exam 4- 05:22:13

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BIO 10100 Exam 4 Version 1 Prior to completing this exam, please read the following statement and sign below to indicate that you have read and understood the statement. No exam will be graded unless the pledge is signed. “I pledge that I have neither given nor received unauthorized assistance on this exam. I understand that if I am charged with cheating, my name will be submitted to the Dean for disciplinary action.” __________________________________________________ _____________________ Name Date Please make sure you have all 14 pages of the exam. Please read each question carefully. If you do not understand the meaning or intention of a question, please ask. Please put away your cell phones, blackberries, iPods, laptop computers, etc., otherwise you will receive a grade of 0. Please separate the last sheet from the rest of the exam. Write the answer to only the last question on the last sheet. The last sheet is graded separately from the rest of the exam. Note that a correct statement is not necessarily the best answer to a multiple-choice question. Each multiple-choice question is worth 2 points. If a question states “Select all that apply” then select all of the correct choices. For each multiple-choice question write the letter for the best answer in the space provided. Note: circling your answer, a , or an x won’t count. You will receive ZERO credit for any question for which we cannot tell what your final answer is. If you cross out an answer, please be clear as to what your final answer is. Please pace yourself and use your time efficiently; allow time for the essay type questions. If a question states “Answer either a or b, not both”: if you answer both, then we will read only the answer to a, regardless of how much better your answer to b might be. Write your answers using a pen. Answers written using pencils will not be graded. 2 Bonus Points: Write your name and lab section on all sheets. 1. The nucleotide sequence of a DNA piece being transcribed is

5’GGATGAGGCGGTGTTGATCCTATA3’ 3’CCTACTCCGCCACAACTAGGATAT5’ A messenger RNA molecule with complementary codons is transcribed from the template strand of this piece of DNA. a. 1 point. Which strand is the template strand?

a. Top b. Bottom Answer: b

b. 3 points. What is the nucleotide sequence of the mRNA produced from the template strand from start to stop? Only include the sequence from the start codon to the stop codon. Do not include the upstream and downstream sequences (before the start codon

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or after the stop codon). Regardless of which direction the template strand is, write the messenger RNA below 5’ to 3’ from left to right. 5’AUGAGGCGGUGUUGA3’

c. 3 points. Write the Amino Acid sequence of the polypeptide that would be produced from this mRNA if the ribosome starts translating from the start codon. Met-Arg-Arg-Cys

Consider different types of point mutations: silent, nonsense, missense, and frame-shift.

d. 4 points. Pick one of the codons in the mRNA you wrote above. Write the simplest mutation in the original DNA sequence that would turn this mRNA codon to a stop codon. Original mRNA codon: example: UGU DNA sequence from ACA to ACT This type of mutation is called: nonsense

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BIO 10100 Exam 4 Version 1 Prior to completing this exam, please read the following statement and sign below to indicate that you have read and understood the statement. No exam will be graded unless the pledge is signed. “I pledge that I have neither given nor received unauthorized assistance on this exam. I understand that if I am charged with cheating, my name will be submitted to the Dean for disciplinary action.”

__________________________________________________ _____________________

Name Date Please make sure you have all 12 pages of the exam. Please read each question carefully. If you do not understand the meaning or intention of a question, please ask. Please pace yourself and use your time efficiently. Please put away your cell phones, blackberries, iPods, laptop computers, etc. Write your answers using a pen. Answers written in pencil will not be accepted. Each multiple-choice or True/False question is worth 2 points. Note that a correct statement is not necessarily the best answer to a multiple-choice question. For each multiple-choice question, circle the best answer. You will receive ZERO credit for any question for which we cannot tell what your final answer is. If you cross out an answer, please make it clear as to which answer is your final answer. Please pace yourself and use your time efficiently; allow time for the essay type questions. If a question states “answer either a or b, not both”: if you answer both, then we will read only the answer to a, regardless of how much better your answer to b might be. 2 points: Write your name and lab section on all sheets. 1. A new DNA strand elongates only in the 5' to 3' direction because

a) DNA polymerase begins adding nucleotides at the 5' end of the template. b) Okazaki fragments prevent elongation in the 3' to 5' direction. c) the polarity of the DNA molecule prevents addition of nucleotides at the 3' end. d) replication must progress toward the replication fork. e) DNA polymerase can only add nucleotides to the free 3' end.

2. The leading and the lagging strands differ in that a) the leading strand is synthesized in the same direction as the movement of the replication

fork, and the lagging strand is synthesized in the opposite direction. b) the leading strand is synthesized by adding nucleotides to the 3' end of the growing

strand, and the lagging strand is synthesized by adding nucleotides to the 5' end. c) the lagging strand is synthesized continuously, whereas the leading strand is synthesized

in short fragments that are ultimately stitched together. d) the leading strand is synthesized at twice the rate of the lagging strand.

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3. What is the role of DNA ligase in the elongation of the lagging strand during DNA

replication? a) synthesize RNA nucleotides to make a primer b) catalyze the lengthening of telomeres c) join Okazaki fragments together d) unwind the parental double helix e) stabilize the unwound parental DNA

4. The start codon also codes for the amino acid methionine (Met). But not all polypeptides have methionine as their first amino acid. That is possible because a) some mRNAs have a start codon other than the AUG b) the first methionine (Met) may be cleaved later in processing c) other polypeptides are added to the beginning later in processing d) the methionine is folded in the 3-D structure so it’s not considered “first”

5. The DNA of telomeres has been found to be highly conserved throughout the evolution of

eukaryotes. What does this most probably reflect? a) the inactivity of this DNA b) the low frequency of mutations occurring in this DNA c) that new evolution of telomeres continues d) that mutations in telomeres are relatively advantageous e) that the critical function of telomeres must be maintained

6. The nucleotide sequence of a DNA piece being transcribed is

5’…CCTTTAAGATATC CTCC TCATAAT…3’ 3’…GGAAATTCTATAGGAGGAGTATTA…5’ While both strands could potentially code for a messenger RNA, only one strand from this particular piece of DNA will code for a messenger RNA that will lead to the synthesis of a polypeptide. An mRNA molecule that will code for a polypeptide is transcribed from this piece of DNA. Note that the mRNA has to be read from 5’ to 3’. Also note that the mRNA

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is always longer than the sequence that is translated; it has untranslated regions on both sides of the translated region. a) 4 points. Which DNA strand, the top or the bottom, from this piece of DNA is the coding

strand for the mRNA? How do you know? The top strand, because it will code for a start codon (TAC in DNA, corresponding to AUG in mRNA).

b) 4 points. What is the nucleotide sequence of the mRNA corresponding to the sequence of the coding strand of DNA? Regardless of whether the top or the bottom strand is used, the mRNA is written 5’ to 3’. 5’ AUUAUGAGGAGGAUAUCUUAAAGG 3’ Start

c) 2 points. Mark the start of the reading frame (the start codon) on the mRNA. d) 4 points. Write out the Amino Acid sequence of the polypeptide that would be produced

from this mRNA if the ribosome starts translating from the start codon. Met-Arg-Arg-Ile-Ser (stop)

7. In humans there is one gene that controls the curliness of hair: the individuals with straight hair are homozygous for the straight hair allele (SS), the individuals with curly hair are homozygous for the curly hair allele (ss), and individuals with wavy hair are heterozygous for the two alleles (Ss). Suppose a second gene can lead to absence (dominant, L) or presence (recessive, l) of hair. The two genes are on separate chromosomes. Consider the cross shown below. Write the genotype of the children in the F1 generation. What is the phenotype of the F1 generation? Use a Punnett square to figure out the genotypes and the

