Attached assignment has 19 questions to be answeredMAC May 2005 Paper MOLECULES, GENES AND CELLS 1...

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Attached assignment has 19 questions to be answeredMAC May 2005 Paper MOLECULES, GENES AND CELLS 1 Question Paper 	 This question paper provides you with the data and questions for sections A and B.   MOLECULES, GENES AND CELLS 1 Section A This section will be marked out of 100. Answer all ten questions [A1 - A10]. 1  105 yeast cells were added to a culture flask in medium containing adequate nutrients for maximal growth. The cells were harvested 12 hours later and the number of cells counted. At this point the number of cells was 6.4  106. A1.  Using the information given above, show that the time taken for the cells to double in number (the population doubling time) is 2 hours.  (10 marks) A2.  The population doubling time of cells is often used as an estimate of the cell cycle time. Give two possible reasons why this calculation might give an inaccurate estimate of the cell cycle time, and explain why.  (10 marks) A3.  What is meant by the terms ‘G1’, ‘S’, ‘G2’, ‘M’ and ‘cytokinesis’ when referring to phases of the cell cycle?  Briefly explain the events occurring at each stage.  (10 marks) For the rest of Section A, you should assume that the population doubling time is equal to the cell cycle time. Yeast cells are easy to grow and observe under the microscope. An experiment was set up to measure the volume of yeast cells in culture over a long period. The average cell volume was then plotted as a function of time. The graph obtained is shown below. Average cell volume (arbitrary units) A4.  Describe what happens to the average cell volume over time. Suggest a reason for the periodic sharp drops in cell volume.  (10 marks) To investigate how nutrients affect the cell cycle, the researcher then measured the cell volume over time in medium that contained a reduced supply of nutrients. The graph of average cell volume against time now looks like this: Average cell volume (arbitrary units) A5.  Describe what has happened to the rate of cell growth and the cell cycle time compared with growth with adequate nutrients.  (5 marks) The researcher has several strains of yeast available, including some that may contain mutations in genes affecting the cell cycle. In one of these strains, named mutant strain 1, the researcher notices the following growth pattern in reduced nutrient conditions: Average cell volume (arbitrary units) A6.  Describe what happens to cell volume and cell cycle time in this mutant strain. What aspect of the cell cycle might the mutation affect?  (10 marks) A7.  What adverse consequences on cell viability might result from this mutation?  (10 marks) To investigate the regulation of the cell cycle in mutant strain 1, the researcher extracts all protein from wild type and mutant cells at both the G1 and G2 phases of the cell cycle. He runs the extracted proteins on an SDS-polyacrylamide gel, and treats the gel such that it shows the presence of a single protein, which he names ‘protein X’. The resulting gel is shown below: Wild type Strain G1 Wild type strain G2 Mutant strain 1 G1 Mutant strain 1 G2 Molecular weight (kDa) 120 -                 100 - A8.  What is the size of Protein X in the cells indicated below? Complete the table on the answer sheet.  (10 marks) 	Cells in G1 	Cells in G2  [wild type] 	Cells in G2  [mutant strain 1] A9.  What, in molecular terms, could account for this size difference?   (10 marks) A10.  Based on the information given, and your knowledge of the proteins that regulate the cell cycle, suggest a possible identity for Protein X, and what its activity might be.  (15 marks) 	 MOLECULES, GENES AND CELLS 1 Section B This section will be marked out of 100. Answer all nine questions [B1-B9]. Collagen is the most abundant protein in humans; it is the major fibrous protein in the extracellular matrix and in connective tissue, providing structural support and great tensile strength. It is the main component of bone and teeth; along with keratin it is also responsible for providing strength and elasticity to our skin. The amino acid sequence of collagen largely contains repeats of the three amino acids —Gly—X—Pro— where X can be any amino acid. Post-translational modification of several amino acids is required for the formation of mature collagen molecules. One such modification is the hydroxylation of proline residues to 4-hydroxyproline. This reaction is catalysed by the enzyme prolyl 4-hydroxylase and requires two cofactors, Fe2+ and ascorbate (vitamin C). prolyl hydroxylase (Fe2+, ascorbate)      proline                      2-oxoglutarate                            4-hydroxyproline            succinate The influence of co-factors on prolyl 4-hydroxylase function Experiment 1 This experiment investigates the effect of two cofactors, ascorbate and Fe2+, on the hydroxylation of proline residues in proteins. Three separate cultures of HeLa cells derived from human tissue were grown in medium containing 14C-proline (radio-labelled proline), which will be incorporated into newly synthesised proteins, and either: Culture X:  0.5 mM 2.2′-bipyridine (used to remove any free Fe2+ in solution) Culture Y:  0.1 mM sodium ascorbate Culture Z:  No 2.2′-bipyridine or sodium ascorbate added (control sample) After one hour, 1.0 ml aliquots of each of the X, Y and Z cell suspensions were sonicated to lyse the cells then treated with ammonium sulphate to precipitate all of the proteins. The samples were then centrifuged, the supernatants removed and discarded, and the precipitated proteins were treated with HCl to hydrolyse the peptide bonds, resulting in free amino acids. The amino acids in each of the extracts X, Y and Z were then separated by anion exchange chromatography. Samples were collected from the column at one minute intervals. A pure sample of 14C-proline was applied to the same anion exchange column under identical conditions and samples were collected in the same way; this was then repeated with 14C-hydroxyproline (14C-Hyp). Table 1 shows the radioactivity, in counts per minute (cpm) per sample, detected in samples eluted from the column at the times shown. Table 1. Radioactivity levels in samples eluted from the anion exchange column 	Elution Time (min) 	 Radioactivity (cpm) 	 	Extract X 	Extract Y 	Extract Z 	14C-proline 	14C-Hyp 	10 	3

