Dihybrid crosses, pedigrees and blood groups

Dihybrid crosses

A dihybrid cross follows two genes at once. Mendel’s law of independent assortment states that the alleles of different genes separate independently of each other when gametes form (for genes on different chromosomes).

In peas, round seeds (R) are dominant over wrinkled (r), and yellow seeds (Y) over green (y). A plant that is RrYy makes four kinds of gametes in equal numbers: RY, Ry, rY, ry.

✏️ Worked example: RrYy × RrYy

1Gametes from each parent: RY, Ry, rY, ry. Draw a 4 × 4 Punnett square (16 boxes).

2Count phenotypes: 9 round yellow, 3 round green, 3 wrinkled yellow, 1 wrinkled green.

3Phenotypic ratio 9 : 3 : 3 : 1.

A test cross crosses an individual showing the dominant phenotype with a homozygous recessive (rryy). If the offspring appear in a 1 : 1 : 1 : 1 ratio, the parent was RrYy.

Pedigree analysis

A pedigree (family tree) uses squares for males and circles for females; shaded symbols are affected individuals; a horizontal line joins parents; vertical lines lead to children.

Pattern	Clues in the pedigree
Autosomal recessive	Two unaffected parents can have an affected child; the trait can skip generations; males and females equally affected
Autosomal dominant	Every affected person has at least one affected parent; does not skip generations; two affected parents can have an unaffected child
X-linked recessive	Mostly males affected; an affected father passes the allele to all daughters (carriers) but to no sons

✏️ Worked example

1Two unaffected parents have an affected daughter.

2The allele must be recessive, and both parents are carriers (Aa); the daughter is aa.

3It cannot be X-linked recessive, because the father would then be affected (X^aY) for his daughter to be X^aX^a.

Blood groups: multiple alleles

The ABO blood group gene has three alleles: I^A and I^B are codominant, and both are dominant over i.

Blood group	Possible genotypes
A	I^AI^A or I^Ai
B	I^BI^B or I^Bi
AB	I^AI^B
O	ii

✏️ Worked example

1A man with blood group A (I^Ai) and a woman with blood group B (I^Bi) have children.

2Offspring: I^AI^B (AB), I^Ai (A), I^Bi (B), ii (O) — each with a probability of ¼.

Paternity questions: blood groups can rule a man out as a father but cannot prove he is the father. DNA profiling is needed for that.

Answer in your exercise book, then open each memo to mark your work.

Question 1 · (2 marks)

Give all the possible gametes of a plant with genotype TtRr.

Show memo

TR ✓, Tr, tR, tr ✓ (all four needed for 2 marks; two or three correct earn 1).

Question 2 · (6 marks)

In guinea pigs, black fur (B) is dominant over white (b), and short hair (S) is dominant over long hair (s). Two guinea pigs heterozygous for both genes are crossed. Show the cross and give the phenotypic ratio of the offspring.

Show memo

P₁ phenotypes: black short × black short; genotypes BbSs × BbSs ✓
Meiosis — gametes: BS, Bs, bS, bs from each parent ✓
Fertilisation — correct 4 × 4 Punnett square ✓✓
F₁ phenotypic ratio: 9 black short : 3 black long : 3 white short : 1 white long ✓
P₁/F₁ and meiosis/fertilisation labels shown ✓

Question 3 · (4 marks)

In a pedigree, two unaffected parents have an affected son and an unaffected daughter. (a) Is the condition dominant or recessive? Explain. (b) Give the genotypes of the parents and the affected son (use A and a), assuming it is autosomal.

Show memo

(a) Recessive ✓: unaffected parents had an affected child, so both carry a hidden recessive allele ✓.
(b) Parents Aa and Aa ✓; affected son aa ✓.

Question 4 · (4 marks)

A mother has blood group O and her child has blood group A. Two men claim to be the father: man 1 has blood group B, man 2 has blood group AB. Which man could be the father? Explain.

Show memo

Mother ii, so the child received i from her and I^A from the father; the child is I^Ai ✓. The father must carry I^A ✓. Man 1 (group B: I^BI^B or I^Bi) cannot give I^A ✓. Man 2 (I^AI^B) could be the father ✓.