key for exam 3 • biology ii • winter 2013

Version A p. 1 of 12 12 March 2013

Key for Exam 3 • Biology II • Winter 2013

Multiple Choice Questions. Circle the one best answer for each question. (1 point each)

1. Which of the following is not part of the cell theory:

A. All cells come from other cells.

B. Only eukaryotic cells have membrane-bounded organelles.

C. The cell is the smallest living unit of a living thing.

D. All living things are made up of cells.

2. A protein that belongs in the plasma membrane of a eukaryotic cell might be found (at some point) in

which organelle:

A. Golgi apparatus

B. ribosome

C. cytoplasmic reticulum

D. nucleus

E. mitochondrion

3. A phospholipid is composed of:

A. three fatty acids and a molecule of glycerol

B. chains of hydrophobic amino acids

C. two fatty acids, a molecule of glycerol and a highly hydrophilic group

D. amphipathic fatty acids

E. a phosphate, a ribose or deoxyribose sugar, and a fatty acid

4. A membrane would be more permeable if:

A. …its phospholipids had longer tails.

B. …its phospholipids had shorter tails.

C. …it contained more cholesterol.

D. …its phospholipids had more saturated tails.

E. …its proteins were more hydrophobic.

5. An enzyme is working at its Vmax: A. …at the high point of the product vs. time curve.

B. …when its active site is continuously full of substrate.

C. …when the concentration of substrate is equal to its km.

D. …when it is assayed at its optimum temperature and pH.

E. …at the early time points when its slope is the steepest.

6. In an enzyme-catalyzed reaction, the role of the enzyme is to:

A. …increase the available activation energy.

B. …make an endergonic reaction exergonic.

C. …make an exergonic reaction endergonic.

D. …use ATP energy to run a thermodynamically unfavorable reaction.

E. …speed up the rate of the reaction.


7. Choose the correct citation format for the reference listed below.

Sastri, J., C. O’Connor, C. M. Danielson, M. McRaven, P. Perez, F. Diaz-Griffero, and E. M. Campbell. 2010. Identification of residues within the L2 region of rhesus TRIM5α that are required for retroviral restriction and cytoplasmic body localization. Virol. 405:259-266.

A. Campbell, et al., 2010

B. Sastri, et al., 2010

C. Sastri, J., et al., 2010

D. Sastri and Campbell, 2010

E. O’Connor, et al., 2010

8. Which of the following is not true about a recessive genetic trait?

A. Only homozygous recessive individuals show the trait.

B. One allele must be inherited from each parent in order to show the trait.

C. The allele responsible for the trait is usually rare.

D. The allele responsible for the trait usually encodes a non-functional protein.

E. It is possible to have the allele associated with the trait but not show the trait.

9. A possible genetic symbol for an allele that produces red pigment and shows incomplete dominance

would be:

A. CRC

W

B. AQ

C. R

D. a filled circle or square

E. Cr

10. The number of alleles of one gene that can be found in one individual is:

A. 2

B. 1

C. 1 or 2

D. 3

E. limited only by the number of nucleotides that might be mutated

11. Whose experiment provided evidence that the genetic material of a virus is composed of DNA?

A. Griffiths

B. Avery, McCarty and MacLeod

C. Hershey and Chase

D. Watson and Crick

E. Meselson and Stahl

12. Which of the following can add nucleotides to a 5′ end during replication?

A. primase

B. DNA polymerase I

C. DNA polymerase III

D. RNA polymerase II

E. none of the above


13. A mutation converts a T nucleotide within a coding sequence to an A. This mutation is a:

A. …missense mutation.

B. …nonsense mutation.

C. …frameshift mutation.

D. …silent mutation.

E. …there is not enough information to decide.

14. Meiosis I separates:

A. …homologous pairs of chromosomes.

B. …sister chromatids.

C. …the X chromosome from the Y chromosome.

D. …gametes from somatic cells.

E. …the two strands of a DNA molecule.

15. A sex-linked gene is located:

A. …on the same chromosome as another gene.

B. …on the Y chromosome.

C. …on the X chromosome.

D. …on the X or Y chromosome.

E. …in the gamete cells.

16. When an E. coli cell is growing in broth containing glucose and lactose:

A. …the lac repressor and the CAP protein are both bound to DNA.

B. …the CAP protein is bound to DNA, but not the lac repressor.

C. …cAMP is bound to the CAP protein.

D. …lactose is bound to the lac repressor.

E. both A and D.

17. The light reactions of photosynthesis produce:

A. …glucose.

B. …ATP and NADPH.

C. …ATP and NADP+.

D. …three-carbon carbohydrates.

E. …water.

18. We know that a molecule is being oxidized at a particular step in a biochemical pathway when:

A. …a molecule of ATP is produced.

B. …a molecule of NAD+ is produced.

C. …a molecule of FADH2 is produced.

D. …a molecule of pyruvate is produced.

E. …a proton is pumped out by the electron transport system.


19. When a yeast cell ferments pyruvate to form ethanol and CO2, what important purpose does this step

serve?

A. increases the amount of ATP that can be produced from glucose

B. produces the alcohol used for energy generation by yeast cells

C. produces oxidized forms of electron carriers

D. produces ethanol that can be further broken down in the citric acid cycle

E. prevents buildup of toxic pyruvate

20. Adrenaline is a peptide hormone. We would expect to find its receptor:

A. inserted into the cell membrane of target cells

B. in the cytoplasm of target cells

C. in the nucleus of target cells

D. in the cytoplasm of both target and non-target cells

E. bound to the DNA of target cells

21. BLAST is a program that can:

A. …determine the DNA sequence of a gene whose function is unknown.

B. …identify diseases that may be associated with a gene whose function is unknown.

C. …find genes in a database with sequences similar to that of a gene whose function is unknown.

D. …determine the biochemical function of a gene whose function is unknown.

E. …waste the time of students in an introductory biology course.

22. Which of the following enzymes would not be used in cloning a gene by one of the methods described

in class?

A. a DNA polymerase

B. an RNA polymerase

C. reverse transcriptase

D. DNA ligase

E. a restriction endonuclease

23. In order to express a human gene in a bacterial cell, we would have to provide:

A. a Shine-Dalgarno sequence

B. a TATA box

C. introns

D. a 5′ cap

E. human ribosomes

24. A yeast spore consists of:

A. …two diploid cells in the process of mitosis (budding).

B. …four diploid cells surrounded by an ascus.

C. …one resistant cell surrounded by a protective coat.

D. …several yeast cells gathered together for protection.

E. …four haploid cells surrounded by an ascus.


25. Calculating the standard deviations for two sets of measurements can tell you:

A. …how close they are to the actual values.

B. …whether there is a significant difference between the two sets of data.

C. …whether your hypothesis is supported by the data.

D. …how accurate your measurements were.

E. …whether the average measurement is valid or not.

26. When you assayed tyrosinase in potato extracts, you were directly measuring:

A. …absorbance of tyrosinase.

B. …activity of tyrosinase.

C. …concentration of tyrosinase.

D. …production of a quinone.

E. …absorbance of pyrocatechol.

27. When you performed the Ames test on the chemical you wanted to screen as a possible carcinogen,

the positive control was:

A. …the number of colonies on minimal medium plates in the absence of the chemical.

B. …the number of colonies formed on LB plates.

C. …the number of colonies on minimal medium plates for cells treated with sodium azide.

D. …the number of colonies on minimal medium plates when you used the highest concentration.

E. …the number of cells originally present in each spot on the minimal medium plates.

28. In gel electrophoresis, DNA moves toward:

A. …the positive electrode.

B. …the negative electrode.

C. …the smallest fragments.

D. …the DNA ladder.

E. …the dark blue tracking dye.


29. How does each factor below participate in the process of gene expression? Circle all of the choices

that are correct. (1 point for each factor)

a. Promoter Organism: Eukarote Prokaryote

Composed of: DNA RNA Protein

Process: Transcription Translation

b. -10 and -35 Organism: Eukarote Prokaryote



c. Ribosome Organism: Eukarote Prokaryote



d. Stop codon Organism: Eukarote Prokaryote



e. Operator Organism: Eukarote Prokaryote



f. trp Repressor Organism: Eukarote Prokaryote



g. CAP Organism: Eukarote Prokaryote



h. Estrogen receptor Organism: Eukarote Prokaryote




Short Answer Questions. Please be sure to read the questions carefully and be sure you have answered

what is being asked. Think before you write! The best answers are brief and to the point. For genetics

problems, show your work or thinking, diagram crosses (you do not necessarily have to show a

Punnett square) and be sure to define symbols.