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phenotypic ratios (the proportion of children with a given phenotype) in the F2 generation. Show your work. P genotypes: SSll X ssLL 1 points: F1 genotype: SsLl 2 points: F1 phenotype: No hair (bald)

Gametes

SL

Sl

sL

sl

SL

SSLL Bald

SSLl Bald

SsLL Bald

SsLl Bald

Sl

SSLl Bald

SSll Straight hair

SsLl Bald

Ssll Wavy hair

sL

SsLL Bald

SsLl Bald

ssLL Bald

ssLl Bald

sl

SsLl Bald

Ssll Wavy hair

ssLl Bald

ssll Curly hair

8 points: Punnett Square to show the genotypes of the F2 generation 3 points: Phenotypic ratio of F2 generation 12 Bald : 2 Wavy: 1 Straight: 1 Curly

8. 12 points. Answer either a or b, not both.

a) Hershey and Chase wanted to find out what the genetic material was that bacteriophage viruses used when they infected bacteria. To answer this question, they carried out two experiments. They cultured the phages with either radioactive sulfur (Exp 1) or radioactive phosphorus (Exp 2). They mixed the labeled phages with bacteria, allowed the phages to infect the bacterial cells, then agitated the mixture in a blender to separate what was left outside the bacteria cells from the bacteria cells, then centrifuged the mixture so that bacteria formed a pellet at the bottom, and the parts of the phages that had not entered the bacteria cells were in the supernatant, then measured the radioactivity in the pellet and the liquid. What are the alternative hypotheses being tested? What are the experimental predictions based on these hypotheses?

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Exp 1

Exp 2

Alternative hypothesis 1:

DNA is the genetic material used by the bacteriophage viruses.


Proteins are the genetic material used by the bacteriophage viruses.

Experimental prediction for alternative hypothesis 1 for experiment 1:

If the proteins are labeled with radioactive sulfur, and the phages infect bacterial cells and then the cells are separated from what is left outside through agitation, and the cells are collected in a pellet through centrifugation, in the final extraction the radioactive sulfur ends in the supernatant.


If the DNA is labeled with radioactive phosphorus, and the phages infect bacterial cells and then the cells are separated from what is left outside through agitation, and the cells are collected in a pellet through centrifugation, in the final extraction the radioactive phosphorus ends in the pellet.


If the proteins are labeled with radioactive sulfur, and the phages infect bacterial cells and then the cells are separated from what is left outside through agitation, and the cells are collected in a pellet through centrifugation, in the final extraction the radioactive sulfur ends in the pellet.

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Experimental prediction for alternative hypothesis 2 for experiment 2

If the DNA is labeled with radioactive phosphorus, and the phages infect bacterial cells and then the cells are separated from what is left outside through agitation, and the cells are collected in a pellet through centrifugation, in the final extraction the radioactive phosphorus ends in the supernatant.

b) Meselson and Stahl wanted to find out whether the conservative, the semi-conservative, or the dispersive models are the correct way to explain how DNA is replicated. To answer this question, they cultured E. coli bacteria for many generations in a medium containing nucleotides labeled with a heavy isotope of nitrogen. They then transferred the bacteria to a medium with only the light isotope of nitrogen. Two DNA samples were taken from this culture: one at 20 minutes after one round of replication by the bacteria, and a second one at 40 minutes after a second round of replication by the bacteria. They extracted DNA from the bacteria and centrifuged it and could distinguish the density of the extracted DNA through this method. For each of the 3 alternative hypotheses, describe the experimental predictions after both the first and the second rounds of replication. Explain which model is supported by the data.

Experimental predictions for Conservative model:

If bacteria are grown in heavy N for many generations and then allowed to grow in light N for just one generation, then half of the daughter cells will have heavy N DNA (parental) and half of the daughter cells will have light N DNA (all new), leading to 2 bands in the centrifuge tubes. If the cells are allowed to divide for a second time, since the original heavy N DNA would always remain heavy and all of the new DNA would be light, after the second division, following centrifugation again there will be 2 bands in the centrifuge tube.

Semiconservative model:

If bacteria are grown in heavy N for many generations and then allowed to grow in light N for just one generation, then all of the daughter cells will have DNA that has one strand made of heavy nucleotides and one strand made of light nucleotides, leading to 1 band in the centrifuge tube. However, if the bacteria are grown for a second generation, then half of the daughter cells will have DNA made of only light DNA while half will have DNA that has one strand made of heavy nucleotides and one strand made of light nucleotides, leading to 2 bands in the centrifuge tube.

Dispersive model:

If bacteria are grown in heavy N for many generations and then allowed to grow in light N for just one generation, then all of the daughter cells will have DNA molecules made of a mixture of heavy and light nucleotides leading to one band in the centrifuge tube. Every generation in the light N would lead to lighter and lighter DNA, but always one band in the centrifuge tube.

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Correct model? Why?

The results of the first replication eliminate the conservative model (2 bands predicted, only one observed), and the results of the second replication eliminate the dispersive model (one band predicted, but two observed). The semi-conservative model is supported (predictions match observations).

9. True/False. A frequency of recombination of 50% for two genes indicates that the two genes must be on separate chromosomes. They could be on the same chromosome but really far apart.

10. A diploid cell (2N) is going through meiosis. During anaphase 2 there is non-disjunction for one chromosome in one cell. What are the ploidies of the resulting gametes? a) N, N, N+1, N-1 b) N+1, N+1, N-1, N-1 c) N, N, N+1, N+1 d) N, N, N-1, N-1 e) N, N, N, N-1 f) N, N, N, N+1

11. A diploid cell (2N) is going through meiosis. During anaphase 1 there is non-disjunction for

one chromosome. What are the ploidies of the resulting gametes?

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a) N, N, N+1, N-1 b) N+1, N+1, N-1, N-1 c) N, N, N+1, N+1 d) N, N, N-1, N-1 e) N, N, N, N-1 f) N, N, N, N+1

12. In humans, XX = female and XY = male. If gender determination is based on chromosomal

genes alone, then what would be the gender of the following humans: XO, XXX, XXY, and XYY (see next page) a) Female, Female, Male, Male b) Female, Female, Hermaphrodite, Male c) Hermaphrodite, Female, Hermaphrodite, Male d) Hermaphrodite, Female, Male, Male

13. Red-green color blindness is a sex-linked recessive trait in humans: Xn is the normal allele,

Xc is the color blind allele. A man and a woman with normal color vision have a color-blind son. What are the genotypes of the parents? a) XnXn and XcY b) XnXc and XnY c) XnXc and XcY d) XnXn and XnY e) XcXc and XcY f) XcXc and XnY

14. The following is a map of four genes on a chromosome. The units are CentiMorgans.

Between which two genes would you expect the lowest frequency of recombination?

a) A and G b) A and W c) W and E d) E and G e) A and E

15. Hydrangea plants of the same genotype are planted in a large flower garden. Some of the plants produce blue flowers and others pink flowers. This can be best explained by which of the following? a) The allele for blue hydrangea being completely dominant b) The alleles being codominant c) Environmental factors such as soil pH d) The fact that a mutation has occurred e) Acknowledging that multiple alleles are involved

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16. In cats, black fur color is caused by an X-linked allele (Xb); the other allele at this locus causes orange color (Xo). The heterozygote is tortoiseshell (XbXo). In Cats females are XX and males are XY. What kinds of offspring would you expect from the cross of a black female and an orange male? a) Tortoiseshell females; tortoiseshell males b) Black females; orange males c) Orange females; orange males d) Tortoiseshell females; black males e) Orange females; black males XbXb x XoY All female offspring are XbXo (tortoiseshell) while all male offspring are XbY (black).

