Alleles and inheritance patterns

Consider the simple inheritance of height in pea plants. Let $T$ represent the dominant allele for tall plants and $t$ the recessive allele for short plants.

A dominant allele is expressed when just one copy is present. A recessive allele is expressed only when both copies are recessive in this model.

The tall phenotype therefore does not reveal whether a plant is $TT$ or $Tt$. The short phenotype does identify its genotype as $tt$.

From parents to gametes to offspring

Gametes are reproductive cells formed by meiosis. They contain one set of chromosomes rather than the two sets found in diploid cells. In humans, a sperm or egg normally has 23 chromosomes; their fusion at fertilisation produces a zygote with 46 chromosomes.

For a particular gene, each gamete receives only one of the parent's two alleles. A $TT$ pea plant produces gametes carrying $T$, whereas a $Tt$ plant produces gametes carrying either $T$ or $t$, with equal probability. Fertilisation brings together one allele from each parent.

This explains why siblings can inherit different genotypes even when they have the same parents: different gametes can combine.

Building a monohybrid genetic diagram

A monohybrid cross follows one gene. A complete genetic diagram connects parental genotypes, possible gametes and offspring genotypes.

Crossing a homozygous tall plant with a homozygous short plant gives:

• Parents: $TT \times tt$.
• Gametes: $T$ from the tall parent and $t$ from the short parent.
• Offspring: all $Tt$, so all are tall.

Now cross two of these heterozygous plants: $Tt \times Tt$. Each parent can produce $T$ or $t$ gametes. Place one parent's gametes across the top of a Punnett square and the other parent's down the side. Each inner cell combines the allele above it with the allele beside it.

The four cells represent equally likely combinations. There is one $TT$, two $Tt$ and one $tt$, giving a genotype ratio of $1:2:1$. Three combinations are tall and one is short, giving a phenotype ratio of $3:1$ tall to short.

Turning combinations into probabilities

Count the combinations producing the requested outcome, then divide by the total number of equally likely combinations:

$$P(\text{outcome})=\frac{\text{combinations giving that outcome}}{\text{total equally likely combinations}}$$

In the $Tt \times Tt$ cross, the probability of a short plant is $1/4=0.25=25\%$. The probability of a tall plant is $3/4=0.75=75\%$.

A different cross gives different outcomes. For $Tt \times tt$, the offspring combinations are $Tt$, $Tt$, $tt$ and $tt$. The tall-to-short ratio is $1:1$, with a $50\%$ probability of each phenotype.

These are predictions, not a fixed schedule of births. Four offspring from two heterozygous parents need not include exactly one short plant. Larger numbers of offspring are more likely to show proportions close to the predicted ratio.

Reading inheritance through a family pedigree

A family pedigree follows a characteristic through generations. Squares represent males and circles represent females. A horizontal line joins parents, with their children connected beneath them. The key explains shading: it may show affected individuals, carriers or particular genotypes.

For a recessive disorder such as cystic fibrosis, let $F$ represent the dominant allele associated with not having the disorder and $f$ the recessive disorder allele. An affected individual is $ff$. An unaffected individual could be $FF$ or $Ff$; a heterozygous $Ff$ individual is a carrier.

If two unaffected parents have an affected child, the child must have inherited $f$ from each parent. Each parent therefore carries $f$, but must also have $F$ because neither is affected. Their genotypes are both $Ff$.

Using these inferred genotypes in a Punnett square gives $FF$, $Ff$, $Ff$ and $ff$. For each child, the probability of being affected is $1/4$ or $25\%$, and the probability of being a carrier is $2/4$ or $50\%$. The affected-to-unaffected ratio is $1:3$.

Reasoning about a dominant trait

Polydactyly, which involves extra fingers or toes, can illustrate dominant inheritance. Let $P$ represent the dominant polydactyly allele and $p$ the recessive allele. In this simple model, $PP$ and $Pp$ individuals are affected, while $pp$ individuals are unaffected.

An affected parent who has an unaffected child must carry $p$: the child received a recessive allele from each parent. Because that parent also shows the dominant trait, its genotype must be $Pp$.

If the other parent is unaffected, it is $pp$. The cross $Pp \times pp$ gives equal probabilities of $Pp$ and $pp$: a $50\%$ chance of polydactyly and a $1:1$ affected-to-unaffected ratio.

Pedigree analysis therefore works by combining what a person's phenotype tells you with the alleles they must have received or passed on. Where the evidence leaves two possible genotypes, retain both possibilities rather than assuming one.

How fertilisation determines the sex of human offspring

Human body cells normally contain 23 pairs of chromosomes. One pair consists of the sex chromosomes. In the usual pattern, females have $XX$ and males have $XY$. Here, $X$ and $Y$ name chromosomes, not dominant and recessive alleles.

During meiosis, the chromosomes in each pair separate, so each gamete receives just one sex chromosome. Every egg carries an $X$ chromosome. A sperm carries either $X$ or $Y$, with an equal probability of each in this model.

At fertilisation, an $X$-bearing sperm joining an egg produces an $XX$ zygote, which develops as female in the usual pattern. A $Y$-bearing sperm joining an egg produces an $XY$ zygote, which develops as male. The sex chromosome contributed by the sperm therefore determines which combination forms.

The genetic diagram follows the same sequence as other crosses: parents, gametes, then the combinations possible at fertilisation. The two $XX$ outcomes and two $XY$ outcomes give a predicted female-to-male ratio of $1:1$. The probability of either outcome is $2/4=1/2=0.5=50\%$ for each fertilisation. This does not mean that a family must have equal numbers of girls and boys.

Each gamete carries one allele of the gene. Fertilisation combines one allele from each parent in a zygote.

$$P(\text{outcome})=\frac{\text{favourable combinations}}{\text{total equally likely combinations}}$$

Multiply probability by $100$ to express it as a percentage. Each fertilisation is independent.

Pedigree deductions

• Square = male; circle = female. Check the shading key.
• Recessive disorder: affected individuals are $aa$; unaffected individuals may be $AA$ or carriers $Aa$.
• Two unaffected parents with an affected child must both be carriers in the simple recessive model.
• Dominant trait: unaffected individuals are $aa$; affected individuals may be $AA$ or $Aa$.
• An affected parent with an unaffected child must be $Aa$ in the simple dominant model.
• Infer parental genotypes first, then use a Punnett square to calculate offspring probabilities.

Human sex determination

• Usual sex chromosomes: female $XX$; male $XY$.
• Eggs carry $X$; sperm carry $X$ or $Y$.
• Fertilisation: $X+X\rightarrow XX$ (female); $X+Y\rightarrow XY$ (male).
• Predicted female:male ratio $1:1$; probability of each $1/2=0.5=50\%$.