30. When you eat fats, the fatty acids are broken up into two-carbon units that get converted to acetyl

CoA. If you have a fatty acid 18 carbons long, how much ATP could be generated by oxidizing it

completely? (4 pts).

18 carbons = 9 two-carbon units = 9 acetyl CoA. In the citric acid cycle, each acetyl CoA results in

production of 1 GTP = 1 ATP, 3 NADH = 3 ATP each = 9 ATP and 1 FADH2 = 2 ATP. So, 12 ATP

per acetyl CoA, or 108 ATP from just one fatty acid.

31. For each of the following pairs of compounds, draw a circle around the most oxidized form of the

molecule. Then, draw a square around the one molecule (of the eight) with the most potential energy

stored (1 pt each).

NADH NAD+

Glucose Pyruvate

FADH2 FAD

CO2 CH4

32. E. cocacoli is an imaginary bacterium that can live on Diet Coke®, using aspartame as its source of

energy. The diagrams below show the aspartame operon. Add sites, proteins and additional genes to

the diagrams as necessary to show how this operon might be regulated. (5 points)


33. Rotenone is an insecticide. It kills insects by acting on coenzyme Q (CoQ) so that it can’t deliver

electrons from Complex I to Complex III. It’s not very toxic to humans because it is very poorly

absorbed through the skin or digestive tract.

a. When you treat an insect with rotenone, you immediately observe that its cells greatly slow (but

do not stop) their rate of ATP production. Why does this happen, and where is the ATP that is

being made now coming from? (2 points)

The Rotenone stops transfer of electrons in the electron transport chain, so no more ATP can be

made by ATP synthase. However, as long as some glucose is present, a little ATP can be made by

glycolysis alone.

b. After a few minutes, the insect’s cells completely stop making any ATP. What has happened? (2

points)

Once the cell runs out of oxidized electron carriers (NAD+), it won’t be able to do glycolysis

anymore.

c. After the insect dies, the enzymes in its cells remain functional for a while. You suddenly get a

brilliant idea (since you understand the process of respiration very well) and try putting some of

the insect’s cells in a low-pH solution (assume for the purposes of this problem that the solution

will be able to change the pH of the cytoplasm). You know that the cells must have lots of ADP,

and sure enough, in the low-pH solution, you detect a slight decrease in the amount of ADP and a

slight increase in the amount of ATP. What is going on? (2 points)

The actual process of ATP synthesis (chemiosmosis) in the mitochondrial membrane is

accomplished by having a high concentration of protons (H+) outside the mitochondrion and

letting them in through the ATP synthase, which harnesses their energy to make ATP. Lowering

the pH of the cytoplasm means increasing the concentration of H+, and even though this is

artificial, those H+ can enter through ATP synthase, leading to generation of some ATP.

34. In fruit flies, a gene affecting eye shape is located on the X chromosome. The wild-type allele of this

gene produces round eyes, and a second allele produces narrow eyes. However, this gene shows

incomplete dominance, so a heterozygous fly has kidney-shaped eyes. A second gene, this one

autosomal, affects eye color; the recessive allele produces scarlet (bright red) eyes instead of normal

red eyes.

a. Give appropriate symbols for the alleles of these genes. (2 points)

XEN = narrow eyes (must show X chromosome and use a superscript on the gene name to show

incomplete dominance)

XER = round eyes

s = scarlet eyes

S = red eyes

b. A male with round, scarlet eyes is mated with a kidney-eyed female. The female has red eyes, but

her mother had scarlet eyes. If these flies have 300 offspring (not at all unlikely for fruit flies!),

how many will be males with narrow, scarlet eyes? (3 points)

For eye shape, the male is XER Y and the female is X

ER XEN . So, the male offspring’s eye shapes

will be determined by which allele they get from the female, half round and half narrow. For eye

color, the male is ss and the female is Ss (because her mother had scarlet eyes). So, half the

offspring (males and females) will be Ss and red, and half ss and scarlet. So, if half the offspring

are males, half the males have narrow eyes and half the offspring have scarlet eyes, ½ × ½ × ½ =

1/8 males with narrow, scarlet eyes.


