The Dihybrid Cross
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You should be comfortable with the concept of independent assortment from meiosis and able to perform a simple monohybrid cross using a Punnett square. Familiarity with dominant and recessive alleles is assumed.
Why this matters
Understanding how two traits are inherited simultaneously is the cornerstone of classical genetics and predictive breeding.
While a monohybrid cross reveals how a single gene segregates, real organisms inherit thousands of genes at once. The dihybrid cross extends Mendel’s logic to two unlinked loci, demonstrating that the inheritance of one trait does not influence the inheritance of another. This principle of independent assortment allows us to predict complex phenotypic ratios in offspring, providing a mathematical framework for understanding genetic diversity, hybrid vigor in agriculture, and the segregation of linked versus unlinked genes in medical genetics.
The Logic of Two Independent Choices
Imagine you are rolling two standard six-sided dice. The outcome of the first die has absolutely no effect on the outcome of the second. If you want to know the probability of rolling a specific combination, such as a three on the first die and a five on the second, you simply multiply the individual probabilities. Genetics works in a similar way when we look at two genes located on different chromosomes. During meiosis, the chromosomes align randomly at the metaphase plate. This means that the allele a gamete receives for gene A is chosen independently of the allele it receives for gene B. If a parent is heterozygous for both traits, they can produce four types of gametes, each with equal frequency. When two such parents mate, the combinations of these gametes create a grid of sixteen possible zygotes. Because the traits are independent, we can treat the inheritance of the first trait and the second trait as separate, parallel events. We calculate the probability of the first trait’s phenotype, calculate the probability of the second trait’s phenotype, and then multiply them together to find the probability of the combined phenotype. This multiplicative rule is the heart of the dihybrid analysis. It transforms a complex sixteen-square grid into two simple four-square grids, making the prediction of offspring ratios intuitive and manageable without needing to draw the full square every time.
Constructing the Gametes
To predict the offspring, we must first determine the gametes produced by the parents. Consider a cross between two individuals that are heterozygous for two traits, denoted as \( AaBb \). According to the law of independent assortment, the alleles for gene A segregate independently from the alleles for gene B. This means that an \( A \) allele can pair with either a \( B \) or a \( b \) allele with equal probability. Consequently, a single \( AaBb \) individual produces four distinct types of gametes: \( AB \), \( Ab \), \( aB \), and \( ab \). Each of these gamete types is produced in equal proportions, meaning each has a probability of \( \frac{1}{4} \) of being formed. This is a crucial step because it establishes the genetic "currency" that will be exchanged during fertilization. If the genes were completely linked on the same chromosome, we would only see two types of gametes (the parental types), but in the standard dihybrid model, we assume unlinked loci. The formation of these four gamete types is the direct physical consequence of the random orientation of homologous chromosome pairs during metaphase I of meiosis. Understanding this step prevents the common error of assuming that dominant alleles always travel together. The \( A \) allele does not "know" or "care" whether it is paired with \( B \) or \( b \); it is a random shuffle of genetic material.
The Multiplicative Rule of Probability
Instead of drawing a large \( 4 \times 4 \) Punnett square, we can use the product rule of probability to derive the phenotypic ratios. For a single gene, a cross between two heterozygotes (\( Aa \times Aa \)) yields a phenotypic ratio of 3 dominant to 1 recessive. Therefore, the probability of a dominant phenotype for gene A is \( \frac{3}{4} \), and the probability of a recessive phenotype is \( \frac{1}{4} \). The same logic applies to gene B. To find the probability of an offspring showing the dominant phenotype for both traits, we multiply the independent probabilities: \( \frac{3}{4} \times \frac{3}{4} = \frac{9}{16} \). This gives us the first part of the classic ratio. Similarly, the probability of being dominant for A but recessive for B is \( \frac{3}{4} \times \frac{1}{4} = \frac{3}{16} \). The probability of being recessive for A but dominant for B is \( \frac{1}{4} \times \frac{3}{4} = \frac{3}{16} \). Finally, the probability of being recessive for both traits is \( \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} \). Summing these numerators gives us the familiar 9:3:3:1 ratio. This method is not just a shortcut; it is a fundamental demonstration that independent genetic events are statistically independent. It allows us to scale this logic to three or more traits, where the ratios become \( 27:9:9:9:3:3:3:1 \), simply by multiplying the probabilities for each additional locus.
