Codominance and Incomplete Dominance Problems.pdf | Summaries ... - Free Printable

Human Genetics Practice Worksheet #3

#### Explanation and Solutions

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1. Explain the difference between incomplete dominance and codominance.

- Codominance: In codominance, both alleles in a heterozygous individual are fully expressed. The phenotype is a combination of both traits. For example:
- A chicken with black feathers (allele B) bred with a white chicken (allele W) produces offspring that have both black and white feathers (heterozygous individuals express both traits).

- Incomplete Dominance: In incomplete dominance, the heterozygous individual shows a blended phenotype that is intermediate between the two homozygous phenotypes. For example:
- Crossing a white flower (allele W) with a pink flower (allele P) results in pink offspring (the heterozygote expresses a blended phenotype).

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2. Co-Dominance Problems

#### a. What is the genotype for black chickens?
- Black chickens are homozygous for the black allele.
- Genotype: BB

#### b. What is the genotype for white chickens?
- White chickens are homozygous for the white allele.
- Genotype: WW

#### c. What is the genotype for erminette chickens?
- Erminette chickens are heterozygous, expressing both black and white feathers.
- Genotype: BW

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3. If two erminette chickens were crossed, what is the probability that:

#### Parents: BW × BW
We will use a Punnett square to determine the probabilities.

| | B | W |
|-------|-----|-----|
| B | BB | BW |
| W | BW | WW |

#### Phenotypes and Probabilities:
- BB: Black chickens (homozygous black)
- BW: Erminette chickens (heterozygous, black and white spotted)
- WW: White chickens (homozygous white)

From the Punnett square:
- Probability of black chicks (BB): 25%
- Probability of white chicks (WW): 25%
- Probability of erminette chicks (BW): 50%

#### Answers:
- a. They would have a black chick: 25%
- b. They would have a white chick: 25%

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4. A black chicken and a white chicken are crossed. What is the probability that they will have erminette chicks?

#### Parents: BB × WW
We will use a Punnett square to determine the probabilities.

| | B | W |
|-------|-----|-----|
| B | BB | BW |
| W | BW | WW |

#### Phenotypes and Probabilities:
- All offspring will be heterozygous (BW), which means they will all be erminette chickens.

#### Answer:
- Probability of erminette chicks: 100%

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5. Roan is a coat color found in many animals, including horses, cattle, and dogs. It is defined generally as an even mixture of white and red hair. What offspring are expected from mating a roan bull and a roan cow?

#### Assumptions:
- Roan is caused by incomplete dominance.
- Let's assume the alleles are:
- R: Red
- r: White

Roan is the heterozygous phenotype (Rr), which shows a blend of red and white.

#### Parents: Rr × Rr
We will use a Punnett square to determine the probabilities.

| | R | r |
|-------|-----|-----|
| R | RR | Rr |
| r | Rr | rr |

#### Phenotypes and Probabilities:
- RR: Red (homozygous red)
- Rr: Roan (heterozygous, blend of red and white)
- rr: White (homozygous white)

From the Punnett square:
- Probability of red offspring (RR): 25%
- Probability of roan offspring (Rr): 50%
- Probability of white offspring (rr): 25%

#### Answer:
The expected offspring are:
- 25% red, 50% roan, and 25% white.

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6. What phenotypes would you expect from a cross between a red bull and a white cow?

#### Assumptions:
- Let's assume the alleles are:
- R: Red
- r: White

#### Parents: RR × rr
We will use a Punnett square to determine the phenotypes.

| | R | r |
|-------|-----|-----|
| R | RR | Rr |
| r | Rr | rr |

#### Phenotypes:
- All offspring will be heterozygous (Rr), which means they will all be roan (a blend of red and white).

#### Answer:
All offspring will be roan.

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7. Two short-tailed (Manx) cats are bred together. They produce three kittens with long tails, five short tails, and two without any tails. From these results, how do you think tail length in these cats is inherited? Show the genotypes for both the parents and the offspring to support your answer.

#### Analysis:
- The observed phenotypes are:
- Long tails
- Short tails
- No tails
- This suggests that the inheritance pattern involves at least two alleles and possibly a dominant-recessive relationship with a lethal effect.

#### Hypothesis:
Let’s assume:
- T: Allele for normal tail length (dominant)
- t: Allele for short tail (recessive)
- tt: Homozygous recessive results in no tail (lethal or non-viable)

#### Genotypes:
- Parents: Both short-tailed cats (heterozygous, Tt)
- Offspring:
- Long tails: TT (homozygous dominant)
- Short tails: Tt (heterozygous)
- No tails: tt (homozygous recessive, lethal)

#### Punnett Square:
| | T | t |
|-------|-----|-----|
| T | TT | Tt |
| t | Tt | tt |

#### Phenotypes and Probabilities:
- TT: Long tails (homozygous dominant)
- Tt: Short tails (heterozygous)
- tt: No tails (homozygous recessive, lethal)

From the Punnett square:
- Expected ratio: 1 TT : 2 Tt : 1 tt
- Observed ratio: 3 long tails, 5 short tails, 2 no tails (approximately matches the expected ratio).

#### Conclusion:
- The inheritance of tail length follows a dominant-recessive pattern with a lethal effect for the homozygous recessive condition (tt).

#### Genotypes:
- Parents: Tt × Tt
- Offspring:
- Long tails: TT
- Short tails: Tt
- No tails: tt (lethal)

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Final Answers:

1. Difference between incomplete dominance and codominance: Explained above.
2. a. Genotype for black chickens: BB
b. Genotype for white chickens: WW
c. Genotype for erminette chickens: BW
3. a. Probability of black chicks: 25%
b. Probability of white chicks: 25%
4. Probability of erminette chicks: 100%
5. Expected offspring: 25% red, 50% roan, 25% white
6. Phenotypes of offspring: All roan
7. Inheritance pattern: Dominant-recessive with lethality
- Parents: Tt × Tt
- Offspring:
- Long tails: TT
- Short tails: Tt
- No tails: tt (lethal)

Final Answer: \boxed{See detailed solutions above}