Complete Dominance Mendelian Genetics Worksheet Answer Key

The principles of Mendelian genetics, established by Gregor Mendel in the 19th century, lay the foundation for understanding inheritance patterns. While Mendel's laws, including the law of segregation and the law of independent assortment, provide a strong framework, genetic inheritance often presents complexities beyond simple dominant and recessive relationships. Complete dominance, where one allele completely masks the expression of another, is just one piece of the puzzle. This article delves into the concept of complete dominance, explores its application in Mendelian genetics problems, and provides a detailed answer key to a comprehensive worksheet designed to reinforce understanding of these principles.

Understanding Complete Dominance in Mendelian Genetics

Complete dominance occurs when the phenotype of the heterozygote (an individual with two different alleles for a trait) is identical to the phenotype of one of the homozygotes (an individual with two identical alleles for a trait). In simpler terms, if allele A is completely dominant over allele a, then individuals with genotypes AA and Aa will exhibit the same phenotype. The a allele's trait will only be expressed in individuals with the aa genotype.

Key Concepts:

• Allele: A variant form of a gene.
• Genotype: The genetic makeup of an individual, represented by the alleles they possess for a specific trait (e.g., AA, Aa, aa).
• Phenotype: The observable characteristics of an individual, determined by their genotype and environmental factors (e.g., tall, short, blue eyes, brown eyes).
• Homozygous: Having two identical alleles for a trait (e.g., AA or aa).
• Heterozygous: Having two different alleles for a trait (e.g., Aa).
• Dominant Allele: An allele that masks the expression of the recessive allele in a heterozygote.
• Recessive Allele: An allele whose expression is masked by the dominant allele in a heterozygote.

Example:

Consider pea plants, which Mendel famously studied. Let's say T represents the allele for tallness and t represents the allele for shortness. If T is completely dominant over t:

• TT genotype = Tall phenotype
• Tt genotype = Tall phenotype (because the T allele masks the t allele)
• tt genotype = Short phenotype

Punnett Squares: A Tool for Predicting Inheritance

Punnett squares are visual tools used to predict the possible genotypes and phenotypes of offspring from a given cross (mating) between two parents. Each parent's alleles are written along the top and side of the square, and the possible combinations of alleles in the offspring are filled in the boxes.

Example:

If we cross a heterozygous tall pea plant (Tt) with another heterozygous tall pea plant (Tt), the Punnett square would look like this:

From this Punnett square, we can see the following genotypic and phenotypic ratios:

• Genotypic Ratio: 1 TT : 2 Tt : 1 tt
• Phenotypic Ratio: 3 Tall : 1 Short

Complete Dominance Mendelian Genetics Worksheet: Problems and Solutions

This section presents a comprehensive worksheet with problems related to complete dominance in Mendelian genetics. Each problem is followed by a detailed solution to help solidify understanding.

Worksheet Problems:

1. Problem: In pea plants, purple flowers (P) are dominant to white flowers (p). If a heterozygous purple-flowered plant is crossed with a white-flowered plant, what are the predicted genotypic and phenotypic ratios of the offspring?

2. Problem: In guinea pigs, black fur (B) is dominant to white fur (b). If two heterozygous black-furred guinea pigs are crossed, what is the probability that an offspring will have white fur?

3. Problem: In humans, the ability to taste PTC (phenylthiocarbamide) is dominant (T) to the inability to taste PTC (t). If a taster (heterozygous) marries a non-taster, what proportion of their children would be expected to be tasters?

4. Problem: A farmer breeds two types of chickens. Black feathers (B) are dominant to white feathers (b). He crosses a homozygous black chicken with a homozygous white chicken. What will be the phenotype of the F1 generation? If he then crosses two F1 chickens, what will be the genotypic and phenotypic ratios of the F2 generation?

5. Problem: In a certain species of beetle, long antennae (L) are dominant to short antennae (l). You cross a beetle with long antennae but unknown genotype with a beetle with short antennae. You observe that half of the offspring have long antennae and half have short antennae. What is the genotype of the beetle with long antennae?

6. Problem: Cystic fibrosis is a recessive genetic disorder. If two parents, both carriers (heterozygous) for the cystic fibrosis allele (Cc), have a child, what is the probability that the child will have cystic fibrosis (cc)? What is the probability that the child will be a carrier (Cc)?

7. Problem: In dogs, brown eyes (B) are dominant to blue eyes (b). A dog breeder has a brown-eyed dog of unknown genotype. She breeds the dog with a blue-eyed dog. The litter contains 7 brown-eyed puppies and 6 blue-eyed puppies. What is the genotype of the brown-eyed parent?

8. Problem: In fruit flies, gray body color (G) is dominant to ebony body color (g). A gray-bodied female is crossed with an ebony-bodied male. The offspring consist of 142 gray-bodied flies and 138 ebony-bodied flies. What are the probable genotypes of the parents?

