Genetics & Heredity — Science

  1. Grade 7 — DNA, Genes & Chromosomes
  2. Grade 7 — Cell Division: Mitosis
  3. Grade 7 — Inheritance & Punnett Squares
  4. Grade 7 — Inherited vs. Acquired Traits & Mutations
  5. Grade 10 — DNA Structure & Replication

    ↑ Deepens: from 'DNA carries traits' to the molecular structure that lets it copy itself.

  6. Grade 10 — Protein Synthesis
  7. Grade 10 — Mitosis & the Cell Cycle
  8. Grade 10 — Meiosis & Sexual Reproduction
  9. Grade 10 — Mendelian Genetics

Apply it in a new setting

Problem 1

In pea plants, tall (T) is dominant to short (t), and purple flowers (P) are dominant to white (p), and the two genes assort independently. A plant heterozygous for both traits (TtPp) is test-crossed with a short, white plant (ttpp). What percent of the offspring are expected to be BOTH tall and purple?
Hints

A test cross with a fully recessive parent lets you treat each gene separately, then multiply the probabilities.

For height, Tt × tt gives what fraction tall? For flower colour, Pp × pp gives what fraction purple?

Multiply the two independent probabilities together.

Answer and explanation

25 %

Height: Tt × tt → ½ Tt (tall) and ½ tt (short), so P(tall) = ½.

Colour: Pp × pp → ½ Pp (purple) and ½ pp (white), so P(purple) = ½.

Independent traits multiply: ½ × ½ = ¼ = 25%.

Problem 2

In pea plants, round (R) is dominant to wrinkled (r) and yellow (Y) is dominant to green (y), assorting independently. Two plants heterozygous for both genes (RrYy) are crossed. What percent of the offspring are expected to be BOTH wrinkled and green (showing both recessive traits)?
Hints

Handle each gene on its own, then combine.

Rr × Rr gives ¼ wrinkled (rr); Yy × Yy gives ¼ green (yy).

Multiply the two independent fractions.

Answer and explanation

6.25 %

Seed shape: Rr × Rr → ¼ rr (wrinkled).

Seed colour: Yy × Yy → ¼ yy (green).

Both recessive: ¼ × ¼ = 1/16 = 6.25%.

Problem 3

Cystic fibrosis is caused by a recessive allele (f). Two unaffected parents who are both carriers (Ff × Ff) have children. Considering only their UNAFFECTED children, what percent of those unaffected children are expected to be carriers (Ff)?
Hints

First write the full genotype ratio from Ff × Ff.

'Unaffected' removes the ff group — you are now looking only within the FF and Ff children.

Carriers ÷ (all unaffected) = 2 ÷ (1 + 2).

Answer and explanation

66.67 %

Ff × Ff → 1 FF : 2 Ff : 1 ff. The ff child is affected; the other three are unaffected.

Among the three unaffected children, two are carriers (Ff).

2 ÷ 3 = 0.6667 = 66.67% (about two-thirds).

Problem 4

In the ABO blood system the A and B alleles are codominant and the O allele is recessive. A father has type A blood with genotype AO, and a mother has type B blood with genotype BO. What percent of their children are expected to have a blood type different from BOTH parents — that is, either type O or type AB?
Hints

Set up the cross AO × BO and list the four equally likely allele combinations.

Remember A and B are codominant (together give type AB), and OO is type O.

Add the fraction that is type AB to the fraction that is type O.

Answer and explanation

50 %

AO × BO → AB (type AB), AO (type A), BO (type B), OO (type O) — each ¼.

Type AB = ¼ and type O = ¼; the type A and type B children match a parent.

Different from both parents (O or AB) = ¼ + ¼ = ½ = 50%.

Problem 5

In snapdragons, flower colour shows incomplete dominance: RR is red, rr is white, and the heterozygote Rr is pink. Two pink snapdragons (Rr) are crossed and produce 200 offspring. How many of the offspring are expected to look DIFFERENT from their pink parents?
Hints

With incomplete dominance the heterozygote has its own colour, so the phenotype ratio is 1 : 2 : 1, not 3 : 1.

Rr × Rr → ¼ RR (red) : 2/4 Rr (pink) : ¼ rr (white). Which of these differ from pink?

Add the red and white fractions, then take that fraction of 200.

Answer and explanation

100 offspring

Rr × Rr → 1 RR (red) : 2 Rr (pink) : 1 rr (white).

Different from the pink parents = red + white = ¼ + ¼ = ½.

½ × 200 = 100 offspring.

Problem 6

Three genes assort independently. Two individuals heterozygous for all three (AaBbCc × AaBbCc) are crossed. What percent of the offspring are expected to be homozygous recessive for all three genes (aabbcc)?
Hints

Each gene behaves independently, so find the chance for one gene and extend the pattern.

For a single gene, Aa × Aa gives ¼ aa. The same ¼ applies to each of the three genes.

Multiply: ¼ × ¼ × ¼.

Answer and explanation

1.5625 %

For each gene, heterozygote × heterozygote → ¼ homozygous recessive.

Three independent genes: ¼ × ¼ × ¼ = 1/64.

1/64 = 0.015625 = 1.5625%.

Problem 7

A purple-flowered plant (purple, P, is dominant to white, p) is crossed with a white-flowered plant. The offspring come out very close to 51 purple : 49 white — about a 1 : 1 ratio. What were the most likely genotypes of the two parent plants?
  1. PP × pp
  2. Pp × pp
  3. Pp × Pp
  4. PP × Pp
Hints

The white parent must be pp (a recessive trait only shows when homozygous). Focus on what the purple parent must be.

A 1 : 1 purple : white ratio means half the offspring are white (pp). Where do a white offspring's two p alleles come from?

For half the offspring to be pp, the purple parent must be able to pass a p — so it is heterozygous.

Answer and explanation

Pp × pp

White is recessive, so both the white parent and every white offspring are pp.

Getting ½ white offspring requires the purple parent to donate p half the time, so it is Pp: Pp × pp → ½ Pp (purple) : ½ pp (white).

A 1 : 1 ratio matches Pp × pp (PP × pp and PP × Pp give all purple; Pp × Pp gives 3 : 1), therefore B.

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