Incomplete Dominance And Codominance Practice Problems

Let's dive into the fascinating world of genetics, specifically focusing on incomplete dominance and codominance. These concepts offer a nuanced understanding of how traits are inherited, moving beyond the simple dominant-recessive relationships often introduced in basic biology. By working through practice problems, we'll solidify your understanding and sharpen your ability to predict genotypes and phenotypes in various scenarios.

Understanding Incomplete Dominance

Incomplete dominance occurs when neither allele is fully dominant over the other. The resulting heterozygote phenotype is a blend of the two homozygous phenotypes. Think of it as mixing paint: red and white don't produce red, but a shade of pink.

• Key Indicator: A heterozygous phenotype that is intermediate between the two homozygous phenotypes.
• Example: Flower color in snapdragons.

Understanding Codominance

Codominance is a situation where both alleles in a heterozygote are fully expressed. Unlike incomplete dominance, there is no blending. Instead, you see both traits expressed simultaneously.

• Key Indicator: A heterozygous phenotype that displays both homozygous phenotypes.
• Example: Human blood type AB.

Practice Problems: Incomplete Dominance

Let's tackle some practice problems to cement your understanding of incomplete dominance.

Problem 1: Snapdragon Flowers

In snapdragons, flower color is controlled by incomplete dominance. A homozygous plant with red flowers (RR) is crossed with a homozygous plant with white flowers (WW).

a) What is the phenotype of the F1 generation?

b) If two F1 plants are crossed, what are the genotypic and phenotypic ratios of the F2 generation?

Solution:

a) F1 Generation Phenotype:

• Parental genotypes: RR (red) x WW (white)
• Gametes produced: R and W
• F1 genotype: RW

Since this is incomplete dominance, the RW heterozygote will not be red. Instead, it will be pink.

• F1 Phenotype: Pink

b) F2 Generation Genotypic and Phenotypic Ratios:

• F1 cross: RW (pink) x RW (pink)
• Gametes produced: R and W from each parent

We can use a Punnett square to determine the F2 generation:

	R	W
R	RR	RW
W	RW	WW

• Genotypic Ratio: 1 RR : 2 RW : 1 WW
• Phenotypic Ratio: 1 Red : 2 Pink : 1 White

Problem 2: Feather Color in Chickens

Feather color in a certain breed of chickens is controlled by incomplete dominance. The allele for black feathers is represented by B, and the allele for white feathers is represented by W. Heterozygous chickens (BW) have blue feathers.

a) What offspring would you expect from a cross between a blue chicken and a white chicken? Give the genotypic and phenotypic percentages.

b) What cross would produce offspring with a phenotypic ratio of 1 black : 2 blue : 1 white?

Solution:

a) Blue Chicken x White Chicken:

• Parental genotypes: BW (blue) x WW (white)
• Gametes produced: B, W and W

Punnett Square:

	B	W
W	BW	WW
W	BW	WW

• Genotypic Ratio: 2 BW : 2 WW
• Genotypic Percentage: 50% BW, 50% WW
• Phenotypic Ratio: 2 Blue : 2 White
• Phenotypic Percentage: 50% Blue, 50% White

b) Cross for 1 Black : 2 Blue : 1 White:

This phenotypic ratio mirrors the classic 1:2:1 ratio seen in incomplete dominance crosses in the F2 generation. Therefore, the cross must be between two heterozygotes.

• Cross: BW (blue) x BW (blue)

This is the same cross we did in Problem 1 with the snapdragons, just with different alleles and phenotypes.

Problem 3: Tail Length in Lizards

In a species of lizard, tail length is governed by incomplete dominance. Lizards homozygous for the allele L have long tails, lizards homozygous for the allele S have short tails, and heterozygotes (LS) have medium-length tails.

A population of lizards consists of 36 long-tailed lizards, 48 medium-tailed lizards, and 16 short-tailed lizards.

a) What are the allele frequencies for L and S in this population?

b) Assuming the population is in Hardy-Weinberg equilibrium, what would you expect the genotype frequencies to be in the next generation?

