In compliance with the Dartmouth honor principle, all the work you hand in on this exam is to be your own. Please remember to be precise in your wording - scientific descriptions rely on accurate use of specific terms. Also, try to keep your answers concise. If you can say something briefly there is no need to create a lengthy answer just to fill up space. Use the space provided and the point values for each question as indicators of the amount of detail your answer should contain. If you really need extra space for an answer, turn the page over and continue your answer on the back of the same piece of paper. To facilitate grading, we separate the exam by page; if your answer is on a different sheet of paper we will not see it. Please put your name on each page now.

If something is not clear to you, please ask me during the exam. That is why I stay in the room. Good luck. - Prof. Gross

In the video "Burden of Knowledge" that we saw in class, a number of families were followed as they struggled to make difficult decisions. The video centered on whether one wants additional genetic (or other prenatal) knowledge about a future child, and what to do with that knowledge once it becomes available. Discuss why sometimes this knowledge is a burden and other times a benefit. Is it always best to have the knowledge and then make a decision? or are there some situations where it is not appropriate to gain the knowledge? 16 points

(you can use the back of this page if you absolutely must, but please try to keep your answer brief)

What is the difference between genotype and phenotype? 4 points

Genotype is the genetic makeup of the individual, while phenotype is the physical appearance or some other measurable characteristic of the individual.

The figure at the right shows the results of a pulse-chase experiment in which cells were exposed briefly to a radioactive amino acid and then "chased" for the times indicated

a. What is meant by a pulse-chase experiment? 3 points

This means that in an experimental system, there is a brief exposure to some labeling condition (the pulse), followed by removal of the label and observation of the material labeled during the pulse (this is the chase period). This is used to follow the fate of the labeled material.

b. What does the result of this experiment (see figure) say about the path that proteins take on their way to being secreted from the cell? a=after 3 minute pulse with labeled amino acid, b=after 17 minute chase, c=after 117 minute chase. Explain your answer and name the organelles (cell components) involved. 7 points

Initially (after the pulse) the labeling occurs over the rough endoplasmic reticulum. This indicates that it is the site of protein synthesis in this cell. After a 17 minute chase, much of the label has moved to the Golgi complex. Since the cell is not synthesizing any additional labeled protein at this point in time, the label must be due to the original proteins synthesized during the pulse. This indicates that the proteins synthesized in the RER originally move to the Golgi complex next. At 117 minutes of chase, the proteins have moved from the RER through the Golgi and into the secretory vesicles for secretion from the cell. This experiment indicates that the path for protein secretion is from the RER to the Golgi to the secretory vesicles and then out of the cell

You are given a pea plant that has round seeds. How can you test this plant to determine if it is heterozygous or homozygous for round vs. wrinkled peas? 4 points

Since round is dominant, these peas must be either RR, homozygous, or Rr, heterozygous. The way to distinguish these two possibilities is to perform a self-fertilization and examine the offspring. If the plant is RR, non of the offspring will be wrinkled (rr). If the plant is Rr, then approximately 25% of the offspring will be wrinkled (rr).

Why is inbreeding (incest) a genetically bad idea? 4 points

We are all heterozygous for anywhere from 5 to 15 deleterious traits. In a random mating, it is extremely unlikely that our mate would have the same deleterious mutation and therefore the incidence of an affected child is very small. In an incestuous mating, since both parents have the same ancestors, it is likely that they also contain the same set of mutations. This makes it much more likely that any offspring will be affected.

PKU is a genetic disorder that can be treated successfully by special diet. Explain how this special diet works and what unexpected consequences there were as a result of the diet when it was implemented in the 1970s. 6 points

PKU is the result of a deficiency in the ability to metabolize the amino acid phenylalanine (Phe). This results in toxic levels of Phe in the bloodstream and can produce severe and debilitating effects on the PKU individual. PKU can be successfully treated by keeping the patient on a diet very low in Phe. This prevents toxic levels from accumulating, although Phe levels are higher than normal. The success of this treatment resulted in functional adults afflicted with PKU. For PKU women, their increased abilities and more sophisticated social interactions led to pregnancies. All children born to PKU mothers are debilitated because of their exposure to the elevated Phe levels during gestation.