35. The jack jumper ant has the smallest known number of chromosomes: for this organism, n=2. In the

box labeled (A), draw a cell from a jack jumper ant as it would appear during metaphase of mitosis,

then in the box labeled (B), draw a germ cell as it would appear during anaphase I. (2 points each)

A B

36. You are writing a new textbook for introductory biology. When you get to the chapter on DNA

replication, the publisher’s artist produces the beautiful figure below. Unfortunately, it is inaccurate in

two important details. Please correct the figure so that it won’t mislead your future students. Hint:

you’re looking for inaccuracies, not things that might be missing. (3 points)

37. Glucose can be used by almost every organism on earth, because it is such a high-energy molecule.

a. At the end of glycolysis, where is the energy that was originally in the glucose molecule? Be sure

to list all of the places it would be. (2 points)

pyruvate, ATP (2), electrons on NADH (2), and lost as heat

b. After two rounds of the citric acid cycle, where is the energy that was originally in the glucose

molecule? Be sure to list all of the places it would be. (2 points)

ATP (4), electrons on NADH (10), electrons on FADH2 (2), CO2 and lost as heat


38. Hunter syndrome (sorry, Hunter! ☺) is a human genetic disease in which the body fails to properly

break down and recycle a specific kind of polysaccharide known as glycosaminoglycans (GAG). This

disease affects the head and facial features and results in developmental delay and progressive nervous

system disorders.

a. Mutations that cause Hunter syndrome occur in the gene for an enzyme called I2S. Where would

you expect to find this enzyme in a normal cell, and

why? (2 points)

In lysosomes, because that’s where materials are

broken down and recycled.

b. The pedigree at right is for a family where Hunter

syndrome occurs. Is this mutant allele that produces

this disease dominant or recessive? Give specific

evidence to support your answer. (2 points)

It is recessive, because there is an affected child (IV-3) of two unaffected parents (III-3 and III-4).

c. Is the gene for I2S autosomal or sex-linked? Give specific evidence to support your answer. You

may assume that this disease is rare. (2 points)

If it were autosomal, then I-1, II-2, III-4 and III-6 all have to be unrelated individuals who have at

least one allele for this rare disease, which is pretty unlikely. If it’s sex-linked, as shown by the

genotypes above, then the disease allele is introduced only by the obviously affected I-I and III-6,

and the observed pattern of mother-to-son but no father-to-son transmission makes sense.

d. Individual IV-1 plans to marry a man who does not have a Hunter syndrome allele. What is the

probability that they will have a child with Hunter syndrome? (2 points)

IV-1 must have inherited one Hunter allele from her father. This means her sons have a 50-50

chance of getting the disease. Her daughters, however, will not get it because they’ll get a

dominant allele on the X chromosome from the husband.

39. Prednisone (shown at right) is a synthetic hormone widely used to treat

inflammation in diseases ranging from arthritis to asthma to Crohn’s disease. It

suppresses the immune system by turning on expression of genes for

regulatory proteins called lipocortins. Draw a clear diagram to show how

prednisone would act on a target cell to turn on expression of a lipocortin

gene. (4 points)


40. Individuals with glycogen storage disease (GSD) are impaired in the ability to store glucose as

glycogen. The sequence below represents part of a gene involved in GSD, starting at the +1

nucleotide. The top sequence is the normal gene; the bottom one is an allele from a GSD patient.

Normal: GCACCGGAACTTGCTACTACCAGCACCATGCCCTACCAATATCCAGCACTGACCCCGGAG

GSD: GCACCGGAACTTGCTACTACCAGCACCATGCCCTACCAATAGCCAGCACTGACCCCGGAG

a. Patients with GSD tend to be hypoglycemic, which means their blood sugar is abnormally low.