Solution:

a) Allele Frequencies:

• Total number of lizards: 36 + 48 + 16 = 100

• Number of L alleles: (36 LL lizards * 2 L alleles/lizard) + (48 LS lizards * 1 L allele/lizard) = 72 + 48 = 120

• Number of S alleles: (16 SS lizards * 2 S alleles/lizard) + (48 LS lizards * 1 S allele/lizard) = 32 + 48 = 80

• Total number of alleles in the population: 100 lizards * 2 alleles/lizard = 200

• Frequency of L allele (p) = Number of L alleles / Total number of alleles = 120 / 200 = 0.6

• Frequency of S allele (q) = Number of S alleles / Total number of alleles = 80 / 200 = 0.4

• Therefore, p = 0.6 and q = 0.4

b) Hardy-Weinberg Equilibrium:

The Hardy-Weinberg equation states: p² + 2pq + q² = 1, where:

• p² = frequency of the LL genotype
• 2pq = frequency of the LS genotype
• q² = frequency of the SS genotype

Using our calculated allele frequencies:

• p² = (0.6)² = 0.36 (frequency of LL, long-tailed lizards)

• 2pq = 2 * 0.6 * 0.4 = 0.48 (frequency of LS, medium-tailed lizards)

• q² = (0.4)² = 0.16 (frequency of SS, short-tailed lizards)

• Expected genotype frequencies in the next generation: 36% LL, 48% LS, 16% SS

Practice Problems: Codominance

Now, let's move on to codominance with some practice problems.

Problem 1: Roan Cattle

In cattle, coat color is codominant. The allele for red coat is represented by R, and the allele for white coat is represented by W. Heterozygous cattle (RW) have a roan coat (a mixture of red and white hairs).

a) What offspring would you expect from a cross between a roan bull and a white cow? Give the genotypic and phenotypic percentages.

b) A farmer wants to produce only roan cattle. Is this possible? Explain why or why not.

Solution:

a) Roan Bull x White Cow:

• Parental genotypes: RW (roan) x WW (white)
• Gametes produced: R, W and W

Punnett Square:

	R	W
W	RW	WW
W	RW	WW

• Genotypic Ratio: 2 RW : 2 WW
• Genotypic Percentage: 50% RW, 50% WW
• Phenotypic Ratio: 2 Roan : 2 White
• Phenotypic Percentage: 50% Roan, 50% White

b) Producing Only Roan Cattle:

No, it is not possible to produce only roan cattle. Roan cattle are heterozygotes (RW). To get 100% roan offspring, you would need to cross two homozygous roan cattle, but that is not possible because there's no such thing as a homozygous roan cattle (they are, by definition, heterozygous). Any cross will always produce some non-roan offspring (either red or white). The closest one could get would be to cross a red (RR) and a white (WW) cow which will yield 100% Roan offspring, but then the farmer would need to keep the two purebred parents separated from their offspring.

Problem 2: Human Blood Types (ABO System)

The ABO blood group system in humans is an example of both codominance and multiple alleles. There are three alleles: Iᴬ, Iᴮ, and i. Iᴬ and Iᴮ are codominant, and both are dominant over i.

• Genotype IᴬIᴬ or Iᴬi results in blood type A.
• Genotype IᴮIᴮ or Iᴮi results in blood type B.
• Genotype IᴬIᴮ results in blood type AB.
• Genotype ii results in blood type O.

A woman with blood type AB marries a man with blood type O.

a) What are the possible blood types of their children?

b) What are the chances that their child will have blood type A?

Solution:

a) Possible Blood Types:

• Parental genotypes: IᴬIᴮ (AB) x ii (O)
• Gametes produced: Iᴬ, Iᴮ and i

Punnett Square:

	Iᴬ	Iᴮ
i	Iᴬi	Iᴮi
i	Iᴬi	Iᴮi

• Possible Blood Types: A (Iᴬi) and B (Iᴮi)

b) Chance of Blood Type A:

From the Punnett square, we can see that two out of four offspring will have blood type A (Iᴬi).

• Chance of Blood Type A: 50%

Problem 3: Erminette Chickens

In Erminette chickens, the gene for feather color exhibits codominance. The allele for black feathers is represented by B, and the allele for white feathers is represented by W. Chickens that are heterozygous (BW) have both black and white feathers, resulting in a speckled appearance.

