predicted genetic outcomes of a monohybrid cross and a monohybrid test cross

What this dot point is asking

VCAA wants you to predict the offspring of a monohybrid cross (one gene at a time) using a Punnett square, and explain how a test cross with a homozygous recessive parent reveals the unknown genotype of a parent showing the dominant phenotype.

The answer

Setting up a monohybrid cross

A monohybrid cross considers one gene with two alleles.

Steps:

1. Identify the alleles and their relationship (dominant/recessive, codominant, etc.).
2. Assign genotypes to the parents.
3. List the gametes each parent can produce (each gamete carries one allele, from meiosis).
4. Draw a Punnett square with one parent's gametes across the top and the other's down the side.
5. Fill the cells with the genotype produced by each combination.
6. Count genotypes and phenotypes for the ratio.

Mendel used the conventional capital and lowercase letters for dominant and recessive alleles (P, p). For codominance, use superscripts (IA, IB, i). For sex-linked, use the chromosome (X-H, X-h, Y).

The four basic monohybrid crosses (autosomal dominant pattern)

Using P (purple, dominant) and p (white, recessive).

1. Homozygous dominant times homozygous recessive (PP times pp).

Offspring: all Pp. Phenotype: all purple. (This is how Mendel produced his F1 generation.)

2. F1 self-cross (Pp times Pp).

Genotypes: 1 PP : 2 Pp : 1 pp. Phenotypes: 3 purple : 1 white. This is Mendel's famous 3:1 ratio.

3. Heterozygous times homozygous recessive (Pp times pp). This is the test cross.

Genotypes: 1 Pp : 1 pp. Phenotypes: 1 purple : 1 white (1:1).

4. Homozygous dominant times heterozygous (PP times Pp).

Genotypes: 1 PP : 1 Pp. Phenotypes: all purple.

The test cross

A test cross is the cross between an organism showing the dominant phenotype (unknown genotype: PP or Pp) and a homozygous recessive organism (pp).

Why use the homozygous recessive as the tester? Because pp produces only one kind of gamete (p), so the offspring phenotypes directly reveal the alleles in the unknown parent's gametes.

Interpreting results:

• If all offspring show the dominant phenotype: the unknown parent is most likely homozygous dominant (PP), because PP × pp gives all Pp (all dominant).
• If offspring show both phenotypes in a 1:1 ratio: the unknown parent is heterozygous (Pp), because Pp × pp gives 1 Pp : 1 pp (1 dominant : 1 recessive).
• A test cross can also detect linkage between two genes if you cross dihybrids (see linked-and-unlinked-genes).

Sample-size caveat. With small numbers of offspring, a heterozygous parent might by chance produce all-dominant offspring just from random gamete sampling. The more offspring observed, the more confident the conclusion.

Worked example

A pea plant with round seeds (R dominant, r recessive) is found in a wild population. The grower wants to know if it is RR or Rr. They cross it with a known wrinkled pp plant (rr).

• If RR times rr: all offspring Rr, all round. Conclusion: the parent is RR.
• If Rr times rr: 1 Rr : 1 rr offspring, half round and half wrinkled. Conclusion: the parent is Rr.

After producing 20 offspring, the grower sees 11 round and 9 wrinkled. The roughly 1:1 split indicates the parent is heterozygous Rr.

Why monohybrid crosses matter

The 3:1 (and 1:1) ratios are the empirical signature of:

• One gene with two alleles.
• Complete dominance (one allele fully masks the other in heterozygotes).
• Independent segregation of alleles into gametes (Mendel's First Law: the Law of Segregation).

The Law of Segregation says that the two alleles for each gene separate from each other during gamete formation (this is what meiosis does in anaphase I), so each gamete carries only one allele.

When you scale up to two genes at once, the predictable 3:1 becomes 9:3:3:1 (the dihybrid cross; see linked-and-unlinked-genes for when this breaks down).

Beyond Mendel

The same approach works with non-classical dominance:

• Codominance. Two heterozygotes IA IB × IA IB gives 1 IA IA : 2 IA IB : 1 IB IB. Phenotypes: 1 A : 2 AB : 1 B.
• Incomplete dominance. Rr × Rr gives 1 RR : 2 Rr : 1 rr. Phenotypes: 1 red : 2 pink : 1 white.
• Sex-linked. Use sex notation in the Punnett square (see models-of-inheritance worked example).

Examples in context

Example 1. Mendel's peas reimagined with Mornington Peninsula vineyards. Mornington Peninsula viticulturists use a monohybrid test cross to confirm the genotype of a Pinot Noir vine that produces a desirable berry colour. Suppose dark berries (D) are dominant over light berries (d). A vine with dark berries could be DD or Dd. By crossing it with a known light-berry vine (dd), the breeder predicts: if the dark vine is DD, all offspring will be dark; if Dd, the offspring will be 1:1 dark to light. The team at Australian Wine Research Institute typically grows 50 to 100 seedlings to get a statistically clear 1:1 ratio, which reveals the parent as Dd. This is a textbook monohybrid test cross applied to an Australian crop.

Example 2. Albinism in eastern grey kangaroos in Anglesea. Anglesea on the Surf Coast has a small population of eastern grey kangaroos (Macropus giganteus) including occasional albinos. Albinism is autosomal recessive (aa). Two normally pigmented parents producing an albino joey must both be heterozygous carriers (Aa). A monohybrid cross Aa x Aa predicts a 3:1 phenotypic ratio (3 pigmented to 1 albino) and a 1:2:1 genotypic ratio (1 AA: 2 Aa: 1 aa). University of Melbourne researchers tagged the herd and recorded an observed ratio of 28 pigmented to 9 albino over five years, consistent with the predicted 3:1 within statistical confidence intervals.

Try this

Q1. Define a test cross and state its purpose. [2 marks]

• Cue. Cross between an organism of unknown genotype showing the dominant phenotype and a homozygous recessive; reveals whether the unknown is homozygous dominant or heterozygous.

Q2. A dark-berry Pinot Noir vine is crossed with a light-berry vine. Of 80 offspring, 41 are dark and 39 are light. State the genotype of the dark parent and justify with a Punnett square. [3 marks]

• Cue. Approximately 1:1 ratio so dark parent is Dd; cross Dd x dd gives half Dd (dark) and half dd (light).

Q3. A pea plant breeder crosses tall (T) and short (t) pea plants. (a) State the genotype of a true-breeding tall plant. (b) Predict the F1 phenotype if a true-breeding tall is crossed with a true-breeding short. (c) Predict the F2 phenotypic ratio if F1 plants are self-pollinated. [2+2+2 marks]

• Cue. (a) TT. (b) All tall (Tt). (c) 3 tall : 1 short from Tt x Tt cross.