9. Problem: A plant with round seeds is crossed with a plant with wrinkled seeds. All the offspring have round seeds. Explain the genotypes of the parents and offspring, assuming that round seeds (R) are dominant to wrinkled seeds (r).

10. Problem: Polydactyly (having extra fingers or toes) is a dominant trait in humans. A man with polydactyly, whose mother did not have polydactyly, marries a woman without polydactyly. What is the probability that their child will have polydactyly?

Answer Key and Detailed Solutions:

1. Problem: In pea plants, purple flowers (P) are dominant to white flowers (p). If a heterozygous purple-flowered plant is crossed with a white-flowered plant, what are the predicted genotypic and phenotypic ratios of the offspring?

   Solution:

   • Parental genotypes: Pp x pp
   • Punnett Square:

   	P	p
   p	Pp	pp
   p	Pp	pp

   • Genotypic Ratio: 2 Pp : 2 pp (or simplified: 1 Pp : 1 pp)
   • Phenotypic Ratio: 2 Purple : 2 White (or simplified: 1 Purple : 1 White)

   Answer: The genotypic ratio is 1 Pp : 1 pp. The phenotypic ratio is 1 Purple : 1 White.

2. Problem: In guinea pigs, black fur (B) is dominant to white fur (b). If two heterozygous black-furred guinea pigs are crossed, what is the probability that an offspring will have white fur?

   Solution:

   • Parental genotypes: Bb x Bb
   • Punnett Square:

   	B	b
   B	BB	Bb
   b	Bb	bb

   • Genotypes: BB, Bb, bb
   • Probability of white fur (bb): 1 out of 4 (1/4) or 25%

   Answer: The probability that an offspring will have white fur is 25%.

3. Problem: In humans, the ability to taste PTC (phenylthiocarbamide) is dominant (T) to the inability to taste PTC (t). If a taster (heterozygous) marries a non-taster, what proportion of their children would be expected to be tasters?

   Solution:

   • Parental genotypes: Tt x tt
   • Punnett Square:

   	T	t
   t	Tt	tt
   t	Tt	tt

   • Genotypes: Tt, tt
   • Proportion of tasters (Tt): 2 out of 4 (2/4) or 50%

   Answer: 50% of their children would be expected to be tasters.

4. Problem: A farmer breeds two types of chickens. Black feathers (B) are dominant to white feathers (b). He crosses a homozygous black chicken with a homozygous white chicken. What will be the phenotype of the F1 generation? If he then crosses two F1 chickens, what will be the genotypic and phenotypic ratios of the F2 generation?

   Solution:

   • Parental genotypes: BB x bb
   • F1 generation:

   	B	B
   b	Bb	Bb
   b	Bb	Bb

   • F1 genotype: All Bb

   • F1 phenotype: All black

   • F2 generation (crossing two F1 chickens: Bb x Bb):

   	B	b
   B	BB	Bb
   b	Bb	bb

   • F2 Genotypic Ratio: 1 BB : 2 Bb : 1 bb
   • F2 Phenotypic Ratio: 3 Black : 1 White

   Answer: The F1 generation will all have black feathers. The F2 genotypic ratio will be 1 BB : 2 Bb : 1 bb. The F2 phenotypic ratio will be 3 Black : 1 White.

5. Problem: In a certain species of beetle, long antennae (L) are dominant to short antennae (l). You cross a beetle with long antennae but unknown genotype with a beetle with short antennae. You observe that half of the offspring have long antennae and half have short antennae. What is the genotype of the beetle with long antennae?

   Solution:

   • Known: Long antennae is dominant (L), short antennae is recessive (l). One parent is ll (short antennae). The other parent has long antennae, so it could be LL or Ll.
   • Offspring: Half long, half short.
   • If the long-antennaed parent was LL, all offspring would be Ll (long antennae). This doesn't match the observed results.
   • Therefore, the long-antennaed parent must be Ll. Let's check this with a Punnett square:

   	L	l
   l	Ll	ll
   l	Ll	ll

   • This Punnett square shows half Ll (long antennae) and half ll (short antennae), which matches the observation.

   Answer: The genotype of the beetle with long antennae is Ll.

6. Problem: Cystic fibrosis is a recessive genetic disorder. If two parents, both carriers (heterozygous) for the cystic fibrosis allele (Cc), have a child, what is the probability that the child will have cystic fibrosis (cc)? What is the probability that the child will be a carrier (Cc)?

   Solution:

   • Parental genotypes: Cc x Cc
   • Punnett Square:

   	C	c
   C	CC	Cc
   c	Cc	cc

   • Probability of having cystic fibrosis (cc): 1 out of 4 (1/4) or 25%
   • Probability of being a carrier (Cc): 2 out of 4 (2/4) or 50%

   Answer: The probability that the child will have cystic fibrosis is 25%. The probability that the child will be a carrier is 50%.

7. Problem: In dogs, brown eyes (B) are dominant to blue eyes (b). A dog breeder has a brown-eyed dog of unknown genotype. She breeds the dog with a blue-eyed dog. The litter contains 7 brown-eyed puppies and 6 blue-eyed puppies. What is the genotype of the brown-eyed parent?

