WwXxYy: Gamete Types and Probabilities (Mendelian Genetics)
The genotype WwXxYy describes three loci (\(W\), \(X\), and \(Y\)), each heterozygous. Under standard Mendelian assumptions—equal segregation and independent assortment (unlinked loci)—a gamete receives exactly one allele from each locus.
The count comes from \(2^n\) where \(n\) is the number of heterozygous loci.
Step 1: Count heterozygous loci
In WwXxYy, each locus is heterozygous: \(Ww\), \(Xx\), \(Yy\). Therefore \(n = 3\).
Step 2: Use the gamete-count rule
For \(n\) heterozygous loci that assort independently:
\[ \text{number of distinct gamete genotypes} = 2^n \]
\[ 2^3 = 8 \]
Step 3: List the eight gamete genotypes
A gamete must choose one allele from each pair: \((W \text{ or } w)\), \((X \text{ or } x)\), \((Y \text{ or } y)\). Writing the alleles in the order \(W\), \(X\), \(Y\) gives:
| Gamete genotype | Allele choices | Probability |
|---|---|---|
| WXY | \(W\) from \(Ww\), \(X\) from \(Xx\), \(Y\) from \(Yy\) | \((1/2)(1/2)(1/2)=1/8\) |
| WXy | \(W\), \(X\), \(y\) | \(1/8\) |
| WxY | \(W\), \(x\), \(Y\) | \(1/8\) |
| Wxy | \(W\), \(x\), \(y\) | \(1/8\) |
| wXY | \(w\), \(X\), \(Y\) | \(1/8\) |
| wXy | \(w\), \(X\), \(y\) | \(1/8\) |
| wxY | \(w\), \(x\), \(Y\) | \(1/8\) |
| wxy | \(w\), \(x\), \(y\) | \(1/8\) |
Step 4: Justify the probability \(1/8\) using the product rule
Each heterozygous locus contributes a \(1/2\) chance for either allele in a gamete. With independent assortment, the multiplication rule applies:
\[ P(\text{specific gamete}) = P(\text{allele at } W)\cdot P(\text{allele at } X)\cdot P(\text{allele at } Y) = \frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{8} \]
Visualization: Probability tree for gametes from WwXxYy
If the loci are linked (not assorting independently), the gamete probabilities may deviate from \(1/8\), but the equal \(1/8\) result is the standard Mendelian outcome for unlinked loci.