Can you explain why? (2 points)

When blood sugar drops, normally glucagon is released to stimulate the liver to break down

glycogen. GSD patients would be deficient in glycogen so would not be able to raise blood sugar

back to normal levels easily.

b. What are the first four amino acids of the protein encoded by this gene? (Use the three-letter

code.) (1 point)

Met-Pro-Tyr-Gln

c. What mutation has occurred in the GSD patient’s DNA? (1 point)

A T has been mutated to a G at the +42 position.

d. How would we classify this mutation in terms of its effect on the DNA? (1 point)

Substitution.

e. How would we classify this mutation in terms of its effect on the protein? Be as specific as

possible. (2 points)

This mutation changes a TAT codon, encoding tyrosine (Tyr) to a TAG stop codon. This is a

nonsense mutation which will halt translation at this point.

f. Would you expect the GSD patient’s allele to be a dominant or recessive allele? Explain. (1 point)

This mutation will certainly produce a non-functional protein (only four amino acids long!), and

non-functional alleles are usually recessive.

41. The sequence below is the first part of an mRNA from a yeast cell, starting with the +1 nucleotide:

AUGUCGAGCAUCCCAUGGACCAUGAUAGGAGGUCCACAUGGUCCA

a. There are four potential start codons. Circle the one that will actually be used and tell how you

know that this is the correct one. (2 points)

Eukaryotic cells use the first start codon from the 5′ end of the mRNA. The very first AUG,

though, can’t be used because there is nowhere for the ribosome to bind. So, in this case, the

second AUG, in blue above, would be chosen.

b. Suppose this gene were cloned into a bacterial cell and correctly transcribed to produce this same

mRNA. Would the same protein be produced? Why or why not? (2 points)

No. Bacterial cells don’t look for the first AUG but rather for a Shine-Dalgarno sequence

(underlined above) followed by an AUG (red above). Translation would thus start at a different

point and (especially since it’s out of frame) produce a different protein.


42. On the graph shown below, show where we can find the Vmax and Km for this enzyme. If you can’t

figure out one or both of these, make the changes to the graph necessary to find these values. (4 points)

This graph is a rate graph, not a Michaelis-Menten graph.

Therefore in order to get the values you would need to have

the x-axis labeled as substrate concentration, and the y-axis

labeled as Initial Velocity. Then you could find the Vmax and

Km.

43. For the following examples involving aerobic or anaerobic respiration, match the correct process with

the example. There can only be one correct answer. Each answer in the right column can be used

more than once. (1 point each)

C Phosphofructokinase phosphorylating fructose-6-P A. Citric Acid Cycle

B Pyruvate reduced to lactic acid B. Fermentation

A Last carbon of glucose oxidized to CO2 C. Glycolysis

C Phosphorylation of glucose D. beta-oxidation

A Acetyl-CoA added to oxaloacetate E. Electron transport

E Electrons transferred from NADH to Cytochrome complex. F. Gluconeogenesis

44. You decide to develop a gene therapy reagent to make brussel sprouts tasty for everyone! Knowing

that your father LOVES brussel sprouts, and that this is linked to the Glucosinolate Isomerase

(GluIso) gene. The GluIso gene is flanked by two BamHI restriction enzyme sites.

a. Describe two viruses that could be used for this gene therapy, and potential risks associated with

each virus? (4 points)

Adenovirus- can cause massive inflammation in host, if you inject it into the blood stream.

Retrovirus- can randomly insert into host genome. If you insert into a gene that controls cell

division, you could cause that cell to become cancerous.

b. How would you clone your father’s gene into a plasmid that contains a multicloning site with the

following restriction enzyme sites: EcoRI, SacI, BamHI, XhoII, BglIII? Describe how you will get

the insert (the GluIso gene), and how you will put that insert into your plasmid. (5 points)

You would first need to isolate the gene via PCR of the genome or from cDNA copies of mRNA

from your father. It doesn’t matter which source since the ultimate target is eukaryote cells and they

can handle splicing the introns out. Once you have created a lot of copies of your insert, cut both

insert and plasmid with BamHI, and mix the two DNA molecules together. Then, you would add

ligase to this mixture to ensure the insert forms phosphodiester bonds with the plasmid DNA. You

can transform this plasmid into bacteria, grow it up and isolate the plasmid DNA. Then you would

cut the DNA with BamHI to see if the insert is present.

key for exam 3 • biology ii • winter 2013

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