The Cell Cycle

Yet another revision post!

Q1: Since there are about 10^13 cells in an adult human, and about 10^10 cells die and are replaced each day, we become new people every three years. (True/False, explain why)

This almost seems more like a philosophical question than a biological question...

Anyway my answer is false, since not all of the cells in our body die and are replaced. For example, skeletal muscle cells and neurons do not get replaced. The latter is particularly significant because we often think of our brains as making us who we are.

Q2: In order for proliferating cells to maintain a relatively constant size, the length of the cell cycle must match the time it takes for the cell to double in size. (True/False, explain why)

Yes, it is true that everything within the cell has to be doubled before cell division so that all daughter cells will be of a "normal" size.

Q3: While other proteins come and go during the cell cycle, the proteins of the origin recognition complex remain bound to the DNA throughout. (True/False, explain why)

I think this is true. To my understanding, origin recognition complexes remain bound to the DNA throughout, but they require the proteins of the pre-replicative complex (preRC) to allow initiation of replication to take place.

Q4: Chromosomes are positioned on the metaphase plate by equal and opposite forces that pull them toward the two poles of the spindle. (True/False, explain why)

Yes, that is indeed how they are aligned at the spindle. Once at the metaphase plate, chromosomes often oscillate gently as they are tugged back and forth by the two spindle poles.

Q5: Meiosis segregates the paternal homologs into sperm and the maternal homologs into eggs. (True/False, explain why)

This statement is false. During the first meiotic division, the paternal and maternal homologs cross over before being pulled apart into different cells, and hence chromosomes contain a mixture of paternal and maternal DNA. In males, all of the daughter cells in meiosis become spermatids that can undergo spermiogenesis to become sperm, while in females, one daughter cell becomes an egg while the others degenerate. This process takes place regardless of whether the chromosomes within the sperm or eggs originated from the father or the mother.

Q6: If we could turn on telomerase activity in all our cells, we could prevent ageing. (True/False, explain why)

Probably false. In any case, turning on telomerase may instead cause cancer- many cancer cells have the ability to produce telomerase and therefore stop cells from undergoing replicative cell senescence (i.e. not replicating after a certain number of cycles).

Q7: Many cell-cycle genes from human cells function perfectly well when expressed in yeast cells. Why do you suppose that is considered remarkable? After all, many human genes encoding enzymes for metabolic reactions also function in yeast, and no one thinks that is remarkable.

I'm just going to take a wild stab in the dark here. I think that this is possibly due to humans being more complex organisms than yeast. Humans have a variety of different cells that have different requirements with regards to the cell cycle, whereas yeast does not. Hence perhaps it is remarkable that so many human cell-cycle genes are still able to function in yeast.

Q8: Hoechst 33342 is a membrane-permeant dye that fluoresces when it binds to DNA. When a population of cells is incubated briefly with Hoechst dye and then sorted in a flow cytometer, which measures the fluorescence of each cell, the cells display various levels of fluorescence as shown in the figure.

(Description of figure: relative fluorescence per cell on x-axis (0, 1, 2), number of cells on y-axis (no scale). There is a large peak at relative fluorescence of 1 and a smaller peak at 2. There are some cells in between the two as well.)

A. Which cells in the figure are in the G1, S, G2 and M phases of the cell cycle? Explain the basis for your answer.

At a relative fluorescence of 1, there is relatively little DNA in the cell (i.e. the DNA has not replicated yet). Hence I would say that cells with a relative fluorescence of 1 are in the G1 stage.

Between relative fluorescence levels of 1 and 2, there is between the normal amount and twice the amount of DNA in each cell. The cells here are probably in S phase, where the DNA is beginning to replicate but the replication process is not quite complete.

At a relative fluorescence of 2, the DNA has already doubled. Hence, I would say that cells with this fluorescence are in the G2 and M stages.

B. Sketch the sorting distributions you would expect for cells that were treated with inhibitors that block the cell cycle in the G1, S or M phase. Explain your reasoning.

I can't be bothered drawing anything, so I'll just describe here.

If the cell cycle was blocked in the G1 phase, the cell would be stuck with the base amount of DNA. Hence all of the cells would have a relative fluorescence of 1.

If the cell cycle was blocked in S phase, the cell would either be stuck with the base amount of DNA (if it was stuck in the beginning of S phase), twice the amount of DNA (if it was stuck at the end of S phase), or somewhere in between. I'm not entirely sure what the relative fluorescence of the cells would be- maybe most cells would be in between.

If the cell cycle was blocked in M phase, the cell would be stuck with twice as much DNA. Hence all of the cells would have a relative fluorescence of 2.

Q9: The yeast cohesin subunit Scc1, which is essential for sister-chromatid cohesion, can be artificially regulated for expression at any point in the cell cycle. If expression is turned on at the beginning of S phase, all the cells divide satisfactorily and survive. By contrast, if Scc1 expression is turned on only after S phase is completed, the cells fail to divide and they die, even though Scc1 accumulates in the nucleus and interacts efficiently with chromosomes. Why do you suppose that cohesin must be present during S phase for cells to divide normally?

I think that this is because cohesin is required to make sure that the two sister chromatids are closely joined as they are generated. If the chromatids are generated first, the lack of cohesin may cause them to break apart, and adding cohesin to them at this point may result in chromatids from one chromosome joining to chromatids from other chromosomes. Haphazard joining of chromatids would cause cells to be unable to divide normally.

Q10: High doses of caffeine interfere with the DNA damage response in mammalian cells. Why then do you suppose the Surgeon General has not yet issued an appropriate warning to heavy coffee and cola drinkers? A typical cup of coffee (150mL) contains 100mg of caffeine (196g/mole). How many cups of coffee would you have to drink to reach the dose (10mM) required to interfere with the DNA damage response? (A typical adult contains about 40 litres of water.)

40 litres x 10mM = 400 millimoles of caffeine are required to interfere with the DNA damage response.

0.1g caffeine in a cup of coffee / 0.196 g/mmol = 0.51 millimoles of caffeine in a cup of coffee (2 d.p.)

400/0.51 = 784 cups of coffee (nearest whole number).

That's a lot of coffee.

Q11: How many kinetochores are there in a human cell at mitosis?

There are 46 chromosomes in a human cell. At mitosis, each of these is composed of two sister chromatids, each with its own kinetochore. Hence there are 46 x 2 = 92 kinetochores in a human cell at mitosis.

Q12 is just a series of pictures to put in order. It's not something I can type up here so I'll just skip this one.

Q13: Down syndrome (trisomy 21) and Edwards syndrome (trisomy 18) are the most common autosomal trisomies seen in human infants. Does this fact mean that these chromosomes are the most difficult to segregate properly during meiosis?

No, it does not. The fact that Down syndrome and Edwards syndrome are the most common autosomal trisomies seen in human infants may simply mean that these are the autosomal trisomies most compatible with life. Other autosomal trisomies may occur at the same rate, but these foetuses do not make it to full term.

Q14: The human genome consists of 23 pairs of chromosomes (22 pairs of autosomes and one pair of sex chromosomes). During meiosis, the maternal and paternal sets of homologs pair, and then are separated into gametes, so that each contains 23 chromosomes. If you assume that the chromosomes in the paired homologs are randomly assorted to daughter cells, how many potential combinations of paternal and maternal homologs can be generated during meiosis? (For the purpose of this calculation, assume no recombination occurs.)

Well, there are two options for chromosome 1 (paternal or maternal), and then two for chromosome 2, etc. This gives a whopping 2^23 combinations before considering recombination.