Bio 4 Exam#1, Feb 2, 2000
1. At the Stem Cell Symposium on Saturday, there was a great deal of discussion about the use of adult vs. embryonic stem cells. Explain why stems cells are thought to be important medically and why there is concern over embryonic vs. adult stem cell usage in experiments designed to bring stem cell research to medical applications. 15 points
Briefly:
Stem cells are undifferentiated cells that can be caused (under some conditions) to differentiate into adult tissues. The medical potential for tissue repair or replacement is enormous including organ replacement and repair, recovery from trauma, and cancer treatments. Research needs to be done to learn how to cause the stem cells to differentiate into the tissues that are sought.
The basic concern is over the source of the stem cells. Some people believe that it is wrong to use cells from aborted fetuses or non-implanted embryos from IVF. On the other hand, adult stem cells might not be as useful in medical applications.
2. 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.
3. 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
Radioactivity is localized by the appearance of black "grains" over the location in the cell containing the radioactivity. 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 radioactive protein at this point in time, the radioactivity 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.
4. You are given a pea plant that has round seeds. How can you test this plant to determine if it is heterozygous or homozygous? 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).
5. What experiment did Mendel perform that lead him to propose his law of Independent Assortment? 5 points
Mendel conducted a series of dihybrid crosses - crosses between strains breeding true for two characters. Consider round/wrinkle (R/r) and yellow/green (Y/y) seeds. In P generation, the double dominant homozygote (RRYY). is crossed with a double recessive homozygote (rryy). All F1s therefore are RrYy and exhibit round yellow phenotype. The double heterozygotes were then used to make an F2 generation. In the F2 generation, the ratio of round: wrinkled and of yellow:green was 3:1. In other words, the presence of one trait did not impact the appearance of the other trait - they sorted independently.
6. 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.
7. 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. 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.
8. List the five different phases of mitosis and briefly describe what occurs at each stage. 10 points
interphase - chromosomes decondensed, nuclear envelope intact
prophase - nuclear envelope begins to break down and chromosomes begin to condense
metaphase - chromosomes line up on metaphase plate; spindle fibers attach at centromeres and extend to poles of cell
anaphase - centromeres split and chromosomes are moved towards poles of cell
telophase - chromosomes congregate at locations where new nuclei form; nuclear envelopes begin to form around each nucleus; cell begins to pinch in two
8. Briefly describe the roles of the following cellular components. 10 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.
10. Describe two differences between prokaryotes and eukaryotes. 4 points
Many possible answers including: nucleus in euk but not prok, organelles in euk but not prok, chromatin in euk but not prok, circular genome in prok but not euk
11. 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? 7 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.
12. 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 form 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.
13. Explain how the core RNA polymerase interacts with other factor(s) to bring about correct initiation of transcription in prokaryotic cells. 5 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 appropriate start site 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.
14. Name the three major kinds of RNAs involved in protein synthesis and describe their roles in the process. 6 points
tRNA - brings an amino acid into the protein synthesizing complex and donates it at the correct codon.
mRNA - contains the coding sequence used to direct the assembly of amino acids into a protein
rRNA - essential part of the ribosome. It holds the ribosomal proteins together and probably plays a role in formation of the peptide bond.
15. What is a nucleosome? 2 points
A nucleosome is a basic unit of chromatin structure. It contains a core of 8 histones around which is wrapped the DNA.