17. A Barr body is normally found in the nucleus of which kind of human cell? a) Unfertilized egg cells only b) Sperm cells only c) Somatic cells of a female only d) Somatic cells of a male only e) Both male and female somatic cells

18. Adenine makes up 28% of the nucleotides in a sample of DNA from an organism. Approximately what percentage of the nucleotides in this sample will be guanine? a) 12 b) 22 c) 44 d) 56 e) It cannot be determined from the information provided. 28% A, 28% T, 22% C, 22% G add up to 100%.

19. What determines the nucleotide sequence of the newly synthesized strand during DNA replication? a) the particular DNA polymerase catalyzing the reaction b) the relative amounts of the four nucleoside triphosphates in the cell c) the nucleotide sequence of the template strand d) the primase used in the reaction e) the arrangement of histones in the sugar phosphate backbone

20. One type of baldness in humans is called “male pattern baldness”. There are two alleles for

the gene that codes for male pattern baldness: normal (Hn) and bald (Hb). The gene is autosomal (not sex-linked). The expression of the gene is affected by physiological factors so that in men the bald allele is dominant while in women the bald allele is recessive. If a heterozygous bald man and a heterozygous normal woman have children, what is the probability for each of the following? a) 2 points. If they have a son, the son is bald. ¾

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b) 2 points. If they have two sons, both sons are bald. ¾ x ¾ = 9/16

c) 2 points. If they have a daughter, the daughter is bald. ¼

HnHb x HnHb

Gametes

Hn Hb

Hn HnHn Normal for both sexes HnHb Normal female, Bald male

Hb HnHb Normal female, Bald male

HbHb Bald for both sexes

2 points Bonus: List the last names of 2 of the authors of your textbook.

Campbell, Reece, Urry, Cain, Wasserman, Minorsky, Jackson

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21. 20 points. Answer either a or b, not both. You are expected to write a logically constructed

answer; organization does count. You may want to outline your answer first. a) Consider the genetic code: on any mRNA molecule that is transcribed from a gene, each

sequence of 3 nucleotides usually codes for one amino acid. See the genetic code table provided. Describe what is meant by a mutation. What does a mutation do to the mRNA? To the final product? Why are some mutations harmless? Why are some mutations harmful? Why are some more harmful than others? Explain your answer and provide simple examples that illustrate what you describe.

3 points. A mutation is any change in the nucleotide sequence of the DNA. 3 points. Mutations in the DNA sequences may or may not necessarily affect mRNAs. If a mutation occurs in the non-coding regions of the DNA, then it will have no effect on the mRNA. If it occurs in coding regions, then it may affect the mRNA. 2 Alternative points: In a Eukaryote, mutations in introns again will not affect the mRNA, while mutations in exons will change the corresponding nucleotide sequence in the mRNA (unless they affect the splice site: one more alternative point). 6 points. Mutations that change the nucleotide sequence in the mRNA may or may not necessarily affect the final product: the polypeptide. If the mutation is a substitution in the third nucleotide of an mRNA codon, then the chances that it will affect the final product are very low due to the redundancy of the genetic code. For example, if a mutation changes a codon from CUU to CUC, the translated polypeptide will still contain the amino acid Leu (Leucine) at this position so there is no effect on the final product. Such mutations are silent mutations as they cause no changes in the amino acid sequence and they are harmless. It is possible that a mutation in a codon causes a change in the amino acid without affecting the function of the protein as the change may not affect the shape and thus the function of the protein. Such a mutation would be harmless as well. If the change in the amino acid affects the shape of the protein in a way that it interferes with its function, then that would be a harmful mutation. 4 points. Some mutations have the potential to be much more harmful than others because they affect more than just one amino acid. For example, if a mutation changes a codon for an amino acid to a stop codon (for example UAC to UAG), what is called a nonsense mutation (one alternative point if know the name), then the final polypeptide synthesized from this mutated mRNA will be truncated. This could have devastating effects on the organism. 4 points. Another example of a simple mutation that could have devastating effects is the addition or deletion of one or two nucleotides in the sequence. This type of frame-shift mutation changes all of the codons downstream from the point of the mutation and can change the entire sequence of amino acids in the polypeptide. This could result in a completely different polypeptide with very different properties, which could have devastating consequences. For example, the coding sequence could change from UUU UUA CUU … (Phe-Leu-Leu- …) to UUA UUU ACU U… (Leu-Phe-Thr-…) simply by the addition of one nucleotide (A) after the second U. The shift in the reading frame changes the amino acids completely. Alternative point: sometimes frameshift mutations are nonsense mutations (stop codon forms)!

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b) What does the organizational concept of sexuality entail? Explain. Why does the organizational concept fail to explain vertebrate sexuality? Explain. Describe two different modes of sex determination that illustrate why the organizational concept fails to explain vertebrate sexuality. Explain what the evolutionary view of sexuality is as opposed to the organizational concept. Describe one piece of evidence (discussed by David Crews) in support of the evolutionary view.

4 points: According to the organizational concept, the gender and sexual characteristics of the organism are ultimately determined by the chromosomes. The animal’s gonads is determined at the time of conception by the chromosomes inherited from the parents. The gonads produce sex hormones that sculpt the sexual features. 2 points. The organizational concept fails to explain vertebrate sexuality because many vertebrates lack sex chromosomes and depend on non-genetic triggers to determine gender. 4 points. One mode of sex/gender determination that does not rely on sex chromosomes is temperature-dependent sex/gender determination. In these cases, the temperature at which the embryo develops governs the animal’s sex ratio. 1 alternative point: this is an all-or-nothing event; does not lead to formation of hermaphrodites in intermediate temperatures, for example. 2 Alternative points if an example is described such as leopard gecko (low and high T produce females, intermediate yields males) . 4 points. Another mode of sex/gender determination that does not rely on sex chromosomes is behavior-dependent sex/gender determination. In most cases, they are hermaphrodites, and the social environment controls whether the individual takes on a male or female role. Some are sequential hermaphrodites changing from one sex to another (alternative point: but only expressing one sex at any given time). Some are simultaneous hermaphrodites. 2 Alternative points if an example is described such as anemone fish (born male, later develop into females) or certain coral reef fish (start female, later develop into males in response to disappearing dominant males) or simultaneously hermaphroditic butter hamlet fish (alternate between male and female behavioral roles during successive matings). 2 points. In the scenario described by the organizational concept males are the organized sex and the females are the default sex: 2 alternative points: If there are testes, then they produce androgens (male hormones) that lead to male characteristics. If there are no testes, then ovaries develop, leading to female hormones, leading to female characteristics. 2 points: In the evolutionary view of sexuality, males are considered to have evolved only after the evolution of the first self-replicating, female organisms (same as saying the female is the ancestral sex and the male the derived sex). 2 points: Examples of pieces of evidence in support of the evolutionary view: a) Fish species that are born male and become female nevertheless pass through a modified ovarian

stage before developing testes. b) In some species testosterone is converted to estrogen in the brain, and estrogen activates both

copulatory behaviors in males and sexual receptivity in females. c) In songbirds, estrogen originates primarily in the brain, suggesting that it transcends gender

boundaries. d) Male rats and humans show a pronounced daily rhythm in progesterone secretion, peaking at the

onset of night, when copulatory behavior most often occurs. e) Doses of the “female hormone” progesterone can induce castrated rats to resume mounting. f) RU486, which nullifies progesterone (the “female” hormone) reduces sexual behavior in male rats.

