8 Marking scheme: End-of-chapter test (part 1) 1 2 a Anaphase; II b Centromeres divide; chromatids pulled / move to opposite poles; by / along (spindle) microtubules. c Independent assortment; of homologous chromosomes; at metaphase I. Crossing over / chiasma formation; during prophase I; leads to exchange of genetic material between chromatids of homologous pairs. [2] [3] [Max. 4] a Recessive; brothers had thalassaemia but their parents didn’t have disease, so both parents must be heterozygous carriers / equivalent answer. b Let T = normal allele and t = recessive (thalassaemia) allele (accept other suitable symbols); father’s genotype = Tt, mother’s genotype = Tt; gametes = T , t and T , t ; [2] [1] [1] [1] Father’s gametes Mother’s gametes T t T TT Tt t Tt tt correctly drawn Punnett square; possible genotypes of offspring (cannot be tt) = TT, Tt or Tt (or ratio 1 TT to 2 Tt); probability = 23 or 0.67 or 67%; c Probability that both parents are carriers is 2 3 therefore probability of their child being tt is [1] [1] [1] or 0.67; 2 3 × 2 3 = [1] 4 9 or 0.44 or 44%. [1] d A gene carried on the X chromosome; but not on the Y chromosome. 3 [2] a The genotype is (a statement of) the alleles possessed by an organism; the phenotype is the characteristic(s) of the organism. b i short hair; long hair ii male = Ss; female = ss (or other correct symbols); iii short hair : long hair = 1 : 1 c i Let black = B (dominant) and b = white (recessive); male = SSbb, female = ssBB; so all F1 must be SsBb; F1 cross = SsBb × SsBb; gametes = SB , Sb , sB , sb and SB , Sb , sB , sb ; COAS Biology 2 Teacher Resources Original material © Cambridge University Press 2009 [2] [2] [2] [1] [1] [2] [1] [1] 1 8 Marking scheme: End-of-chapter test (part 1) Male’s gametes SB Female’s gametes SB Sb sB sb Sb sB sb SSBB SSBb SsBB SsBb short, black short, black short, black short, black SSBb SSbb SsBb Ssbb short, black short, white short, black short, white SsBB SsBb ssBB ssBb short, black short, black long, black long, black SsBb Ssbb ssBb ssbb short, black short, white long, black long, white correctly drawn Punnett square; ratio (short, black) : (short, white) : (long, black) : (long, white) = 9 : 3 : 3 : 1 (one mark for ratio, one mark for correctly matched phenotypes) ii Carry out test cross with white-haired guinea pig. If there are any white offspring, parent must be heterozygous / if all offspring are black, parent must be homozygous. 4 [1] [2] [1] [1] 1 = Abat; 2 = asas; 4 = atas; 5 = Abas; 9 = Abat or Abas; (both are required for 1 mark) b 7 = Abas and 10 = atas / cross = Abas × atas; gametes = Ab , as and at , as ; a [5] [2] Gametes of dog 7 Ab t as Abat atas black tan Abas as as black spotted a Gametes of dog 10 s a Correctly drawn Punnett square; Ratio = 2 black : 1 tan : 1 white (one mark for ratio, one mark for correctly matched phenotypes) COAS Biology 2 Teacher Resources Original material © Cambridge University Press 2009 [1] [2] 2 8 Marking scheme: End-of-chapter test (part 1) 5 a When two genes interact to affect a characteristic. b AE Ae aE AAEE AAEe AaEE AaEe red red red red AAEe AAee AaEe Aaee red red red red AaEE AaEe aaEE aaEe red red black black AaEe Aaee aaEe aaee red red black red [1] ae AE Ae aE ae c Correct gametes; correct genotypes of offspring; correct phenotypes of offspring. ¼ / 0.25. COAS Biology 2 Teacher Resources Original material © Cambridge University Press 2009 [3] [1] 3
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