulfide bridges = covalent bonds

c) Name the cell type of the immune system which would:

i) bind an antigen floating around in the blood.B cell

ii) recognize an antigenic peptide on a MHC Class II protein displayed by a macrophage.Helper T cell

iii) recognize an antigenic peptide on a MHC Class I protein displayed by a skin cell.Killer T cell

iv) nonspecifically engulf and digest a variety of pathogens (disease-causing organisms).Macrophage


v) secrete large amounts of antibody in response to an infection.Plasma B cell

vi) recognize a viral infected body cell and destroy it.Cytotoxic (killer) T cell

vii) Children immunized against measles virus by injection of viral proteins have long-livedimmunity to measles due to this cell type.Memory cell.

d) Will an individual who does not express MHC Class II proteins on the surfaces ofB cells and macrophages produce a humoral immune response?Without MHC class II protein expression, antigenic peptides cannot be presented to a T helper cell. Asa result, the T helper cells are never activated and neither are the B cells.

e) Will siblings have the exact same antibody repertoire? What about identical twins? Why orwhy not?Neither siblings nor identical twins will produce the same antibodies, because the DNA rearrangementprocess which produces the antibody repertoire is a random event in each B cell.

;Question 13Question 1In 1930, 30 astronaut couples were secretly sent to Titan, where no humans previously resided.However their spacecraft was damaged on landing and they could not return. It took more then 70years to make another spacecraft and reach the newly formed human population on Titan.

The new population was analyzed for some traits and it was compared to the population on earth.Blood samples had been taken of each of the 60 people (astronauts) so the frequencies of these traits inthe Titan colonizers are known, as well as those of current Titan population. See data below.

Trait Earth in 1930 Earth 2004 Astronauts Titan 2004Tongue rolling 0.7 0.7 0.5 0.1O blood type 0.05 0.05 0.1 0.3

a) The frequencies of tongue rolling and blood type traits of the astronauts were different from those of the earthpopulation in 1930. Which of the following phenomena best explains this?

a) genetic drift

b) founder effect

c) mutation


b) Trait frequencies of Titan today are different from those of the astronauts. Given that there was no selectivepressure in the planet for any trait, which of the following is the best reason of these changes?

a) genetic drift

b) founder effect

c) migration into the population



Blood samples from all 30 couples were taken prior to their departure from earth. You sequence a specific region of eachwoman’s mitochondrial DNA or man’s Y chromosome. All were grouped into haplotypes. The frequency of haplotypes werecompared to the current population on Titan.

Woman # Haplotype % of polymorphismsin old population


1,29,18,10,13 A 17 217,3,8,20 B 13.3 102,5,24,21 C 13.3 530,27,11,25 D 13.3 304,19,9,16 E 13.3 1328,6,12,22 F 13.3 2515,14,26,23 G 13.3 37 H 3.2 11

Man # Haplotype % of polymorphismsin old population


1, 8,18 A 10 217,3,8,29 B 13.3 112,4,24,15 C 13.3 812,9,11,23 D 13.3 225,19,27,16 E 13.3 40

14,6,30,22F F 13.3 321,28,26 G 10 4

7,25,10,13H H 13.3 10c) Females having which haplotype (A-H) mated most successfully? Circle one.A B C D E F G H

d) Males having which haplotype (A-H) mated most successfully? Circle one.A B C D E F G H

e) The small subunit ribosomal RNA gene is often used to infer evolutionary relationships between distantlyrelated organisms because… (Circle one.)

i) It evolves slowly.ii) It evolves rapidly.iii) It is maternally inherited.iv) It is a Y-linked gene.v) It is found in eukaryotes, prokaryotes and viruses.

f) While studying the relationships between isolated populations of giant sea turtles (labeled A-F), you usePCR to obtain the following DNA sequences.