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__________________________________________________ _____________________

Name Date

Please make sure you have all 8 pages of the exam. Write your name and lab section on all pages. Please read each question carefully. If you do not understand the meaning or intention of a question, please ask. Please pace yourself and use your time efficiently.

Please put away your cell phones, blackberries, iPods, laptop computers, etc.

Each multiple-choice question is worth 2 points. Note that a correct statement is not necessarily the best answer to a multiple-choice question. For each multiple-choice question, circle the best answer. You will receive ZERO credit for any question for which we cannot tell what your final answer is. If you erase or cross out an answer, please make it clear as to which answer is your final answer.

Please pace yourself and use your time efficiently; allow time for the essay type questions. If a question states “answer a or b, not both”: if you answer both, then we will read only the answer to a, regardless of how much better your answer to b might be.

1. 6 points. Phenylketonuria (PKU) is an inherited disease caused by a recessive allele. If a woman and her husband, who are both carriers, have two children, what is the probability of each of the following? Show your work. a) Both children have the disease. b) At least one child is phenotypically normal.

a) The genotype for carriers is Pp (P = dominant, p = recessive) The probability that one child will be homozygous recessive pp will be ¼ based on the Punnett Square below P p P PP Pp P Pp pp The probability that 2 children by these parents will both be homozygous recessive will be ¼ x ¼ = 1/16 b) The probability that at least one child is phenotypically normal is 1-1/16 = 15/16.

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2. 10 points. Radish flowers may be red, purple, or white. A cross between a red-flowered plant and a white-flowered plant yields all-purple offspring. The part of the radish we eat may be oval or long, with long being the dominant characteristic. A true breeding red flowered – long radish is crossed with a true breeding white flowered – oval radish. Write the genotypes of the radish plants in the P, F1, and F2 generations. What is the phenotype of the F1 generation? Use a Punnett square to figure out the phenotypic ratios (the proportion of radish plants with a given phenotype) in the F2 generation. P genotypes: RRLL X rrll F1 genotype: RrLl F1 phenotype: Purple and Long Punnett Square to show the genotypes of the F2 generation

Phenotypic ratio of F2 generation Purple-Long 6 Red-Long 3 White-Long 3 Purple-Oval 2 Red-Oval 1 White-Oval 1

3. Mendel accounted for the observation that traits which had disappeared in the F1 generation reappeared in the F2 generation by proposing that a) new mutations were frequently generated in the F2 progeny, "reinventing" traits that had

been lost in the F1. b) the mechanism controlling the appearance of traits was different between the F1 and the

F2 plants. c) traits can be dominant or recessive, and the recessive traits were obscured by the

dominant ones in the F1. d) members of the F1 generation had only one allele for each character, but members of the

F2 had two alleles for each character. 4. 4 points. Red-green color blindness is a sex-linked recessive trait in humans. Two people

with normal color vision have a color-blind son and a normal daughter. a) What are the genotypes of the parents?

Father: XY Mother: XCX

b) What is the probability that their daughter would be a carrier? 50%

RL Rl rL rl RL RRLL

Red-Long RRLl Red-Long

RrLL Purple-Long

RrLl Purple-Long

Rl RRLl Red-Long

RRll Red-Oval

RrLl Purple-Long

Rrll Purple-Oval

rL RrLL Purple-Long

RrLl Purple-Long

rrLL While-Long

rrLl White-Long

rl RrLl Purple-Long

Rrll Purple-Oval

rrLl White-Long

Rrll White-Oval

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5. In grasshoppers, sex is determined by the X-0 chromosome scheme. Males are X-0 and females are XX. A recessive lethal allele that causes death of the embryo is sometimes present on the X chromosome. What would be the sex ratio in the offspring of a cross between a female that is heterozygous for the lethal allele and a normal male? a) 2:1 male to female b) 1:2 male to female c) 1:1 male to female d) 4:3 male to female e) 3:1 male to female

6. 4 points. Answer either a or b, not both:

a) If a pair of homologous chromosomes fails to separate during anaphase of meiosis I, what will be the chromosome number of the four resulting gametes with respect to the normal haploid number (n=23)? (Show your work for partial credit) The figure below shows non-disjunction during meiosis I for one pair of homologous chromosomes. It leads to one extra chromosome in half of the gametes and one fewer chromosome in half of the gametes. There are no normal gametes produced. Therefore, for n=23, half of the gametes will have 24 chromosomes and half of the gametes will have 22 chromosomes.

b) Review Figure 15.9. Explain why Morgan’s results did not match either of the predicted

ratios. Because the genes were linked but there was some distance between them on the chromosome so that there was crossing over occurring allowing for recombination. This leads to the formation of the recombinant genotypes which is different than what is expected if the genes are perfectly linked, but at lower frequency than expected if the genes were on separate chromosomes or behaved as if they were on separate chromosomes.

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7. 10 points. Choose either Figure a or b not both. Then answer the following questions: What are the alternative hypotheses being tested? What are the experimental predictions based on these hypotheses? Which hypothesis is supported?

a)

Alternative hypotheses being tested:

1. DNA is the genetic material used by the bacteriophage viruses.

2. Proteins are the genetic material used by the bacteriophage viruses.

Experimental Predictions:

1. For hypothesis 1: If the DNA is labeled with radioactive phosphorus, and the phages infect bacterial cells, in the final extraction the radioactive phosphorus ends in the pellet inside the cells.

2. For hypothesis 2: If the proteins are labeled with radioactive sulfur, and the phages infect bacterial cells, in the final extraction the radioactive sulfur ends in the pellet inside the cells.

The experimental results support the first hypothesis because the radioactive label ended up in the pellet only when radioactive phosphorus was used to label the DNA.

b)

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Alternative Hypotheses:

1. Conservative model: when DNA is replicated, the parent DNA molecule remains intact while one copy is made completely with new material.

2. Semi-conservative model: when DNA is replicated, each strand of the double helix acts as a template and each daughter DNA molecule contains one parental strand and one new strand.

3. Dispersive model: when DNA is replicated, new nucleotides are incorporated into the DNA as two new copies are made so that the new daughter DNA molecules are a mixture of new and parental nucleotides.

Experimental predictions:

For hypothesis 1: If bacteria are grown in heavy N for many generations and then allowed to grow in light N for just one generation, then half of the daughter cells will have heavy N DNA (parental) and half of the daughter cells will have light N DNA (all new), leading to 2 bands in the centrifuge tubes. Every generation, the original heavy N DNA would always remain heavy, and all of the new DNA would be light, so there will always be 2 bands in the centrifuge tube.

For hypothesis 2: If bacteria are grown in heavy N for many generations and then allowed to grow in light N for just one generation, then all of the daughter cells will have DNA that has one strand made of heavy nucleotides and one strand made of light nucleotides, leading to 1 band in the centrifuge tube. However, if the bacteria are grown for a second generation, then half of the daughter cells will have DNA made of only light DNA while half will have DNA that has one strand made of heavy nucleotides and one strand made of light nucleotides, leading to 2 bands in the

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centrifuge tube.