Briefly describe the roles of the following cellular components. 5 points

a. mitochondria

organelle that makes energy for the rest of the cell (in the form of ATP)

b. ribosomes

organelle that is the site of protein synthesis; it brings together the components of protein synthesis

c. lysosomes

vesicles that contain digestive enzymes; these vesicles break down food into smaller subunits that can be recycled within the cell.

d. rough endoplasmic reticulum

RER carries out a number of post-translational modifications to proteins.

e. cell membrane

serves as a barrier between the inside and outside of the cell. As such it regulates what enters and leaves the cell and protects the cell from its environment.

Usually, a single region of DNA contains coding information on only one strand. In a few (rare) cases, typically in viruses, there might be overlapping coding regions, one on each strand. Give one reason why this might ba an advantage to the virus. Give a reason why it might be a disadvantage. 6 points

The advantage is that the virus can store more genes in a smaller amount of DNA. Thus it does not have to replicate as much DNA and the additional DNA can make it more "fit" to survive.

The disadvantage is that a single mutation can disrupt two genes at once. Another possible answer is that it would be difficult for RNA polymerases to transcribe both genes at the same time (they would interfere with each other).

Griffith discovered a "transforming principle" in pneumonia bacteria in 1928. How were Avery, MacLeod, and McCarty able to identify the composition of Griffith's transforming principle in 1944? 5 points

The transforming principle was found in lysates of smooth pneumonia bacteria. These lysates could transform rough pneumonia bacteria into the infectious smooth bacteria. To determine what the factor was, Avery et al treated the lysate with either a protease to destroy proteins or with a DNAase to destroy DNA in the lysate. They observed that protease treatment had no effect on the transforming ability, but DNAase eliminated it. This indicated that DNA was the transforming principle.

Describe how it was determined that DNA replication is semi-conservative (you may use a labeled diagram to illustrate your answer if you wish). 8 points

This was done using a pulse-chase approach. DNA was labeled with heavy nitrogen for many generations and then chased in light nitrogen. At the start, all DNA consisted of two heavy strands. After one generation the DNA had a density halfway between heavy and light which ruled out a conservative model of replication. In the second generation, the DNA consisted of three quarters all light and one quarter at a density of 50% light - 50% heavy. This ruled out a dispersive model and left the semi-conservative model as the only viable explanation.

Explain (or diagram) how the prokaryotic core RNA polymerase interacts with other factor(s) to bring about correct initiation of transcription in prokaryotic cells. 6 points

Core RNA polymerase cannot initiate transcription efficiently or accurately by itself. Another factor, sigma, interacts with the core to form a holoenzyme which can then recognize the promoter for transcription and bring the core enzyme to that location. Once bound to the DNA, the core enzyme begins transcription and the sigma factor is released.

Draw a basic tRNA structure and indicate where the anticodon and amino acid binding sites are located. 4 points

a cloverleaf structure with the amino acid attached to the 3' end of the RNA and the anticodon loop at the "opposite" end of the cloverleaf.

Explain how it is possible to have two different DNA sequences code for an identical protein sequence. 4 points

The genetic code is redundant so it is possible to have more than one codon specify the same amino acid. Thus, it is possible for two different DNAs to specify the same protein by using different synonymous codons to specify the same amino acid.

Antibiotics often work by interfering with the process of translation. Erythromycin, for example blocks chain elongation, while Linezolid blocks initiation. You are doing an experiment to follow the rate at which amino acids are being incorporated into proteins within a cell. You divide your bacteria into three test tubes - test-tube A is a control (nothing added), test-tube B has erythromycin added, and test-tube C has linezolid added. I have shown you a plot for test-tube A. Add a curve to this graph for what you would observe in test-tubes B and C and explain your plots. 8 points

B: erythromycin blocks elongation so all synthesis stops immediately.

C: linezolid blocks initiation, but those ribosomes already enganged in protein synthesis will continue to synthesis protein until they reach the ends of the mRNAs they are translating. Thus, protein synthesis will gradually taper off as the ribosomes finish their jobs but cannot reinitiate synthesis.

DNA replication in eukaryotes proceeds at approximately 100 nts/sec.

a. Why is this about 10 times slower than prokaryotic replication? 2 points

Eukaryotic DNA is in the form of chromatin. The chromatin proteins will interfere with the ability of the replication machinery to rapidly replicate the DNA.

b. How can an entire human genome of 6 x 109 basepairs be replicated in an hour? 2 points

Instead of a single origin of replication (as found in prokaryotes), there are thousands of origins. The DNA is therefore replicated in parallel at thousands of locations in the genome.

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last modified Wed, Apr 28, 2004; 8:58 PM

Exam#1 Answer Key