Use mathematics and Punnett squares to predict the genotype and phenotype ratios and probabilities of monohybrid crosses (North Carolina Standard Course of Study, Biology, LS.Bio.7).

What this topic is asking

The North Carolina LS.Bio.7 standards ask you to use mathematics and Punnett squares to predict the outcomes of genetic crosses. For the Biology EOC that means being comfortable with alleles, genotype and phenotype, dominant and recessive, and predicting ratios and probabilities from a monohybrid cross. Many items are technology-enhanced, asking you to drag alleles into a Punnett square or read a probability from one, so the mechanics matter as much as the vocabulary.

Alleles, genotype, and phenotype

Alleles are written as letters: a capital for the dominant allele and the same letter lowercase for the recessive one. For pea-plant height, $T$ is tall (dominant) and $t$ is short (recessive). An organism with two of the same allele ($TT$ or $tt$) is homozygous; with two different alleles ($Tt$) it is heterozygous. The EOC frequently asks you to give the genotype or the phenotype, so read which the question wants.

Dominant and recessive

This masking explains carriers and pedigrees. An organism showing a recessive trait must be homozygous recessive ($tt$); an organism showing the dominant trait could be either $TT$ or $Tt$.

Punnett squares: predicting a cross

A Punnett square sets out the alleles each parent can pass and combines them. To use one: write each parent's possible gametes (each gamete carries one allele, because alleles separate during meiosis), place one parent's gametes along the top and the other's down the side, then fill each box by combining the row and column allele. Counting the boxes gives the expected ratio and probability of each genotype and phenotype.

For a cross between two heterozygotes ($Tt \times Tt$), the four boxes are $TT$, $Tt$, $Tt$, $tt$: a genotype ratio of $1:2:1$ and a phenotype ratio of 3 tall to 1 short. Each offspring has a $\frac{3}{4}$ probability of being tall and a $\frac{1}{4}$ probability of being short. A cross of a heterozygote with a recessive ($Tt \times tt$) instead gives $Tt$, $Tt$, $tt$, $tt$: a 1:1 ratio.

Try this

Q1. A cross of $Tt \times Tt$ is carried out. State the genotype ratio and the phenotype ratio. [2]

• Cue. Genotype ratio $1\,TT : 2\,Tt : 1\,tt$; phenotype ratio 3 dominant to 1 recessive.

Q2. An organism shows a recessive trait. What must its genotype be, and why? [2]

• Cue. Homozygous recessive ($tt$), because a recessive trait appears only when no dominant allele is present.