For hypothesis 3: If If bacteria are grown in heavy N for many generations and then allowed to grow in light N for just one generation, then all of the daughter cells will have DNA molecules made of a mixture of heavy and light nucleotides leading to one band in the centrifuge tube. Every generation in the light N would lead to lighter and lighter DNA, but always one band in the centrifuge tube.

The results of the first replication eliminate the conservative model, and the results of the second replication eliminate the dispersive model. The semi-conservative model is supported.

8. The DNA of telomeres has been found to be highly conserved throughout the evolution of eukaryotes. What does this most probably reflect?

a) the inactivity of this DNA b) the low frequency of mutations occurring in this DNA c) that new evolution of telomeres continues d) that mutations in telomeres are relatively advantageous e) that the critical function of telomeres must be maintained

9. 4 points. Review the figure below. In a couple of sentences explain what the problem is with the lagging strand.

The problem is that DNA must be replicated in the 5’ to 3’ direction and so in the lagging strand DNA polymerase III must work in the direction away from the replication fork. This means the lagging strand will be synthesized discontinuously.

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10. Why does the DNA double helix have a uniform diameter? a) Purines pair with pyrimidines. b) C nucleotides pair with A nucleotides. c) Deoxyribose sugars bind with ribose sugars. d) Nucleotides bind with nucleosides. e) Nucleotides bind with nucleoside triphosphates.

11. If A, B, and C are all required for growth, a strain mutant for the gene encoding enzyme B

would be capable of growing on which of the following media? a) minimal medium b) minimal medium supplemented with "A" only c) minimal medium supplemented with "B" only d) minimal medium supplemented with "C" only e) minimal medium supplemented with nutrients "A" and "B"

12. 8 points. The nucleotide sequence of a DNA piece being transcribed is 5’ATATCCTACATATCCTCCTCT…3’. A messenger RNA molecule with complementary codons is transcribed from the DNA. In the process of protein synthesis in the ribosome (translation), tRNAs pair with the mRNA codons. a) What is the nucleotide sequence of the mRNA corresponding to the sequence provided?

If you read it from left to right: UAUAGGAUGUAUAGGAGGAGA If you read it from right to left: AGAGGAGGAUAUGUAGGAUAU I’ll give you credit for both (even though only one is correct)

b) Mark the start of the reading frame (the start codon) on the mRNA. Again depends on whether you read it from right to left or left to right. Marked in bold

c) What is the Amino Acid sequence of the polypeptide that would be produced from this mRNA if the ribosome starts translating from the start codon? If you read it from right to left: Met-Tyr-Arg-Arg-Arg If you read it from left to right: Met-Stop!

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13. A mutant bacterial cell has a defective aminoacyl synthetase that attaches a lysine to tRNAs with the anticodon AAA instead of a phenylalanine. The consequence of this for the cell will be that a) none of the proteins in the cell will contain phenylalanine. b) proteins in the cell will include lysine instead of phenylalanine at amino acid positions

specified by the codon UUU. c) the cell will compensate for the defect by attaching phenylalanine to tRNAs with lysine-

specifying anticodons. d) the ribosome will skip a codon every time a UUU is encountered.

14. The start codon AUG also codes for the amino acid methionine (Met). But not all

polypeptides have methionine as their first amino acid. That is possible because a) some mRNAs have a start codon other than the AUG b) the first methionine (Met) may be cleaved later in processing c) other polypeptides are added to the beginning later in processing d) the methionine is folded in the 3-D structure so it’s not considered “first”

15. Biologists trying to study the structure of the eukaryotic ribosome in yeast reduced the level of glucose available to the yeast from which they intended to obtain ribosomes. This step a) inhibited the process of translation producing a homogeneous population of empty

ribosomes to study b) promoted translation of enzymes needed to break down other sugars, thus increasing

available active ribosomes c) activated the lac operon in the yeast, thus increasing the chance that active ribosomes

could be harvested d) stopped the replication of yeast cells so that they would be arrested in the G1 phase

16. In trying to study the effects of telomeres and telomerase on aging, biologists have

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engineered mice to lack the enzyme, called telomerase, which then become prematurely decrepit. But the mice bounced back to health when the enzyme was replaced. One of the criticisms of this study is that a) knocking out the enzyme telomerase is not a good simulation of the regular aging

process. b) adding the enzyme telomerase causes rapid replication of the cells and accelerates the

aging process. c) the mice were pre-cancerous due to the rapid rate of cell division in the presence of

telomerase. d) it does not delve into how telomerase actually carries out its function

17. See the figure below. The lactose operon is likely to be transcribed when

a) there is more glucose in the cell than lactose. b) the cyclic AMP levels are low. c) there is glucose but no lactose in the cell. d) the cAMP and lactose levels are both high in the cell. e) the cAMP level is high and the lactose level is low.

18. T/F: If there is no lactose (and hence allolactose) present, then the level of cAMP present makes no difference in the regulation of the expression of the genes in this operon.

19. Acetylation of histones (adding –COCH3) masks the positive charges of the lysines on the tail end of the histones. This means the histone tails no longer bind to neighboring nucleosomes. This phenomenon: a) Exposes the cytosine in DNA for methylation b) Allows easy access for transcription factors c) Promotes folding of chromatin into heterochromatin d) Activates the enhancer sequences in the 5’ UTR

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20. 15 points. Answer a or b, but not both) of the following questions. You are expected to

write a logically constructed answer; organization does count. You may want to outline your answer first. a) Consider the genetic code: on any mRNA molecule that is transcribed from a gene, each

sequence of 3 nucleotides usually codes for one amino acid. The same amino acid may be coded by more than one combination of 3 nucleotides. Describe what is meant by a mutation. What does a mutation do to the mRNA? To the final product? Why are some mutations harmless? Why are some harmful? Why are some more harmful than others? Explain your answer.

3 points. A mutation is any change in the nucleotide sequence of the DNA. 3 points. Mutations in the DNA sequences may or may not necessarily affect mRNAs. If a mutation occurs in the non-coding regions of the DNA, then it will have no effect on the mRNA. If it occurs in coding regions, then it may affect the mRNA. Extra points: In a Eukaryote, mutations in introns again will not affect the mRNA (unless they affect the splice site: more extra points!), while mutations in exons will change the corresponding nucleotide sequence in the mRNA. 6 points. Mutations that change the nucleotide sequence in the mRNA may or may not necessarily affect the final product: the polypeptide. If the mutation is a substitution in the third nucleotide of an mRNA codon, then the chances that it will affect the final product are very low due to the redundancy of the genetic code. For example, if a mutation changes a codon from CUU to CUC, the translated polypeptide will still contain the amino acid Leu (Leucine) at this position so there is no effect on the final product. Such mutations are silent mutations as they cause no changes in the amino acid sequence and they are harmless. It is possible that a mutation in a codon causes a change in the amino acid without affecting the function of the protein as the change may not affect the shape and thus the function of the protein. Such a mutation would be harmless as well. If the change in the amino acid affects the shape of the protein in a way that it interferes with its function, then that would be a harmful mutation. 4 points. Some mutations have the potential to be much more harmful than others because they affect more than just one amino acid. For example, if a mutation changes a codon for an amino acid to a stop codon (for example UAC to UAG), what is called a nonsense mutation (extra point if they know the name), then the final polypeptide synthesized from this mutated mRNA will be truncated. This could have devastating effects on the organism. 4 points. Another example of a simple mutation that could have devastating effects is the addition or deletion of one or two nucleotides in the sequence. This type of frame-shift mutation changes all of the codons downstream from the point of the mutation and can change the entire sequence of amino acids in the polypeptide. This could result in a completely different polypeptide with very different properties, which could have devastating consequences. For example, the coding sequence could change from UUU UUA CUU … (Phe-Leu-Leu- …) to UUA UUU ACU U… (Leu-Phe-Thr-…) simply by the addition of one nucleotide (A) after the second U. The shift in the reading frame changes the amino acids completely.