   Solution:

   • Known: Brown eyes are dominant (B), blue eyes are recessive (b). The blue-eyed dog must be bb. The brown-eyed dog is either BB or Bb.
   • Since some puppies are blue-eyed (bb), they must have inherited a b allele from each parent. Therefore, the brown-eyed parent must have a b allele to pass on.
   • This means the brown-eyed parent's genotype is Bb.

   Answer: The genotype of the brown-eyed parent is Bb.

8. Problem: In fruit flies, gray body color (G) is dominant to ebony body color (g). A gray-bodied female is crossed with an ebony-bodied male. The offspring consist of 142 gray-bodied flies and 138 ebony-bodied flies. What are the probable genotypes of the parents?

   Solution:

   • Known: Gray body (G) is dominant, ebony body (g) is recessive. The ebony-bodied male must be gg. The gray-bodied female could be GG or Gg.
   • Since some offspring are ebony-bodied (gg), they must have inherited a g allele from each parent. Therefore, the gray-bodied female must have a g allele to pass on.
   • This means the gray-bodied female's genotype is Gg.

   Answer: The genotype of the gray-bodied female is Gg. The genotype of the ebony-bodied male is gg.

9. Problem: A plant with round seeds is crossed with a plant with wrinkled seeds. All the offspring have round seeds. Explain the genotypes of the parents and offspring, assuming that round seeds (R) are dominant to wrinkled seeds (r).

   Solution:

   • Known: Round seeds (R) are dominant, wrinkled seeds (r) are recessive. The plant with wrinkled seeds must be rr. The plant with round seeds could be RR or Rr.
   • Since all the offspring have round seeds, the round-seeded parent must be RR. If it were Rr, some offspring would have wrinkled seeds (rr).
   • Therefore, the cross is RR x rr.
   • The offspring genotype is all Rr.

   Answer: The genotype of the round-seeded parent is RR. The genotype of the wrinkled-seeded parent is rr. The genotype of all the offspring is Rr.

10. Problem: Polydactyly (having extra fingers or toes) is a dominant trait in humans. A man with polydactyly, whose mother did not have polydactyly, marries a woman without polydactyly. What is the probability that their child will have polydactyly?

    Solution:

    • Known: Polydactyly (P) is dominant, normal number of digits (p) is recessive. The woman without polydactyly must be pp. The man has polydactyly, so he is either PP or Pp.
    • The man's mother did not have polydactyly, meaning she was pp. Therefore, the man must have inherited a p allele from his mother. Since he has polydactyly, his genotype must be Pp.
    • The cross is Pp x pp.
    • Punnett Square:

    	P	p
    p	Pp	pp
    p	Pp	pp

    • The probability of their child having polydactyly (Pp) is 2 out of 4 (2/4) or 50%.

    Answer: The probability that their child will have polydactyly is 50%.

Beyond Complete Dominance: A Glimpse into Other Inheritance Patterns

While complete dominance provides a foundational understanding of inheritance, many genetic traits exhibit more complex patterns. These include:

• Incomplete Dominance: The heterozygote exhibits a phenotype intermediate between the two homozygous phenotypes. For example, in snapdragons, a cross between a red-flowered plant (RR) and a white-flowered plant (WW) results in pink-flowered plants (RW).
• Codominance: Both alleles in the heterozygote are fully expressed. For example, in human blood types, the A and B alleles are codominant. An individual with the AB genotype expresses both A and B antigens on their red blood cells.
• Multiple Alleles: Some genes have more than two alleles in the population. Human blood type is also an example of multiple alleles, with the A, B, and O alleles.
• Sex-linked Inheritance: Genes located on sex chromosomes (X and Y in humans) exhibit different inheritance patterns in males and females.
• Polygenic Inheritance: Traits controlled by multiple genes, often resulting in a continuous range of phenotypes (e.g., human height, skin color).
• Epistasis: One gene masks or modifies the expression of another gene.

Conclusion

Understanding complete dominance is crucial for grasping the fundamentals of Mendelian genetics. This article has provided a comprehensive overview of this inheritance pattern, along with a detailed worksheet and answer key to reinforce these concepts. While complete dominance is a simplification of the complexities of inheritance, it serves as a critical stepping stone to understanding more intricate genetic phenomena like incomplete dominance, codominance, and polygenic inheritance. By mastering these fundamental principles, students and enthusiasts can gain a deeper appreciation for the fascinating world of genetics and its impact on the diversity of life. The practice problems provided in this worksheet are designed to solidify understanding and build confidence in applying Mendelian genetics principles to real-world scenarios. Remember to always carefully consider the genotypes of the parents, construct accurate Punnett squares, and interpret the resulting genotypic and phenotypic ratios to arrive at the correct answer. Through consistent practice and a solid understanding of the underlying principles, you can successfully navigate the complexities of Mendelian genetics and beyond.