Monohybrid Mice Practice Problems For Monohybrid Crosses

Inheritance patterns are fascinating, especially when exploring how traits are passed down from parents to offspring. One fundamental concept in genetics is the monohybrid cross, where we focus on the inheritance of a single trait determined by a single gene. Using mice as a model organism, we can delve into monohybrid cross practice problems to understand this principle thoroughly.

Understanding Monohybrid Crosses

A monohybrid cross is a genetic cross between homozygous individuals but with different alleles for a single gene of interest. This cross helps determine the dominance relationship between alleles and understand how traits segregate in offspring. The offspring of the parental generation are called the first filial generation (F1), and the offspring of crosses among the F1 generation are called the second filial generation (F2).

Key Terminology

Before diving into practice problems, let's define some essential terms:

• Gene: A unit of heredity that determines a particular trait.
• Allele: Different forms of a gene (e.g., A or a).
• Homozygous: Having two identical alleles for a gene (e.g., AA or aa).
• Heterozygous: Having two different alleles for a gene (e.g., Aa).
• Genotype: The genetic makeup of an organism (e.g., AA, Aa, or aa).
• Phenotype: The observable characteristics of an organism (e.g., black fur or white fur).
• Dominant Allele: An allele that expresses its phenotype even when heterozygous.
• Recessive Allele: An allele that only expresses its phenotype when homozygous.
• Punnett Square: A diagram used to predict the genotypes and phenotypes of offspring in a genetic cross.

Example: Coat Color in Mice

Coat color in mice is a classic example used in genetics. Suppose we have a gene that determines coat color, where black (B) is dominant over white (b). This means:

• BB: Black fur
• Bb: Black fur (since black is dominant)
• bb: White fur

Practice Problem 1: Homozygous Cross

Problem: Cross a homozygous black mouse (BB) with a homozygous white mouse (bb). What are the genotypes and phenotypes of the F1 generation?

Solution:

1. Set up the cross: BB x bb

2. Gametes: The black mouse (BB) can only produce B gametes, and the white mouse (bb) can only produce b gametes.

3. Punnett Square:

   	B	B
   b	Bb	Bb
   b	Bb	Bb

4. Results:

   • Genotype: All offspring are heterozygous (Bb).
   • Phenotype: All offspring have black fur because black (B) is dominant over white (b).

5. Conclusion: The F1 generation consists of all heterozygous black mice (Bb).

Practice Problem 2: Heterozygous Cross

Problem: Cross two heterozygous black mice (Bb x Bb). What are the genotypes and phenotypes of the F2 generation?

Solution:

1. Set up the cross: Bb x Bb

2. Gametes: Each parent can produce two types of gametes: B or b.

3. Punnett Square:

   	B	b
   B	BB	Bb
   b	Bb	bb

4. Results:

   • Genotypes:
     • BB: 1/4
     • Bb: 2/4
     • bb: 1/4
   • Phenotypes:
     • Black fur (BB and Bb): 3/4
     • White fur (bb): 1/4

5. Conclusion: In the F2 generation, the genotypic ratio is 1 BB: 2 Bb: 1 bb, and the phenotypic ratio is 3 black mice: 1 white mouse.

Practice Problem 3: Test Cross

A test cross is used to determine whether an individual showing a dominant trait is homozygous or heterozygous. This involves crossing the individual with a homozygous recessive individual.

Problem: A black mouse has an unknown genotype. You cross it with a white mouse (bb). The offspring are 50% black and 50% white. What is the genotype of the black mouse?

Solution:

1. Possible genotypes of the black mouse: BB or Bb

2. Set up the test crosses:

   • If the black mouse is homozygous (BB): BB x bb
   • If the black mouse is heterozygous (Bb): Bb x bb

3. Punnett Squares:

   • BB x bb:

     	B	B
     b	Bb	Bb
     b	Bb	Bb

   • Bb x bb:

     	B	b
     b	Bb	bb
     b	Bb	bb

4. Results:

   • If the black mouse is BB, all offspring will be Bb (black).
   • If the black mouse is Bb, 50% of offspring will be Bb (black), and 50% will be bb (white).