Genotypic vs Phenotypic Ratios
While the phenotypic ratio of 9:3:3:1 is the most commonly cited result, the genotypic ratio is more complex and provides deeper insight into the genetic makeup of the offspring. In the phenotypic analysis, we group \( A\_B\_ \) together, ignoring the difference between \( AABB \), \( AABb \), \( AaBB \), and \( AaBb \). However, these genotypes have different probabilities. For instance, the genotype \( AABB \) occurs with a probability of \( \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} \). The genotype \( AaBb \) occurs with a probability of \( \frac{1}{2} \times \frac{1}{2} = \frac{4}{16} \). This distinction is vital in breeding programs where specific genotypes are desired, such as creating a true-breeding line. The 9:3:3:1 ratio assumes complete dominance. If incomplete dominance or codominance is present, the phenotypic ratios change significantly, adding extra classes: 3:6:3:1:2:1 when one gene is incompletely dominant, and nine classes (1:2:1:2:4:2:1:2:1) when both are. Interactions between the genes (epistasis) give other modified ratios, such as 9:7, 9:6:1 or 12:3:1. Furthermore, if the two genes are located on the same chromosome and are close together, they will not assort independently. This linkage will skew the ratios away from 9:3:3:1, producing an excess of parental phenotypes and a deficit of recombinant phenotypes. Recognizing when the 9:3:3:1 ratio fails is just as important as knowing when it applies, as it signals the presence of linkage, epistasis, or other non-Mendelian interactions.
The concept
The dihybrid cross
Predicting the 9:3:3:1 ratio of two traits.
A common misconception is that the 9:3:3:1 ratio implies that the dominant phenotype is "stronger" or more likely to be expressed in a way that affects the other gene. In reality, the ratio is purely a statistical outcome of independent segregation. The "9" does not mean the dominant traits interact; it simply means there are nine combinations out of sixteen that result in at least one dominant allele for both loci. Students often confuse this with epistasis, where one gene masks the expression of another, which would alter the ratio entirely (e.g., to 9:7 or 12:3:1). Always verify that the traits are unlinked and that there is no gene interaction before applying the standard dihybrid ratio. C
Worked example
Problem. In a certain plant species, flower colour is determined by gene \( C \), where \( C \) (purple) is dominant to \( c \) (white), and petal shape is determined by gene \( P \), where \( P \) (long) is dominant to \( p \) (short). Two plants that are heterozygous for both traits (\( CcPp \)) are crossed. Calculate the expected number of offspring with purple flowers and short petals in a sample of 160 plants.
- Identify the genotypes of the parents: Both are \( CcPp \).
- Determine the probability of the purple flower phenotype (\( C\_ \)): In a \( Cc \times Cc \) cross, the probability is \( \frac{3}{4} \).
- Determine the probability of the short petal phenotype (\( pp \)): In a \( Pp \times Pp \) cross, the probability is \( \frac{1}{4} \).
- Apply the product rule for independent assortment: \( P(\text{purple, short}) = P(C\_) \times P(pp) = \frac{3}{4} \times \frac{1}{4} = \frac{3}{16} \).
- Calculate the expected number in the sample: \( 160 \times \frac{3}{16} = 10 \times 3 = 30 \).
Units check. The result is a count of individuals, which is dimensionless but expressed as a number of plants, consistent with the input sample size.
Three quick tests
- What is the probability of obtaining a homozygous recessive genotype for both traits in a dihybrid cross between two double heterozygotes?
Answer
The probability is \( \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} \). - If a dihybrid cross yields a 9:7 ratio instead of 9:3:3:1, what genetic phenomenon is likely occurring?
Answer
This suggests complementary gene action, a form of epistasis where the dominant allele of one gene is required for the expression of the dominant phenotype of the other. - How many different genotypes are possible in the offspring of a cross between two \( AaBb \) individuals?
Answer
There are 9 possible genotypes: \( AABB, AABb, AAbb, AaBB, AaBb, Aabb, aaBB, aaBb, aabb \).
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Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.
Concepts cited: T-014 · Threads: information, evolution