(‘-‘ indicates an identical nucleotide)A: CCGTATCGAAACGTAB: -G-A-CG--------C: -G--C-----T-T--D: -G-A-C--C------E: -G---A----T-T--F: -G-----C--T---C

Which phylogenetic tree is consistent with the sequence data? Circle one. THE SECOND ONE. U C A G

U UUU phe (F)UUC phe (F)UUA leu (L)UUG leu (L)

UCU ser (S)UCC ser (S)UCA ser (S)UCG ser (S)

UAU tyr (Y)UAC tyr (Y)UAA STOPUAG STOP

UGU cys (C)UGC cys (C)UGA STOPUGG trp (W)

UCAG

C CUU leu (L)CUC leu (L)CUA leu (L)CUG leu (L)

CCU pro (P)CCC pro (P)CCA pro (P)CCG pro (P)

CAU his (H)CAC his (H)CAA gln (Q)CAG gln (Q)

CGU arg (R)CGC arg (R)CGA arg (R)CGG arg (R)

UCAG

A AUU ile (I)AUC ile (I)AUA ile (I)AUG met (M)

ACU thr (T)ACC thr (T)ACA thr (T)ACG thr (T)

AAU asn (N)AAC asn (N)AAA lys (K)AAG lys (K)

AGU ser (S)AGC ser (S)AGA arg (R)AGG arg (R)

UCAG

G GUU val (V)GUC val (V)GUA val (V)GUG val (V)

GCU ala (A)GCC ala (A)GCA ala (A)GCG ala (A)

GAU asp (D)GAC asp (D)GAA glu (E)GAG glu (E)

GGU gly (G)GGC gly (G)GGA gly (G)GGG gly (G)

UCAG


CC

OO

HNH3

CH2CH2CH2CH2 NH3+

ALANINE(ala)

ARGININE(arg)

ASPARAGINE(asn)

+

-

ASPARTIC ACID(asp)

CYSTEINE(cys)

-

+

C

N CH2

CH2

CH2

H

H

H

-C

OO

+

GLUTAMIC ACID(glu)

GLUTAMINE(gln)

GLYCINE(gly)

HISTIDINE(his)

ISOLEUCINE(ile)

LEUCINE(leu)

LYSINE(lys)

METHIONINE(met) PHENYLALANINE

(phe) PROLINE(pro)

SERINE(ser)

THREONINE(thr)

TRYPTOPHAN(trp) TYROSINE

(tyr)

VALINE(val)

STRUCTURES OF AMINO ACIDS at pH 7.0

C

N

H

H

H

H

HH

CC

OO

HNH3

CH3

-

+

CC

OO

HNH3

CH2CH2CH2 NH

CNH2

NH2

-

+

CC

OO

HNH3

CH2 CO

NH2

-

+

CC

OO

HNH3

CH2 CO

O

-

+

CC

OO

HNH3

CH2 SH

-

+

CC

OO

HNH3

CH2CH2O

OC

-

+

CC

OO

HNH3

CH2CH2

OC

NH2

-

+

CC

OO

HNH3

H

-

+

CC

OO

HNH3

CH2

C N

CNH

H

H

H

-

+

CC

OO

HNH3

CH

CH3

CH2CH3

-

+

CC

OO

HNH3

CH2 CH

CH3

CH3

-

+

-

+

CC

OO

HNH3

CH2CH2 S CH3

-

+

CC

OO

HNH3

CH2

H H

H

HH

-

+

CC

OO

HNH3

CH2 OH

-

+

CC

OO

H

NH3

CH

OHCH3

-

+

CC

OO

HNH3

CH2

-

+CC

OO

HNH3

CH2 OH

H H

HH

-

+

CC

OO

H

NH3

CH

CH3

CH3

-

+


List of terms for Question 1.

You may detach this page from the exam.

A. alleleB. amino acidsC. autosomal geneD. carbohydrateE. cDNAF. competitive inhibitorG. diploidH. endoplasmic reticulumJ. eukaryoteK. G proteinL. genotypeM. haploidN. heterozygoteO. homozygoteP. KMQ. mitochondriaR. non-competitive inhibitorS. nucleotidesT. DNA ligaseU. phenotypeW. phospholipidsX. plasma membraneY. plasmidZ. polymerase chain reactionAA. primerBB. prokaryoteCC. DNA polymeraseDD. replicationEE. repressor proteinFF. sex-linked geneGG. transcriptionHH. translation

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Final Exam Practice - Massachusetts Institute of Technology