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b) The genetic code is thought to be universal; the same codons code for the same amino acids from bacteria such as E. coli to humans. Nevertheless, if you take a human gene such as that for Troponin T for example and put it in a bacterium such as E. coli, even if you manage to get the gene transcribed correctly, the protein produced by the bacterium using that gene will not be anything like the protein produced by the human cell. Explain why.

5 points. The promoter sequences in bacteria and eukaryotes are different. Therefore for the gene to be transcribed at all, it must have a correct bacterial promoter sequence upstream of the coding sequence that is recognized by the bacterial RNA polymerase. So, we must make sure there is an appropriate bacterial promoter sequence upstream of our gene, with the start point of transcription as part of the promoter sequence. 5 points. In addition, in bacterial genes, there is a specific sequence that signals the end of transcription, the terminator. This is not the case in eukaryotic genes. Therefore, we must add this terminator sequence to the 3’ end of our gene so that transcription of the gene occurs normally in the bacterial cells. The promoter and the terminator then define the transcription unit, the sequence of DNA that will be transcribed into an RNA molecule. 5 points. While eukaryotic genes have introns and exons, bacterial genes do not. So, if this gene as is shown in the diagram is transcribed by the bacterial cells as is, the mRNA synthesized will include ALL of the DNA sequence between the start point and the terminator sequence, and all of that RNA sequence will be used for translation. Since there is no RNA splicing in bacteria, only one final polypeptide will be produced from this one gene. This is not what we want. 5 points. Therefore, instead of using the DNA sequence of the eukaryotic gene, we must make 2 separate DNA sequences for our transcription units that we then place in between bacterial promoter and terminator sequences. In effect, we need “2 genes” in order to make the two polypeptides. Each “gene” must have only the DNA exons (no DNA introns) that will be used to code for the amino acids for the polypeptide of interest. And each gene with its own promoter and terminator sequence can then be used by the bacterial cell to make the polypeptide for which its DNA sequence codes. !

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e. 4 points. Pick one of the codons in the mRNA you wrote above. Write the simplest mutation in the original DNA sequence that would turn this mRNA codon to another codon that would code for the same amino acid as before. Original mRNA codon: example: UGU DNA sequence from ACA to ACG This type of mutation is called: silent

2. 4 points. A cell has 2N = 6 chromosomes. Write the correct numbers below. a. How many chromosomes does it have during Prophase I of meiosis?

Answer: 6 b. During Anaphase I of meiosis?

Answer: 6 c. How many chromosomes does each daughter cell have during Prophase II of meiosis?

Answer: 3 d. During Anaphase II of meiosis?

Answer: 6

3. In garden peas the recessive allele p leads to red flowers when homozygous (pp). The dominant allele leads to the wild-type Purple color. Another gene has a recessive allele l that produces round pollen grains when homozygous (ll). The dominant allele leads to Long pollen grains. Consider a cross between a PPLL female and a ppll male. a. 2 points. What is the genotype of the F1 generation? PpLl b. 2 points. What is the phenotype of the F1 generation for flower color and pollen grain

shape? Flower Color: Purple , Pollen Grain Shape: Long c. 4 points. If you allow the F1 progeny to mate with each other, what will be the genotypes

and the phenotypic ratios in the F2 generation if the genes are linked (on the same chromosome)? Assume there is no crossing over and there is no nondisjunction. You may use a Punnett square if you wish.

Egg\Sperm PL pl PL PPLL PpLl pl PpLl ppll

Phenotypic ratio: 3 purple and long: 1 red and round

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d. 8 points. If you allow the F1 progeny to mate with each other, what will be the genotypes and the phenotypic ratios in the F2 generation if the genes are unlinked (on separate chromosomes)? Use the Punnett square below to show the genotypes for the sperm (top row) and egg (first column) and the F2 generation (squares with dark borders). Assume there is no nondisjunction.

Phenotypic ratio: 9 purple and long : 3 purple and round : 3 red and long : 1 red and round

Egg\Sperm PL Pl pL pl PL PPLL PPLl PpLL PpLl Pl PPLl PPll PpLl Ppll pL PpLL PpLl ppLL ppLl pl PpLl Ppll ppLl ppll

e. 4 points. There are 400 individuals in the F2 generation. The observed data are shown in

the table below. Based on the data are the two genes linked or unlinked? What process in meiosis can easily explain the difference between the numbers observed in the F2 generation below and the expected phenotypic ratios you wrote above?

The data suggested the genes are linked as there were a lot more parental phenotypes than recombinants, but there had to be a mechanism to break the linkage to produce the recombinant types. Crossing over in meiosis can explain this.

f. 5 points. Draw a diagram to illustrate how the process you mentioned in e above works to

affect the chromosomes and thus the gametes formed in the F1 generation.

For example

Gene for flower color

Centromeres

Gene for pollen shape

Maternal Chromosome 1 Paternal Chromosome 1

Phenotype Purple Long Purple Round Red Long Red Round Number Observed

290 10 10 90

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Recombinants:

Gene for flower color p P

Gene for pollen shape L a

4. 6 points. For the following questions, match the key event of meiosis with the stages listed

below by writing the correct Capital Letter (A-H) in the spaces provided. A. Prophase I B. Metaphase I C. Anaphase I D. Telophase I E. Prophase II F. Metaphase II G. Anaphase II H. Telophase II a. Tetrads of chromosomes are aligned at the equator of the spindle; alignment determines

independent assortment. Answer: B

b. Synapsis of homologous pairs occurs; crossing over may occur. Answer: A

c. Nuclear envelopes reform. Each daughter nucleus has half as many chromosomes as the original cell. Each chromosome has 2 sister chromatids attached at the centromere. Answer: D

d. Cohesins holding the centromeres together are cleaved. Sister chromatids separate. Answer: G

e. Individual chromosomes are aligned at the equator of the spindle. Each chromosome has 2 sister chromatids attached at a centromere. Answer: F

f. Cohesins holding homologous pairs are cleaved. Homologous chromosomes move apart. Answer: C

5. Suppose you are provided with an actively dividing culture of E. coli bacteria that has been growing with only the light isotope of N. You then add cytosine made with heavy isotope of N to the culture. What would happen if an E. coli cell replicates once in the presence of this heavy N cytosine?

a. One of the daughter cells, but not the other, would have heavy DNA. b. Neither of the two daughter cells would have heavy DNA. c. All four bases of the DNA would have heavy N. d. Heavy N cytosine would pair with heavy N guanine. e. DNA in both daughter cells would have heavy N. Answer: e

6. 2 points. What would be the expected answer to the above question if DNA actually replicated using the conservative method?

l

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Answer: a

7. Review the experiment by Avery, McCarty, and MacLeod. a. 2 points. What is the question being asked? What is the transforming principle made of?

b. 3 points. What are the three alternative hypotheses? Hypothesis 1: The transforming principle is made of DNA. Hypothesis 2: The transforming principle is made of RNA. Hypothesis 3: The transforming principle is made of proteins.

c. 2 points. Pick one of the hypotheses you wrote above.

