Practice Problems
1. Flower color in snap dragons is governed by two alleles that show incomplete dominance. Determine the genotypes and the phenotypes in the offspring of the following crosses:
a. Red x White
b. Pink x Pink
c. Pink x White
d. Pink x Red
2. ABO blood type is governed by a multiple allele system. Identify the blood types among the offspring if you cross a male with type AB blood with a female with type O blood.
3. A rooster with gray feathers is mated with a hen of the same phenotype. Among their offspring, 15 chicks are gray, 6 are black and 8 are white. What is the simplest explanation for the inheritance of these colors in chickens? What offspring would you predict from the mating of a grey rooster and a black hen?
4. A man with type A blood marries a woman with type B blood. Their child has type O blood. What are the genotypes of these individuals? What other genotypes, and in what frequencies, would you expect in offspring from this marriage?
Answers to follow...
Wednesday, July 25, 2007
Thursday, July 19, 2007
Harry Potter's Earlobes
Another genetics problem:
In humans, attached earlobes is recessive to free or unattached earlobes. James, who has free earlobes, married, Lily, who has attached earlobes. Their son, Harry, has attached earlobes.
a. Based on the information above, what could be James', Lily's and Harry's genotype?
b. If James and Lily had not died, leaving Harry an orphan, what would be the probability that they will have a daughter with free earlobes?
Answers:
a. James = Aa, Lily = aa and Harry = aa
It is already given that both Lily and Harry have attached earlobes, which is the recessive trait. So you know for sure that they are both homozygous for the recessive trait (aa), since the recessive trait can only be expressed in the absence of the dominant allele. James, on the other hand, exhibits the dominant trait (free earlobes) but that means that he can either be homozygous (AA) or heterozygous (Aa) for that trait. But you already know that Harry's genotype is aa, which means that each of his parents contributed a recessive allele (a). Thus James must be a heterozygote (Aa).
b. 1/4
If you perform the cross between James (Aa) and Lily (aa), 1/2 of the expected offspring would have free earlobes (Aa) and the other 1/2 would have attached earlobes (aa). So the probability of a child with free earlobes is 1/2. However, the problem stipulates that this child is a daughter, and the probability for that is again 1/2 (chance of getting an X chromosome from the father). Since we're looking at events that have to occur in combination, i.e., the child has free earlobes
AND is a girl, we need to get the product of the probabilities for each of those independent events. Hence, 1/2 x 1/2 = 1/4.
It took quite a while to make up this problem. I got the idea from an earlier one that was used in previous years, dealing with the inheritance of hair color in the Skywalker family (Padme, Anakin, Luke and Leia). It needed a bit of updating though, and since I'm a big Harry Potter fan, I wanted it to be about him. Unfortunately, the hair color problem was a bit tricky... James and Harry both had the dominant trait (dark hair), even though Lily might pass for blonde (which is recessive). Fortunately, pictures of the actors were readily accessible via google and imdb. But it was quite difficult looking to see whether Daniel Radcliffe's earlobes were indeed attached. After painstakingly looking at pics of the actors with their ears showing, and serious (hahaha) consultations with the other bio teachers, this problem was born.
In humans, attached earlobes is recessive to free or unattached earlobes. James, who has free earlobes, married, Lily, who has attached earlobes. Their son, Harry, has attached earlobes.
a. Based on the information above, what could be James', Lily's and Harry's genotype?
b. If James and Lily had not died, leaving Harry an orphan, what would be the probability that they will have a daughter with free earlobes?
Answers:
a. James = Aa, Lily = aa and Harry = aa
It is already given that both Lily and Harry have attached earlobes, which is the recessive trait. So you know for sure that they are both homozygous for the recessive trait (aa), since the recessive trait can only be expressed in the absence of the dominant allele. James, on the other hand, exhibits the dominant trait (free earlobes) but that means that he can either be homozygous (AA) or heterozygous (Aa) for that trait. But you already know that Harry's genotype is aa, which means that each of his parents contributed a recessive allele (a). Thus James must be a heterozygote (Aa).
b. 1/4
If you perform the cross between James (Aa) and Lily (aa), 1/2 of the expected offspring would have free earlobes (Aa) and the other 1/2 would have attached earlobes (aa). So the probability of a child with free earlobes is 1/2. However, the problem stipulates that this child is a daughter, and the probability for that is again 1/2 (chance of getting an X chromosome from the father). Since we're looking at events that have to occur in combination, i.e., the child has free earlobes
AND is a girl, we need to get the product of the probabilities for each of those independent events. Hence, 1/2 x 1/2 = 1/4.
It took quite a while to make up this problem. I got the idea from an earlier one that was used in previous years, dealing with the inheritance of hair color in the Skywalker family (Padme, Anakin, Luke and Leia). It needed a bit of updating though, and since I'm a big Harry Potter fan, I wanted it to be about him. Unfortunately, the hair color problem was a bit tricky... James and Harry both had the dominant trait (dark hair), even though Lily might pass for blonde (which is recessive). Fortunately, pictures of the actors were readily accessible via google and imdb. But it was quite difficult looking to see whether Daniel Radcliffe's earlobes were indeed attached. After painstakingly looking at pics of the actors with their ears showing, and serious (hahaha) consultations with the other bio teachers, this problem was born.