A breeder crosses two speckled Erminette chickens.

a) What are the expected genotypes and phenotypes of the offspring, and in what proportions?

b) If the breeder wants to produce only speckled chickens, what cross should they perform? Is it possible to achieve this goal consistently?

Solution:

a) Speckled Chicken x Speckled Chicken:

• Parental genotypes: BW (speckled) x BW (speckled)
• Gametes produced: B, W and B, W

Punnett Square:

	B	W
B	BB	BW
W	BW	WW

• Genotypic Ratio: 1 BB : 2 BW : 1 WW
• Phenotypic Ratio: 1 Black : 2 Speckled : 1 White

b) Producing Only Speckled Chickens:

To produce only speckled chickens, the breeder would ideally want to cross two parents that only produce the BW genotype. However, because the speckled phenotype is the heterozygous state, a cross that yields only speckled offspring is impossible. Crossing a black chicken (BB) with a white chicken (WW) will produce 100% speckled (BW) offspring. However, to maintain this, the breeder would need to keep the purebred black and white chickens separate from their speckled offspring. Crossing two speckled chickens will always result in the appearance of black and white offspring as well.

Codominance and Incomplete Dominance: Key Differences

Feature	Incomplete Dominance	Codominance
Heterozygote	Blending of phenotypes (intermediate phenotype)	Both phenotypes are fully expressed
Expression of Alleles	Neither allele is fully dominant	Both alleles are fully expressed
Example	Flower color in snapdragons (red, white, pink)	Roan cattle (red hairs, white hairs)
Another Example	Tail length in lizards (Long, short, medium)	Human blood type AB (both A and B antigens present)

Additional Practice Problems

Here are a few more problems to test your knowledge.

Problem 1: Plant Height

In a certain plant species, height is determined by incomplete dominance. Plants with the genotype TT are tall (20 cm), plants with the genotype tt are short (10 cm), and plants with the genotype Tt are of medium height (15 cm).

What are the expected heights of the offspring if a medium height plant is crossed with a short plant?

Problem 2: Blood Type Inheritance

A child has blood type O. Her mother has blood type A. What are the possible blood types of the father?

Problem 3: Spotted Fur

In a newly discovered mammal, fur color is codominant. FF individuals have solid brown fur, WW individuals have solid white fur, and FW individuals have fur with distinct brown and white spots.

A population of these mammals is observed. 64% have spotted fur, and the rest are either solid brown or solid white. Assuming Hardy-Weinberg equilibrium, what are the frequencies of the F and W alleles in this population?

Explanations and Additional Insights

• Why is understanding these concepts important? Incomplete dominance and codominance provide a more accurate representation of inheritance patterns than simple dominant-recessive models. Many traits in real organisms, including humans, are influenced by these mechanisms. Understanding these concepts is crucial for genetic counseling, predicting traits in breeding programs, and understanding the complexities of evolutionary processes.
• Beyond Mendelian Genetics: These inheritance patterns are examples of non-Mendelian genetics, highlighting that inheritance is not always straightforward. Factors like epistasis, linked genes, and environmental influences can further complicate the expression of traits.
• Real-World Applications: Understanding incomplete dominance and codominance has practical applications in agriculture (e.g., breeding plants with desirable traits) and medicine (e.g., predicting the risk of inheriting certain genetic conditions).
• Hardy-Weinberg Equilibrium Assumptions: Remember that the Hardy-Weinberg equilibrium relies on specific assumptions, including no mutation, random mating, no gene flow, no genetic drift, and no selection. Deviations from these assumptions can lead to changes in allele frequencies over time.

Conclusion

Incomplete dominance and codominance are fascinating examples of how genes interact to determine phenotypes. By understanding these concepts and working through practice problems, you can gain a deeper appreciation for the complexities of inheritance. Continue to practice and explore different scenarios to solidify your understanding and enhance your problem-solving skills in genetics. Remember that these are just a few examples, and the possibilities for genetic variation are endless! Keep exploring, keep learning, and keep questioning. The world of genetics is waiting to be discovered.