State which one you pick: # 2 for example What is the null hypothesis for the alternative hypothesis you picked? The transforming principle is not made of RNA.

d. 2 points. What is the experimental prediction based on the alternative hypothesis you

picked above? If S strain bacteria are heat-killed and homogenized, and the homogenate is filtered, and the filtrate is divided into 3 set of test tubes and set 1 is treated with RNase, set 2 is treated with Protease, and set 3 is treated with DNAse, and the treated samples are added to cultures of R strain bacteria, the bacteria in sets 2 and 3 will become virulent but those in set 1 will not.

e. 1 point. Do the data support the hypothesis you picked? Yes/No.

8. Red-green color blindness is a sex-linked recessive trait in humans. A couple has a color-

blind daughter. Which of the following applies to the parents? a. The father must be color blind; the mother is either color blind or a carrier b. The mother must be color blind; the father is either color blind or a carrier c. Both parents must be carriers d. Both parents must be color blind Answer: a

9. Review the Figures for the tryptophan (trp) operon. The tryptophan operon is a repressible operon that is a. Turned on when the correct activator proteins bind to the enhancer. b. Turned on only when tryptophan is present in the growth medium. c. Turned off only when tryptophan is absent in the growth medium. d. Turned off whenever tryptophan is added to the growth medium. Answer: d

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10. Select all that apply: Cytosine and Guanine together make up 38% of the nitrogenous bases in a sample of DNA from an organism. In this sample: a. Adenine makes up 31% of the bases b. Cytosine makes up 19% of the bases c. Adenine makes up 62% of the bases d. Thymine makes up 62% of the bases Answer: a and b

11. 10 points. Answer either a or b, not both: a. Explain what the problem is with the lagging strand during DNA replication and how the

enzymes involved act to overcome this problem. DNA polymerase III can only add nucleotides to the free 3’ carbon of an existing nucleotide. This means that each new DNA strand is formed in the 5’ to 3’ direction. To make the new complementary strands, primase adds a short RNA primer, and complementary nucleotides are added by DNA polymerase III to the free 3’ carbon of the primer to make the complementary strand. In the lagging strand the replication fork moves in the opposite direction as the direction of DNA polymerase III. Therefore, primase has to jump back continuously to add a new primer to a newly available section of DNA due to the opening of the replication fork, and then DNA polymerase III has to jump back and add nucleotides to that primer until it reaches a previously synthesized segment. Thus, the lagging strand will be synthesized discontinuously in small fragments that have to be joined together later by the enzyme ligase after the RNA pieces have been replaced with DNA nucleotides by DNA polymerase I.!

b. In your own words describe the sequence of events involved in each of the phases of translation: initiation, elongation, and termination.

Initiation: The mRNA binds to the small ribosomal subunit, the first tRNA with the attached Met comes in with a complementing anticodon for the start codon, the large subunit attaches so that the initiator tRNA is in the P site. Elongation: The next tRNA with an anticodon complementing the codon in the open A Site comes in and brings the next Amino Acid. The ribosome attaches the new amino acid to the growing polypeptide and releases the tRNA from the P Site through the E Site. The tRNA in the A site and the mRNA move to the P Site. There is a new codon in the open A site. The process repeats as long as there are codons to be read. Termination: When a stop codon is present in the A site, a release factor enters the A site. It helps the ribosome release the polypeptide from the last tRNA, which then exits through the E site. Then the ribosome assembly comes apart and the mRNA is released.

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Some potentially useful figures:

Griffith had demonstrated that something in the virulent (disease causing) S strain of the bacteria he studied could transform nonvirulent R strain bacteria into a lethal form, even when the S strain bacteria had been killed by high temperatures. The experiment by Avery, MacLeod and McCarty is shown below.

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12. 20 points: Please write a short essay in response to either a or b, not both. You are expected to write a logically constructed answer; organization does count.

a. Explain how each of the following processes helps regulate gene expression in eukaryotes: DNA methylation, and histone phosphorylation, methylation, and acetylation. Explain why these modifications may change during development and cell differentiation in eukaryotes. Refer to the diagram of the chromosomes from normal and cancer cells. Explain the relationships between the histone modifications illustrated, the regions of the chromosomes where they are found, and the final outcome: normal versus cancer cell. You need to include 20 of the following points. 4 points. Chromatin modification is very important in regulation of gene expression in eukaryotes. Methylation, whether of cytosines in DNA or of histones, leads to condensation of the chromosomes. This is because methyl groups are non-polar and hydrophobic and in an aqueous environment will clump together, thus condensing the chromatin. 3 points. Histone acetylation leads to masking the positive charges of lysines on histones, thus leading to a net negative charge, which leads to decondensation of the chromatin as negative charges repel each other. 3 points. Histone phosphorylation increases the net negative charge of the chromatin and also leads to chromatin decondensation similar to acetylation. 5 points. During development and cell differentiation different genes have to be turned on and off at different times. If chromatin is condensed, it is less available to RNA polymerase and transcription of any genes in the condensed chromatin is inhibited. If chromatin is decondensed, it is more available to RNA polymerase and transcription of any genes in the decondensed region is more likely. Thus changing the condensation status of a region of the chromosome can lead to more or less expression of genes in that region. 3 points. In the example of the chromosome in the normal cell provided there are gene-rich regions of the chromosome that are acetylated and have fewer methyl groups and are thus decondensed, while other regions are more methylated and have fewer acetyl groups and are more condensed. 3 points. In the cancer cell, the gene-rich regions have more methyl groups and are deacetylated and thus are more condensed, while the other regions that are more condensed in the normal cell are less condensed because they have fewer methyl groups and more acetyl groups. 2 points. Thus genes that are normally expressed in the normal cell are less available to the RNA polymerase and thus turned off in the cancer cell and vice versa. This might contribute to turning this cell into a cancer cell.

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b. What is transcription? In your own words describe the basic steps in the process by which DNA transcription occurs in eukaryotes. In your answer include the role of the promoter, the processes of initiation, elongation, and termination, as well as RNA processing (including the changes to the 5’ and 3’ ends as well as in between) and its significance to the mRNA. The genetic code is thought to be universal; the same codons code for the same amino acids from bacteria such as E. coli to humans. If you take a human gene and put it in E. coli, you may manage to get the gene transcribed correctly, but the protein produced by the bacterium using that gene will not be anything like the protein produced by the human cell. Explain why. You need to include 20 of the following points. 1 point. Transcription is the synthesis of RNA under the direction of DNA: turning the DNA code into an RNA version. It has 3 stages: initiation, elongation, and termination. 3 points. Each gene has a promoter region that signals the initiation of RNA synthesis. Transcription factors mediate the binding of RNA polymerase to the promoter and the initiation of transcription. 1 point. A promoter region called a TATA box is crucial in forming the initiation complex in eukaryotes 3 points: During elongation RNA polymerase pries the DNA strands apart and hooks together the RNA nucleotides. This uses the same base-pairing rules as DNA, except uracil substitutes for thymine. 2 points: In eukaryotes termination occurs once a sequence called a polyadenylation signal is transcribed. 3 points. The RNA is then processed as follows. A cap of modified GTP is added to the 5’ end. The polyadenylation signal is recognized and the 3’ tail of the RNA is cut and a poly A tail is added. 3 points. These modifications help protect the mRNA against hydrolytic enzymes (prevent breakdown prematurely), help direct the mRNA for export out of the nucleus, and help align the mRNA correctly on the ribosome. 3 points. Introns are cut and exons are joined together through splicing. Alternative splicing may allow multiple mRNAs with different exon combinations and thus multiple polypeptides to form from one pre-mRNA. 5 points. It is possible to put a human gene in E. coli and have the bacterium transcribe the gene and make a protein. However, if the entire gene sequence is used, there will be a problem. Since bacteria don’t have introns, E. coli will not carry out any splicing and thus the entire sequence including the introns will be read as codons and used to make a protein. Therefore the protein made this way will be very different than the protein made form this gene in a human cell.

