PAPER 3 BIOLOGY EXAM QS Fully Solved Accurate For Guaranteed Pass Graded A+.

Q1 a] The onion plant, Allium cepa, is grown as a food crop around the world. The table below contains statements about the root cells of an onion. State in the boxes in the table whether the statements are true or false. - correct answer Statement about onion root cells - contain chloroplasts: *false* - contain mitochondria: *true* - contain 70S ribosomes in the cytoplasm: *false* - have pili: *false* - have cellulose cell walls: *true* Q1 b] Fig. 1 shows a cross section of the root of an onion plant. [refer to the PMT June 2018 paper 3 QP for Fig. 1] Identify the tissues shown at M and N. - correct answer M = xylem N = phloem Q2 [info] - correct answer The colour of onion bulbs is determined by two genes, A/a and B/b. • A is a dominant allele and codes for the production of a red pigment. • Onion bulbs that are homozygous for the recessive allele, a, produce no pigment and are white. • B is a dominant allele that inhibits the expression of allele A. • The recessive allele, b, allows the production of the red pigment. A white onion plant was cross-pollinated with a red onion plant. All 15 offspring had the genotype AaBb. Q2 a] Identify the genotype of the white onion plant - correct answer aaBB Q2 b] Identify the genotype of the red onion plant - correct answer AAbb Q2 c] Identify the phenotype of the offspring - correct answer All are white onion plants - their genotypes are all AaBb - the dominant allele B inhibits the dominant allele A - they do not produce the red pigment Q2 d] State the type of gene interaction shown by the genes A/a and B/b. - correct answer Dominant epistasis Q2 e] Suggest how allele B inhibits the expression of allele A [2 marks] - correct answer - B codes for the repressor protein - The repressor protein binds to the promoter (of A) - The repressor protein stops the transcription and translation of allele A - Synthesis of the red pigment does not occur Q3 ATP can be produced in various ways. Each stage of respiration contributes to the production of ATP. Describe the production of ATP by substrate-level phosphorylation in different stages of respiration with reference to the number of ATP molecules produced [6 marks] - correct answer - ATP is produced in glycolysis - The glucose is phosphorylated by 2 molecules of ATP to form hexose-1,6-bisphosphate - hexose-1,6-bisphosphate then splits into two molecules of triose phosphate - the two molecules of triose phosphate are then dehydrogenated [each one reducing two molecules of NAD to NADH] and dephosphorylated [each one phosphorylating 2 ADP molecules to ATP] - There is a total production of 4x ATP molecules and a net production of 2x ATP molecules - ATP is produced in the Krebs Cycle - 1 ATP produced per turn of the Krebs cycle - This happens when the 5-carbon compound is converted to the 4-carbon compound by decarboxylation and dephosphorylation Q4 [info] - correct answer Glucose and other carbohydrates are present in respiring cells. The concentrations of carbohydrate molecules vary between tissues. A student conducted tests on three tissues, A, B and C. Table 2 shows the results of these tests [refer to the PMT June 2018 paper 3 QP for the table] Q4 Two of the tissues were known to be phloem tissue and liver tissue. Use the evidence in Table 2 to identify which tissue, A, B or C, is phloem and which tissue is liver. Explain your answer [3 marks] - correct answer Tissue B must be phloem because: - there is starch present - there are non reducing sugars [sucrose] - low amounts of reducing sugars present, as sucrose, hydrolysed to monosaccharides Tissue C must be liver because: - there are reducing sugars [glucose] present - there are non reducing sugars [glycogen] present - but no starch present as it is an animal cell Q5 a] Cells can use fatty acids instead of carbohydrates as respiratory substrates. A process called beta oxidation is used to break down fatty acids to acetyl CoA for use in respiration. Fig. 2 shows a simplified example of beta oxidation. [refer to the PMT June 2018 paper 3 QP for Fig. 2] Using the information in Fig. 2, calculate the percentage of carbon atoms in the fatty acid that are able to enter the Krebs cycle. - correct answer START [pre beta - oxidation]: [CH3(CH2)14 COOH] there are 16x C in this fatty acid END = Acetyl CoA 1. Acetyl CoA = 2C 2. 2 / 16 = 1 / 8 = 0.125 3. 0.125 x 100 = *12.5%* Q5 b] The percentage of carbon atoms that a reaction makes available for use in the Krebs cycle can be described as the efficiency of the reaction. Calculate the efficiency of the link reaction. Using your answer to part a], state whether the link reaction is more, less or equally efficient when compared to the reactions described in Fig. 2. Show your working. - correct answer START = pyruvate [3C] END = Acetyl CoA [2C] 1. 2 / 3 = 0.6666667 2. 0.666667 x 100 = 66.666666% 3. 3sf, so *66.7 %* The link reaction is more efficient than the reaction in Fig. 2. Q5 c] Fig. 2 shows the role of coenzyme A in beta oxidation. Suggest a role for coenzymes other than coenzyme A in beta oxidation [1 mark] - correct answer NAD accepts hydrogen atoms and becomes NADH