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Learn/ 12th Class Biology/ Unit 18 /Numericals

Unit 18: Inheritance — Numericals

12th Class Biology · Unit 18: Inheritance

Q2.15.i.Solve the following genetic problem: A plant of genotype AABbCC is self-pollinated. Calculate the phenotypic ratio of F2 generation.

Data Parent genotype = AABbCC (self-pollinated). Genes A and C are homozygous (AA, CC) so they do not segregate — all offspring remain AA and CC for those genes. Only gene B is heterozygous (Bb) and will segregate. Self-cross: Bb × Bb → 1 BB : 2 Bb : 1 bb (genotypic), which gives a 3 (B_) : 1 (bb) phenotypic ratio for that trait. Result: The F2 phenotypic ratio is 3 : 1 (3/4 showing the dominant B-trait phenotype : 1/4 showing the recessive b-trait phenotype), with the A and C traits unchanged in all offspring (no segregation occurs for homozygous genes).

Q2.15.ii.In four-o'clock plant, the red and white flower colours show incomplete dominance. Calculate the ratios in crosses between (i) two red-flowered plants; (ii) a red-flowered and a pink-flowered plant; (iii) a pink-flowered and a white-flowered plant.

Alleles R (red) and W (white) show incomplete dominance; RR = red, RW = pink, WW = white.
(i) RR × RR → all offspring RR → all red flowers (ratio: all red, 1:0:0).
(ii) RR (red) × RW (pink) → gametes R/R from red parent and R/W from pink parent → offspring 1/2 RR (red) : 1/2 RW (pink) → ratio 1 red : 1 pink.
(iii) RW (pink) × WW (white) → gametes R/W from pink parent and W/W from white parent → offspring 1/2 RW (pink) : 1/2 WW (white) → ratio 1 pink : 1 white.

Q2.15.iii.A man with blood group A marries a woman who has blood group AB. What blood groups are expected among their children?
Q2.15.iv.Two white sheep produce a black offspring. What are the parents' genotypes for the colour? What is the probability that their next offspring will be black?

Data White × White → one black offspring appeared, so white must be the dominant phenotype (W) and black the recessive phenotype (w), with both white parents being heterozygous carriers (Ww), since a cross of two homozygous dominant (WW) parents could never produce a recessive (black) offspring.
Parents' genotypes: Ww × Ww.
Punnett square: Ww × Ww → 1 WW : 2 Ww : 1 ww → phenotypic ratio 3 white : 1 black.
Result: Probability that the next offspring will be black = 1/4 (25%).

Q2.15.v.Albinism (lack of pigment) in man is caused by a recessive gene. If normal parents have an albino child, what is the probability that their next child will be normal for colour?

Data Let A = normal (dominant), a = albino (recessive). Both parents are phenotypically normal but had an albino (aa) child, so both parents must be heterozygous carriers: Aa × Aa.
Punnett square: Aa × Aa → 1 AA : 2 Aa : 1 aa → phenotypic ratio 3 normal (AA or Aa) : 1 albino (aa).
Result: Probability that the next child will be normal for colour = 3/4 (75%).

Q4.2.Use a dice to calculate how many times out of 100 throws can you get sixes.

Data A fair six-sided die has 6 equally likely outcomes, so the probability of rolling a six on any one throw = 1/6. Number of throws = 100.
Formula: Expected number of sixes = probability × number of trials = (1/6) × 100.
Calculation: (1/6) × 100 = 16.67.
Result: One would expect to get a six approximately 17 times out of 100 throws (theoretical expectation ≈ 16–17 times); the actual experimental count may vary somewhat around this value due to chance.