Nov 17 H2 P3 ans print updated 2023

Raffles Institution Nov 2017 (H2 Biology) Paper 3 2023 Nov 2017 H2 Bio Paper 3 N17P3Q1 (a) (i) State what is meant by a genetic disease and explain how genetic diseases are caused. [4] 1. Disease is a health / physiological impairment as a result of; 2. Mutation* in the DNA; 3. which could have occurred in the individual with the disease , or inherited from parents; 4. The mutation can be in the form of a nucleotide substitution, insertion or deletion; 5. which would cause a change in the mRNA codon and subsequently, a change in the amino acid in the protein resulting in a loss-of-function; 6. Or could have resulted in a premature stop codon, producing a truncated protein; 7. Mutations could also be at the chromosomal level, where there is a change in the number of chromosomes or chromosomal translocation, duplication, deleti on or inversion; (ii) Justify the claim that the PKU phenotype is the result of genotype and the environment acting together. [2] 1. The PKU phenotype would only be observed in individuals that are homozygous recessive for the disease allele (genotype); 2. and when phenylalanine is present in the diet of the individual (environment); (iii) Describe two ways in which these molecules are similar in structure and explain why these features are important for their function. [4] Both molecules have 1. an amino group (NH2/NH3+); 2. a carboxyl group (COOH/COO-) Importance to function: 3. to participate in condensation reaction; 4. to form peptide bonds in the production of proteins; (b) (i) Suggest, in outline, a procedure that could determine whether or not a PKU allele is present. [4] 1. Extract genomic DNA from cells obtained in the blood / from a mouth swab; 2. Carry out polymerase chain reaction (PCR)* using primers* complementary to regions flanking the PKU allele; 3. Cut the amplified fragment using restriction enzyme*; 4. Carry out gel electrophoresis* to separate the fragments according to size, 5. Carry out Southern blot* and use radioactive probes complementary to the PKU allele to detect specific fragments 6. Carry out RFLP analysis: compare with restriction patterns for known genotypes to determine unknown genotype; A: after PCR – carry out DNA sequencing & compare with known DNA sequences of the normal allele and the PKU allele

Raffles Institution Nov 2017 (H2 Biology) Paper 3 2023 (ii) Predict and explain the effect of this treatment on the result of the Guthrie assay for a baby that is homozygous for the PKU allele. [3] 3. The results of the test would be a false negative , showing wrongly that the baby does not have PKU; 4. No colony / reduced colony diameter would be observed; 5. Despite high levels of phenylalanine in baby’s blood, which should have allowed for bacterial growth; 6. Bacteria growth was inhibited / bacteria was killed by the antibiotics present in the blood; (iii) Suggest how β-2-thienylalanine inhibits bacterial growth. [3] 1. β-2-thienylalanine competes with phenylalanine for the active site* of; 2. Aminoacyl-tRNA synthetase*, that joins phenylalanine* to its tRNA; 3. β-2-thienylalanine is incorporated into polypeptides instead of phenylalanine, resulting in a change in the primary structure / amino acid sequence; 4. Tertiary structure / 3D conformation of the protein is thus changed; 5. producing non -functional bacterial proteins which resulted in bacterial growth being inhibitied; (c) With reference to Fig. 1.1, and the information given, explain how loss of function of enzyme Q causes poor intellectual and behavioural development as well as very fair skin and hair. [4] 1. Loss of function of enzyme Q stops conversion of phenylalanine* to tyrosine*; Less tyrosine results in: 2. less thyroxine* which leads to poor brain development; 3. less melanin* which causes fair skin and hair; 4. less dopamine* which results in poor transmission of nerve impulses; (d) (i) In Scandinavia, 1 in 140 people are carriers of the PKU allele. Complete Table 1.1 by calculating the frequency of newborn babies positive for PKU that would be expected in Scandinavia. You should show your working. [2] P(PKU) = P(Carrier parent 1) X P(Carrier parent 2) X P(child affected) =1/140 x 1/140 x ¼ = 1/ 78400 or one in 78 400 (ii) Explain how evolution could have resulted in PKU being more common in the Northern European population compared to that of sub-Saharan Africa. [4] 1. The frequency of PKU in cold and wet, Northern Europe is the highest in the world where 1 in10 000 are sufferers while sub-Saharan Africa which is hot and dry is lower by 10 times; (Quote data) 2. This is because fungi thrive in cold and wet climates in North Europe but not in sub- Saharan Africa which is hot and dry. This means increased likelihood of renal cancer in North Europe as the fungi produces, ochratoxin A; (link climate to fungi to cancer) 3. Homozygotes for the normal PKU alleles are selected against in North Europe as they are not protected against ochratoxin A; 4. Homozygotes of PKU alleles are selected against because they have metabolic deficiencies that result in poor intellectual and behavioural development; 5. When both selection pressures are applied, as happens in North Europe, the heterozygotes are selected for as they resistant to ochratoxin A while they do not suffer the developmental deficiencies of PKU; 6. This is called heterozygote advantage *, a form of balancing selection where both alleles are selected for under specific circumstances found in North Europe but not found in Africa; N17P3Q2

Raffles Institution Nov 2017 (H2 Biology) Paper 3 2023 (a) Explain the normal function of blood stem cells. [4] 1. Blood stem cells are unspecialized cells which are multipotent*; (Reject- toti/ pluri) 2. Ability to self-renew* and proliferate* 3. to maintain a constant pool of stem cells; 4. Ability to differentiate* into specialized cells in the blood, eg red blood cell, ly mphocytes (name 1 eg) 5. to replace dead cells that died; (b) Outline the roles of the named T lymphocytes in fighting infection. [4] 1. The named T lymphocytes are T helper cells / lymphocytes; 2. A particular naïve T cell* will have T cell receptors (TCR) that can specifically recognise the peptide of peptide:MHC complex on the antigen presenting cell (APC; 3. The APC secretes cytokines that will activate the naïve T cells which will undergo clonal expansion and differentiation to form effector and memory T cells ; 4. T helper cells* secrete cytokines that stimulate/activate specific naïve B cells* to become antibody-secreting plasma cells*; 5. Memory T cells* when re-exposed to the same pathogen/antigen, will recognize it and mount a faster and stronger secondary immune response; (c) Comment on the ethical aspects of this new therapy. [2] 1. Unforeseen circumstances from T cells lacking CCR5 / immune response might be affected if T cells do not have CCR5 receptor; 2. Unintended changes in DNA as a result of gene editing may contribute to development of cancer; 3. Blood stem cells are multipotent and hence cannot form a whole organism, hence no debate of whether or not a life is “sacrificed”; 4. Potential to cure patients completely so patients no longer have to be on lifetime anti -viral therapy; 5. AVP eg. Accessibility to treatment N17P3Q3 (a) Outline how photosynthesis produces triose phosphate. [4] 1. During carbon fixation stage, CO 2 is combined with ribulose bisphosphate (RuBP )*, catalysing this reaction