Wednesday, July 18, 2007
Genetics Problems: Mendelian Inheritance
Try your hand at solving these problems on Mendelian inheritance:
Monohybrid Cross
1. Mendel crossed pea plants heterozygous for the height gene (Tt) and obtained the monohybrid phenotypic ratio of 3:1 and the genotypic ratio 1:2:1. Calculate the genotypic and phenotypic ratios for the following crosses:
a. Homozygous dominant crossed to homozygous recessive
b. Homozygous dominant crossed to heterozygous
c. Homozygous recessive crossed to heterozygous
2. Bob and Joan know from a blood test that they are each heterozygous for the autosomal recessive gene that causes sickle cell disease. If their first three children are healthy, what is the probability that their fourth child will have the disease?
3. In garden pea plants, purple flowers are dominant over white flowers.
a. A heterozygous purple flower is allowed to self-pollinate. What are the probable genotypic and phenotypic ratios in the offspring of this plant?
b. Pollen from a pea plant with white flowers is used to fertilize the ovules (female gametes) of a heterozygous plant. What are the possible phenotypes in the offspring of this cross?
c. You have a pea plant with purple flowers. Design a cross to determine if this plant is homozygous or heterozygous. Use a Punnet square to show all possible crosses.
4. Use a Punnett squares to illustrate test crosses to determine whether a black male guinea is homozygous or heterozygous for black. The black coat is produced by the dominant allele (B).
5. Assume that 50 percent of 10,000 pea plant offspring are short. Use a Punnett square to
show the probable genotypes of the parents and the offspring. Let T stand for the dominant allele, and t for the recessive allele.
Dihybdrid/Trihybrid Cross
1. In the fruit fly Drosophila, wings (A) are dominant over lack of wings (a), and red eyes (E) are dominant over sepia (brownish) eyes (e). A wingless fly that is heterozygous for eye color is crossed with a fly that is heterozygous for both eye color and the presence of wings.
a. What are the genotypic and phenotypic ratios for this cross?
b. What fraction of the offspring from this cross will be wingless and have sepia eyes?
c. What fraction will have the genotype AaEe?
2. A man and a woman each have dark eyes, dark hair and freckles. The genes for these traits assort independently. The woman is heterozygous for each of these autosomal trait, but the man is homozygous. The dominance relationships of the alleles are as follows:
B = dark eyes b = blue eyes
H = dark hair h = blond hair
F = freckles f = no freckles
a. What is the probability that their child sill have the same phenotype as the parents?
b. What is the probability that their child will have the same genotype as each parent?
Use probability or a Punnett square to obtain your answers.
Monohybrid Cross
1. Mendel crossed pea plants heterozygous for the height gene (Tt) and obtained the monohybrid phenotypic ratio of 3:1 and the genotypic ratio 1:2:1. Calculate the genotypic and phenotypic ratios for the following crosses:
a. Homozygous dominant crossed to homozygous recessive
b. Homozygous dominant crossed to heterozygous
c. Homozygous recessive crossed to heterozygous
2. Bob and Joan know from a blood test that they are each heterozygous for the autosomal recessive gene that causes sickle cell disease. If their first three children are healthy, what is the probability that their fourth child will have the disease?
3. In garden pea plants, purple flowers are dominant over white flowers.
a. A heterozygous purple flower is allowed to self-pollinate. What are the probable genotypic and phenotypic ratios in the offspring of this plant?
b. Pollen from a pea plant with white flowers is used to fertilize the ovules (female gametes) of a heterozygous plant. What are the possible phenotypes in the offspring of this cross?
c. You have a pea plant with purple flowers. Design a cross to determine if this plant is homozygous or heterozygous. Use a Punnet square to show all possible crosses.
4. Use a Punnett squares to illustrate test crosses to determine whether a black male guinea is homozygous or heterozygous for black. The black coat is produced by the dominant allele (B).
5. Assume that 50 percent of 10,000 pea plant offspring are short. Use a Punnett square to
show the probable genotypes of the parents and the offspring. Let T stand for the dominant allele, and t for the recessive allele.
Dihybdrid/Trihybrid Cross
1. In the fruit fly Drosophila, wings (A) are dominant over lack of wings (a), and red eyes (E) are dominant over sepia (brownish) eyes (e). A wingless fly that is heterozygous for eye color is crossed with a fly that is heterozygous for both eye color and the presence of wings.
a. What are the genotypic and phenotypic ratios for this cross?
b. What fraction of the offspring from this cross will be wingless and have sepia eyes?
c. What fraction will have the genotype AaEe?
2. A man and a woman each have dark eyes, dark hair and freckles. The genes for these traits assort independently. The woman is heterozygous for each of these autosomal trait, but the man is homozygous. The dominance relationships of the alleles are as follows:
B = dark eyes b = blue eyes
H = dark hair h = blond hair
F = freckles f = no freckles
a. What is the probability that their child sill have the same phenotype as the parents?
b. What is the probability that their child will have the same genotype as each parent?
Use probability or a Punnett square to obtain your answers.
MENDELIAN INHERITANCE: Problem Solving Tips
Monohybrid and Dihybrid Crosses
1. Choose a letter to represent the genes in the cross.
2. Indicate the genotypes of the parents.
3. Determine the possible gametes that can be produced by the parents.
4. Enter the possible gametes at the top and left hand side of the Punnett square.
5. Complete the Punnett square by writing the allelic combinations based on the gametes in the appropriate boxes corresponding to genotypes of the offspring.
6. Determine the phenotypes of the offspring.
1. Choose a letter to represent the genes in the cross.
2. Indicate the genotypes of the parents.
3. Determine the possible gametes that can be produced by the parents.
4. Enter the possible gametes at the top and left hand side of the Punnett square.
5. Complete the Punnett square by writing the allelic combinations based on the gametes in the appropriate boxes corresponding to genotypes of the offspring.
6. Determine the phenotypes of the offspring.
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