5. Conclusion: Since the offspring are 50% black and 50% white, the black mouse must be heterozygous (Bb).

Practice Problem 4: Incomplete Dominance

Incomplete dominance occurs when the heterozygous phenotype is intermediate between the two homozygous phenotypes. For example, suppose we have a gene for fur texture in mice, where:

• CC: Curly fur
• CC': Wavy fur (intermediate)
• C'C': Straight fur

Problem: Cross a curly fur mouse (CC) with a straight fur mouse (C'C'). What are the genotypes and phenotypes of the F1 generation?

Solution:

1. Set up the cross: CC x C'C'

2. Gametes: The curly fur mouse (CC) can only produce C gametes, and the straight fur mouse (C'C') can only produce C' gametes.

3. Punnett Square:

   	C	C
   C'	CC'	CC'
   C'	CC'	CC'

4. Results:

   • Genotype: All offspring are heterozygous (CC').
   • Phenotype: All offspring have wavy fur because of incomplete dominance.

5. Conclusion: The F1 generation consists of all heterozygous wavy fur mice (CC').

Practice Problem 5: Codominance

Codominance occurs when both alleles are expressed equally in the heterozygote. For example, let's consider a gene for spot patterns in mice, where:

• RR: Red spots
• WW: White spots
• RW: Red and white spots (both are expressed)

Problem: Cross a mouse with red spots (RR) with a mouse with white spots (WW). What are the genotypes and phenotypes of the F1 generation?

Solution:

1. Set up the cross: RR x WW

2. Gametes: The red-spotted mouse (RR) can only produce R gametes, and the white-spotted mouse (WW) can only produce W gametes.

3. Punnett Square:

   	R	R
   W	RW	RW
   W	RW	RW

4. Results:

   • Genotype: All offspring are heterozygous (RW).
   • Phenotype: All offspring have red and white spots (both are expressed).

5. Conclusion: The F1 generation consists of all mice with both red and white spots (RW).

Practice Problem 6: Lethal Alleles

Lethal alleles are alleles that cause the death of an organism, usually when homozygous. A classic example is the yellow coat color in mice, controlled by the A gene.

• AYAY: Lethal (dies early in development)
• AYA: Yellow coat
• AA: Agouti (brown) coat

Here, AY is dominant for yellow coat color but recessive for lethality.

Problem: Cross two yellow mice (AYA x AYA). What are the genotypes and phenotypes of the offspring?

Solution:

1. Set up the cross: AYA x AYA

2. Gametes: Each parent can produce two types of gametes: AY or A.

3. Punnett Square:

   	AY	A
   AY	AYAY	AYA
   A	AYA	AA

4. Results:

   • Genotypes:
     • AYAY: 1/4 (dies)
     • AYA: 2/4 (yellow)
     • AA: 1/4 (agouti)
   • Phenotypes:
     • Yellow: 2/3 (since AYAY dies, this fraction is out of the surviving offspring)
     • Agouti: 1/3

5. Conclusion: The offspring will be 2/3 yellow and 1/3 agouti. Note that 1/4 of the zygotes (AYAY) do not survive.

Practice Problem 7: Another Coat Color Example

Assume black coat color (B) is dominant to brown coat color (b) in mice.

Problem: A breeder crosses a black mouse with a brown mouse. They produce 8 offspring, all of which are black. What are the possible genotypes of the parents?

Solution:

1. Possible genotypes for the black mouse: BB or Bb
2. Genotype for the brown mouse: bb (since brown is recessive)
3. Analyze the cross:
   • If the black mouse is BB, the cross is BB x bb. All offspring would be Bb (black).
   • If the black mouse is Bb, the cross is Bb x bb. Offspring could be Bb (black) or bb (brown).
4. Conclusion: Since all 8 offspring are black, the black mouse must be homozygous (BB). If it were heterozygous (Bb), we would expect some brown offspring.

Practice Problem 8: Solving for Unknown Genotypes

Problem: You have two black mice, and when you cross them, they produce 24 black offspring and 8 brown offspring. What are the genotypes of the parents?