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__________________________________________________ _____________________

Name Date Please make sure you have all 12 pages of the exam. Please read each question carefully. If you do not understand the meaning or intention of a question, please ask. Please pace yourself and use your time efficiently. Please put away your cell phones, blackberries, iPods, laptop computers, etc. Write your answers using a pen. Answers written in pencil will not be accepted. Each multiple-choice or True/False question is worth 2 points. Note that a correct statement is not necessarily the best answer to a multiple-choice question. For each multiple-choice question, circle the best answer. You will receive ZERO credit for any question for which we cannot tell what your final answer is. If you cross out an answer, please make it clear as to which answer is your final answer. Please pace yourself and use your time efficiently; allow time for the essay type questions. If a question states “answer either a or b, not both”: if you answer both, then we will read only the answer to a, regardless of how much better your answer to b might be. 2 points: Write your name and lab section on all sheets. 1. A new DNA strand elongates only in the 5' to 3' direction because

a) DNA polymerase begins adding nucleotides at the 5' end of the template. b) Okazaki fragments prevent elongation in the 3' to 5' direction. c) the polarity of the DNA molecule prevents addition of nucleotides at the 3' end. d) replication must progress toward the replication fork. e) DNA polymerase can only add nucleotides to the free 3' end.

2. The leading and the lagging strands differ in that a) the leading strand is synthesized in the same direction as the movement of the replication

fork, and the lagging strand is synthesized in the opposite direction. b) the leading strand is synthesized by adding nucleotides to the 3' end of the growing

strand, and the lagging strand is synthesized by adding nucleotides to the 5' end. c) the lagging strand is synthesized continuously, whereas the leading strand is synthesized

in short fragments that are ultimately stitched together. d) the leading strand is synthesized at twice the rate of the lagging strand.

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3. What is the role of DNA ligase in the elongation of the lagging strand during DNA

replication? a) synthesize RNA nucleotides to make a primer b) catalyze the lengthening of telomeres c) join Okazaki fragments together d) unwind the parental double helix e) stabilize the unwound parental DNA

4. The start codon also codes for the amino acid methionine (Met). But not all polypeptides have methionine as their first amino acid. That is possible because a) some mRNAs have a start codon other than the AUG b) the first methionine (Met) may be cleaved later in processing c) other polypeptides are added to the beginning later in processing d) the methionine is folded in the 3-D structure so it’s not considered “first”

5. The DNA of telomeres has been found to be highly conserved throughout the evolution of

eukaryotes. What does this most probably reflect? a) the inactivity of this DNA b) the low frequency of mutations occurring in this DNA c) that new evolution of telomeres continues d) that mutations in telomeres are relatively advantageous e) that the critical function of telomeres must be maintained

6. The nucleotide sequence of a DNA piece being transcribed is

5’…CCTTTAAGATATC CTCC TCATAAT…3’ 3’…GGAAATTCTATAGGAGGAGTATTA…5’ While both strands could potentially code for a messenger RNA, only one strand from this particular piece of DNA will code for a messenger RNA that will lead to the synthesis of a polypeptide. An mRNA molecule that will code for a polypeptide is transcribed from this piece of DNA. Note that the mRNA has to be read from 5’ to 3’. Also note that the mRNA

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is always longer than the sequence that is translated; it has untranslated regions on both sides of the translated region. a) 4 points. Which DNA strand, the top or the bottom, from this piece of DNA is the coding

strand for the mRNA? How do you know? The top strand, because it will code for a start codon (TAC in DNA, corresponding to AUG in mRNA).

b) 4 points. What is the nucleotide sequence of the mRNA corresponding to the sequence of the coding strand of DNA? Regardless of whether the top or the bottom strand is used, the mRNA is written 5’ to 3’. 5’ AUUAUGAGGAGGAUAUCUUAAAGG 3’ Start

c) 2 points. Mark the start of the reading frame (the start codon) on the mRNA. d) 4 points. Write out the Amino Acid sequence of the polypeptide that would be produced

from this mRNA if the ribosome starts translating from the start codon. Met-Arg-Arg-Ile-Ser (stop)

7. In humans there is one gene that controls the curliness of hair: the individuals with straight hair are homozygous for the straight hair allele (SS), the individuals with curly hair are homozygous for the curly hair allele (ss), and individuals with wavy hair are heterozygous for the two alleles (Ss). Suppose a second gene can lead to absence (dominant, L) or presence (recessive, l) of hair. The two genes are on separate chromosomes. Consider the cross shown below. Write the genotype of the children in the F1 generation. What is the phenotype of the F1 generation? Use a Punnett square to figure out the genotypes and the

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phenotypic ratios (the proportion of children with a given phenotype) in the F2 generation. Show your work. P genotypes: SSll X ssLL 1 points: F1 genotype: SsLl 2 points: F1 phenotype: No hair (bald)

Gametes

SL

Sl

sL

sl

SL

SSLL Bald

SSLl Bald

SsLL Bald

SsLl Bald

Sl

SSLl Bald

SSll Straight hair

SsLl Bald

Ssll Wavy hair

sL

SsLL Bald

SsLl Bald

ssLL Bald

ssLl Bald

sl

SsLl Bald

Ssll Wavy hair

ssLl Bald

ssll Curly hair

8 points: Punnett Square to show the genotypes of the F2 generation 3 points: Phenotypic ratio of F2 generation 12 Bald : 2 Wavy: 1 Straight: 1 Curly

8. 12 points. Answer either a or b, not both.

a) Hershey and Chase wanted to find out what the genetic material was that bacteriophage viruses used when they infected bacteria. To answer this question, they carried out two experiments. They cultured the phages with either radioactive sulfur (Exp 1) or radioactive phosphorus (Exp 2). They mixed the labeled phages with bacteria, allowed the phages to infect the bacterial cells, then agitated the mixture in a blender to separate what was left outside the bacteria cells from the bacteria cells, then centrifuged the mixture so that bacteria formed a pellet at the bottom, and the parts of the phages that had not entered the bacteria cells were in the supernatant, then measured the radioactivity in the pellet and the liquid. What are the alternative hypotheses being tested? What are the experimental predictions based on these hypotheses?

Name: ___________________________Lab Section __________________ 5/23/11

! 5!

Exp 1

Exp 2


DNA is the genetic material used by the bacteriophage viruses.


Proteins are the genetic material used by the bacteriophage viruses.


If the proteins are labeled with radioactive sulfur, and the phages infect bacterial cells and then the cells are separated from what is left outside through agitation, and the cells are collected in a pellet through centrifugation, in the final extraction the radioactive sulfur ends in the supernatant.


If the DNA is labeled with radioactive phosphorus, and the phages infect bacterial cells and then the cells are separated from what is left outside through agitation, and the cells are collected in a pellet through centrifugation, in the final extraction the radioactive phosphorus ends in the pellet.


If the proteins are labeled with radioactive sulfur, and the phages infect bacterial cells and then the cells are separated from what is left outside through agitation, and the cells are collected in a pellet through centrifugation, in the final extraction the radioactive sulfur ends in the pellet.