Punnett Squares, Probability, Test Crosses and Dihybrid Crosses

A monohybrid cross follows one gene.

• Write each parent's gametes: $AA$ gives only $A$; $Aa$ gives $A$ and $a$ in equal numbers.
• Combine every gamete with every gamete (the Punnett square).
• $Aa \times Aa$ gives genotypes $1\ AA : 2\ Aa : 1\ aa$ and phenotypes $3$ dominant $: 1$ recessive.
• $AA \times aa$ gives all $Aa$ (the $F_1$ generation); crossing two $F_1$ gives the $F_2$ with its 3 : 1 ratio.
• Ratios are probabilities for each child, not guarantees for a small family.

• Product rule: for independent events, $P(\text{A and B}) = P(\text{A}) \times P(\text{B})$.
• Sum rule: for events that cannot happen together, $P(\text{A or B}) = P(\text{A}) + P(\text{B})$.
• The sex of a child is independent of an autosomal gene: $P(\text{boy}) = \tfrac12$.
• "At least one" is easiest as $1 - P(\text{none})$.
• "Exactly one of two children" can happen in two orders (first only, or second only): add both.

A test cross (backcross to the recessive) crosses an individual showing the dominant phenotype with a homozygous recessive individual.

• If the unknown is $AA$: all offspring show the dominant phenotype.
• If the unknown is $Aa$: offspring are dominant and recessive in a 1 : 1 ratio.
• A single recessive offspring proves the unknown parent is $Aa$. All dominant offspring only makes $AA$ likely, more so with more offspring.

For genes that assort independently (on different chromosomes) with complete dominance:

• $AaBb \times AaBb$ gives phenotypes 9 : 3 : 3 : 1 ($\tfrac34 \times \tfrac34$, $\tfrac34 \times \tfrac14$, $\tfrac14 \times \tfrac34$, $\tfrac14 \times \tfrac14$).
• The 16 boxes hold 9 different genotypes but only 4 phenotypes. The most frequent single genotype is $AaBb$ ($\tfrac14$).
• $AaBb \times aabb$ (dihybrid test cross) gives 1 : 1 : 1 : 1.
• An individual heterozygous for $n$ genes makes $2^n$ kinds of gamete.
• Two parents both heterozygous for $n$ genes give $3^n$ genotypes and $2^n$ phenotypes.