Varun · Accepted Answer

MAC May 2005 Paper
 
Section A
A1 [10 marks]
	
Ans. Given, No [Initial number cells] = 1× 105
	         Nt [Final number cells after time t] = 6.4 × 105
                   t = 12 h
The doubling time formula is used formula to justify the given situation
 	     Nt = No [1/2]t/td
                    6.4 × 105 = 1× 105 [1/2]12/td 
                    6.4 × 105 / 1× 105 = [1/2]12/td 
                [1/2]6 = [1/2]12/td
	    6 = 12/td
                td = 2
Hence, proved that the doubling time is equal to 2
A2 [10 marks]
	
Ans 2. More often the doubling time of the population is used to estimate the accurate cell cycle time. The molecular state of the cell while going through the cell cycle can be predicted based on a doubling time calculation. The number of actively dividing cells and cells completed the cell cycle can be calculated using doubling time. The main reason for this estimation are discussed herewith:
Auxiliary distribution: The main auxiliary distributions for the cell volume of diving cells and new-born cells can be determined. It also provides an accurate and detailed evaluation of growth rate, growth pattern, and physiological factors affecting the cell cycle which further can be employed to infer the stage of the dividing cells.
Unbiased distribution of size: Doubling time uses apertures that are too small to function properly. The use of these short apertures can prevent serious loss of resolutions.
A3 [10 marks]
	
The cell cycle, or cell-division cycle, is the series of occasions that happen in a cell that causes it to separate into daughters cells. The eukaryotic cell cycle comprises four distinctive stages:  G1 stage, S stage, G2 stage (aggregately known as interphase), and M stage (mitosis and cytokinesis). M stage is itself made out of two firmly coupled cycles: mitosis, in which the cell's core partitions, and cytokinesis, in which the cell's cytoplasm separated framing daughter’s cells. Initiation of each stage is reliant upon the finishing of the past event. Cells that have not separated are said to have entered a condition of resting called the G0 stage. 
The different stages of the cell cycle are:
G-1: It is the first stage of the interphase also called the growth phase. In this phase cell stores energy by attaining the required number of mitochondria and other cell organelles.
S-phase: It is also called the synthesis phase, in phase protein synthesis and DNA replication take place.
G-2: After DNA synthesis, cells rapidly grow in this phase to attain the perfect size to divide. DNA repair mechanism occurred in this phase.
M-phase: It is also called Karyokinesis, it is a short phase including five sub-stages
Prophase
Prometaphase
Metaphase
Anaphase
Telophase
Cytokinesis: In this phase nucleus, cytoplasm, organelle, cell membrane are divided into two equal cells.
A4 [10 marks]
	
Ans 4. The presented graph stated the sharp drop of cell volume after a certain interval. The cell volume increased exponentially up to 2 hours and afterward sharp drop was observed in cell volume following again an increase in cell volume up to 4 hours. The probable reason for such kind of growth pattern is discussed herewith:
Reduced supply of growth medium:

Sun	Mon	Tue	Wed	Thu	Fri	Sat
30	31	1	2	3	4	5
6	7	8	9	10	11	12
13	14	15	16	17	18	19
20	21	22	23	24	25	26
27	28	29	30	1	2	3

MAC May 2005 Paper MOLECULES, GENES AND CELLS 1 Question Paper This question paper provides you with the data and questions for sections A and B. MOLECULES, GENES AND CELLS 1 Section A This section...

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