interpret and construct genetic diagrams, including Punnett squares, to explain and predict the results of dihybrid crosses that involve autosomal linkage and epistasis (knowledge of the expected ratios for different types of epistasis is not expecte

1. Introduction

Understanding how genes determine an organism’s phenotype is central to genetics. In this unit you will learn to:

• Interpret and construct genetic diagrams such as Punnett squares.
• Predict the outcome of dihybrid crosses that involve autosomal linkage and epistasis.
• Explain why observed ratios may deviate from the classic 9:3:3:1 pattern.

2. Genes, Alleles and Phenotype

A gene is a segment of DNA that codes for a particular trait. Different versions of a gene are called alleles. The combination of alleles an individual carries (its genotype) interacts with the environment to produce the observable characteristics (phenotype).

Key points:

• Alleles can be dominant (expressed in heterozygotes) or recessive (expressed only when homozygous).
• Some genes are located on the same chromosome and can be inherited together – this is linkage.
• When the product of one gene masks the effect of another, the interaction is called epistasis.

3. Constructing a Punnett Square

A Punnett square is a diagram that shows all possible gamete combinations from two parents. The steps are:

1. Determine the genotype of each parent.
2. List the possible gametes each parent can produce (considering segregation and independent assortment).
3. Place one parent’s gametes across the top and the other’s down the side.
4. Fill each cell with the combined genotype of the intersecting gametes.

Example: Monohybrid cross (Aa × Aa)

Genotypic ratio: \$1\:AA : 2\:Aa : 1\:aa\$

Phenotypic ratio (if \$A\$ is dominant): \$3\:dominant : 1\:recessive\$

4. Dihybrid Crosses – Independent Assortment

When two genes are on different chromosomes they assort independently. A classic dihybrid cross involves parents heterozygous for both traits (e.g., \$AaBb \times AaBb\$).

Standard 9:3:3:1 Punnett Square

Genotypic and phenotypic ratios (assuming complete dominance for both genes):

• Genotypic: 9 \$A\!-\!B\!-\$, 3 \$A\!-\!bb\$, 3 \$aaB\!-\$, 1 \$aabb\$
• Phenotypic: 9 dominant for both, 3 dominant for first only, 3 dominant for second only, 1 recessive for both.

5. Autosomal Linkage

When two genes are located on the same chromosome they tend to be inherited together. The degree of linkage depends on the distance between the genes (measured in map units or centimorgans).

Key Concepts

• Parental (non‑recombinant) gametes are the most common because the original chromosome arrangement is preserved.
• Recombinant gametes arise from crossing‑over during meiosis; their frequency reflects the map distance.
• Linkage reduces the number of phenotypic classes compared with independent assortment.

Example: Linked genes A and B (distance 10 cM)

Parental genotypes: \$AB/ab\$. Gamete frequencies:

• Parental \$AB\$ and \$ab\$: each \$45\%\$
• Recombinant \$Ab\$ and \$aB\$: each \$5\%\$

Cross \$AB/ab \times AB/ab\$:

Notice the reduction in recombinant phenotypes compared with the 9:3:3:1 expectation.

6. Epistasis

Epistasis occurs when the expression of one gene (the epistatic gene) masks or modifies the effect of another (the hypostatic gene). In dihybrid crosses involving epistasis, the phenotypic ratios differ from the classic 9:3:3:1.

Types of Epistasis (brief description)

• Recessive epistasis – a homozygous recessive genotype at one locus hides the effect of the other locus (e.g., \$aa\$ masks \$B\$).
• Dominant epistasis – a single dominant allele at one locus masks the other locus (e.g., \$A\$ masks \$b\$).
• Other forms (duplicate, complementary, etc.) are not required for this syllabus.

Example: Recessive epistasis

Consider two genes controlling flower colour:

• Gene \$C\$: \$C\$ (purple) is dominant to \$c\$ (white).
• Gene \$W\$: \$W\$ (pigment) is required for colour; \$ww\$ results in white regardless of \$C\$.

Cross \$CcWw \times CcWw\$ (independent assortment assumed for illustration). The expected phenotypic classes are:

• Purple (both \$C\$ and \$W\$ present)
• White (either \$cc\$ or \$ww\$)

Resulting ratio: \$9\$ purple : \$7\$ white (derived from \$9\$ \$C\!-\!W\!-\$, \$3\$ \$C\!-\!ww\$, \$3\$ \$ccW\!-\$, \$1\$ \$ccww\$; the \$3+3+1\$ white classes combine to \$7\$).

7. Worked Example – Dihybrid Cross with Linkage and Recessive Epistasis

Genes:

• \$A\$ – seed shape (round \$A\$ dominant to wrinkled \$a\$).
• \$B\$ – seed colour (yellow \$B\$ dominant to green \$b\$).
• Both genes are on the same chromosome, 15 cM apart.
• Gene \$E\$ – epistatic; \$ee\$ produces colourless seeds regardless of \$A\$ or \$B\$.

Parental genotype: \$AB/ab \; Ee \times AB/ab \; Ee\$.

Step 1 – Gamete frequencies for linked \$A\$ and \$B\$

• Parental \$AB\$ and \$ab\$: \$42.5\%\$ each.
• Recombinant \$Ab\$ and \$aB\$: \$7.5\%\$ each.

Step 2 – Combine with \$E\$ (heterozygous, independent)

Gametes for \$E\$: \$E\$ (50 %) and \$e\$ (50 %).

Step 3 – Build a 4 × 4 Punnett square (simplified representation)

After filling the square, count phenotypes:

• Colourless seeds (\$ee\$) – 1/4 of all offspring.
• Among the \$EE\$ and \$Ee\$ progeny, apply linkage ratios to determine round‑yellow, round‑green, wrinkled‑yellow, wrinkled‑green classes.

Resulting phenotypic proportion (illustrative):

\$\$

\begin{aligned}

\text{Colourless} &= \frac{1}{4} \\

\text{Round‑Yellow} &= \frac{9}{16}\times0.85 \\

\text{Round‑Green} &= \frac{3}{16}\times0.85 \\

\text{Wrinkled‑Yellow} &= \frac{3}{16}\times0.85 \\

\text{Wrinkled‑Green} &= \frac{1}{16}\times0.85

\end{aligned}

\$\$

Exact numbers depend on the recombinant frequency (0.15 in this example).

8. Practice Questions

1. Construct a Punnett square for a cross between \$AaBb\$ (genes on different chromosomes) and \$AaBb\$. List the expected phenotypic ratio.
2. Two genes \$C\$ and \$D\$ are 20 cM apart on the same chromosome. A plant heterozygous \$CD/cd\$ is crossed with \$cd/cd\$. Calculate the expected proportion of recombinant offspring.
3. In peas, gene \$R\$ (round) is dominant to \$r\$ (wrinkled) and gene \$Y\$ (yellow) is dominant to \$y\$ (green). The \$r\$ allele is epistatic to \$Y\$ (i.e., \$rr\$ produces wrinkled seeds irrespective of colour). Predict the phenotypic ratio from a \$RrYy \times RrYy\$ cross.

9. Summary

• Genes encode traits; alleles determine the genotype, which together with the environment gives the phenotype.
• Punnett squares are a systematic way to predict offspring genotypes and phenotypes.
• When genes are linked, recombination frequency modifies expected ratios.
• Epistasis alters phenotypic expression, leading to ratios that differ from the classic 9:3:3:1.
• Accurate prediction requires careful consideration of linkage, recombination, and gene interactions.