Solution:

1. Possible genotypes for black mice: BB or Bb

2. Genotype for brown mice: bb

3. Analyze the cross:

   • Since brown offspring (bb) are produced, both parents must carry the recessive b allele.
   • Therefore, both parents must be heterozygous (Bb).

4. Confirmation with Punnett Square:

   	B	b
   B	BB	Bb
   b	Bb	bb

5. Expected Ratio: The expected ratio is 3 black (BB or Bb) to 1 brown (bb).

6. Comparison to Observed Ratio: The observed ratio is 24 black to 8 brown, which simplifies to 3:1, matching the expected ratio.

7. Conclusion: Both parents are heterozygous (Bb).

Practice Problem 9: Complex Scenario

Problem: In mice, short tails (T) are dominant over long tails (t), and black fur (B) is dominant over brown fur (b). A mouse heterozygous for both traits (BT/bt) is crossed with a mouse with a long tail and brown fur (bt/bt). What are the expected phenotypic ratios in the offspring?

Solution:

1. Set up the cross: BT/bt x bt/bt

2. Gametes from the heterozygous parent: BT and bt

3. Gametes from the homozygous recessive parent: Only bt

4. Punnett Square:

   	BT	bt
   bt	BT/bt	bt/bt
   bt	BT/bt	bt/bt

5. Genotypes and Phenotypes:

   • BT/bt: Short tail, black fur
   • bt/bt: Long tail, brown fur

6. Ratios: 1:1 (50% short tail, black fur; 50% long tail, brown fur)

7. Conclusion: The expected phenotypic ratio in the offspring is 1 short tail, black fur : 1 long tail, brown fur.

Practice Problem 10: Multi-Step Problem

Problem: You have a population of mice. In this population, black fur (B) is dominant to white fur (b). You select a black mouse and cross it with a white mouse. You obtain the following offspring:

• 15 black mice
• 13 white mice

Based on these results:

1. What is the genotype of the black parent?
2. If you cross two of the black offspring, what genotypes and phenotypes would you expect in the F2 generation, and in what ratio?

Solution:

Part 1: Genotype of the Black Parent

1. Analyze the offspring: The presence of white offspring indicates that the black parent must carry the recessive b allele.
2. Possible genotypes: The black parent could be either BB or Bb. Since white offspring (bb) were produced, the black parent must be heterozygous (Bb).
3. Conclusion: The genotype of the black parent is Bb.

Part 2: Cross of Two Black Offspring

1. Identify the cross: Since the black parent was Bb and the white parent was bb, the black offspring must also be Bb.

2. Set up the cross: Bb x Bb

3. Punnett Square:

   	B	b
   B	BB	Bb
   b	Bb	bb

4. Genotypes and Phenotypes:

   • Genotypes:
     • BB: 1/4
     • Bb: 2/4
     • bb: 1/4
   • Phenotypes:
     • Black fur (BB and Bb): 3/4
     • White fur (bb): 1/4

5. Conclusion: In the F2 generation, the expected genotypic ratio is 1 BB: 2 Bb: 1 bb, and the phenotypic ratio is 3 black mice: 1 white mouse.

Advanced Tips for Monohybrid Crosses

• Always start with the phenotype: Identify the dominant and recessive traits.
• Write down possible genotypes: List all possible genotypes for each phenotype.
• Use Punnett Squares consistently: This helps visualize the crosses and predict outcomes.
• Check your results: Ensure your results match the expected ratios.
• Consider alternative inheritance patterns: If the ratios deviate from expected monohybrid ratios, consider incomplete dominance, codominance, or lethal alleles.

Conclusion

Understanding monohybrid crosses is a cornerstone of genetics. By working through various practice problems using mice as a model organism, you can grasp the fundamentals of inheritance and allele interactions. Whether you're dealing with simple dominance, incomplete dominance, codominance, or lethal alleles, the principles remain the same: identify the genotypes, predict the gametes, and use Punnett Squares to determine the potential outcomes. With practice, you’ll be able to solve even the most challenging monohybrid cross problems with confidence. These skills not only enhance your understanding of genetics but also provide a foundation for exploring more complex inheritance patterns in various organisms.