Use Mendel's laws of segregation and independent assortment, with Punnett squares, to predict the genotype and phenotype ratios and probabilities of monohybrid crosses (GSE SB3.b).

What this topic is asking

Standard SB3.b asks you to use Mendel's laws and mathematical models (Punnett squares) to predict patterns of inheritance. For the Georgia Milestones Biology EOC this is the most quantitative genetics topic, so you must be fluent with alleles, genotype and phenotype, dominant and recessive, and predicting ratios and probabilities from a cross. Items almost always involve setting up or reading a Punnett square.

Alleles, genotype, and phenotype

Alleles are written as letters: a capital for the dominant allele and the lowercase for the recessive. For height in pea plants, $T$ 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.

Dominant and recessive

Mendel's two laws

Gregor Mendel's pea-plant experiments gave two laws the EOC expects:

• Law of segregation. The two alleles for a gene separate during gamete formation (meiosis), so each gamete carries only one allele for each gene.
• Law of independent assortment. Alleles for different genes are sorted into gametes independently, so inheriting one trait does not affect another (for genes on different chromosomes).

Segregation is why a Punnett square works: each parent passes one allele, and you combine them to predict the offspring.

Punnett squares: predicting a cross

A Punnett square sets out the alleles each parent can pass and combines them. Write each parent's possible gametes along the top and 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 heterozygous tall plants ($Tt \times Tt$), 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. Define genotype and phenotype. [2 points]

• Cue. Genotype is the alleles an organism has (for example $Tt$); phenotype is the observable trait that results (for example tall).

Q2. A cross of $Tt \times Tt$ is carried out. State the genotype ratio and the phenotype ratio. [2 points]

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