Autosomal Dominant Traits
- typically appears in each generation
- will not occur in offspring unless it also appears in at least one of the parents
- unaffected parents will have only affected offspring
Autosomal Recessive Trait
- if both parents are affected, any child of theirs will be affected
- commonly skips one or more generations
X-linked Recessive Inheritance
- more common in males (although both males and females can be affected)
- all sons of an affected female will be affected
X-linked Dominant Inheritance
- occurs slightly more often in females
- all daughters of an affected male will be affected
- can not be passed on from father to son
Y-linked Inheritance
- all sons of an affected male will be affected
- none of the females are affected
- does not skip generations
Monday, September 3, 2007
Sunday, September 2, 2007
Problems on Sex-related Inheritance
Sex-linked inheritance
1. Illustrate the following crosses among fruit flies and determine genotype and phenotypes of the resulting offspring:
a. true-breeding red-eyed female x white-eyed male
b. heterozygous red-eyed female x white-eyed male
c. white-eyed female x red-eyed male
2. Red- green colorblindness is caused by an X-linked recessive allele. A colorblind man marries a woman with normal vision whose father was colorblind.
a. What is the probability that their daughter will be colorblind?
b. What is the probability that they will have a colorblind son?
3. A man with hemophilia has a daughter of normal phenotype. She marries a man who is normal for the trait.
a. What is the probability that a daughter of this mating will be a hemophiliac?
b. That a son will be a hemophiliac?
4. Given that hypophosphatemia is an X-linked dominant trait. Predict the results of the mating between:
a. An unaffected father and an affected mother
b. An affected father and an unaffected mother
Sex-limited inheritance
5. In domestic fowl, the differences in plumage between males and females is sex-influenced. If the genotypes and phenotypes for plumage are as listed below:
Genotype Female Male
HH hen-feathered hen-feathered
Hh hen-feathered hen-feathered
hh hen-feathered rooster-feathered
Predict the F1 and F2 results of crossing a male that is rooster-feathered and a true-breeding hen-feathered female.
6. In the clover butterfly, all males are yellow, but females may be yellow if they are of the homozygous recessive phenotype (yy) or white if they possess the dominant allele (Y_). What phenotypic proportions are expected from a cross between heterozygous parents?
Sex-influenced inheritance
7. Spotting in cattle is controlled by two alleles where CM is the allele for mahogany and white, while CR is the allele for red and white. This trait is sex-influenced and the different sexes exhibit the two phenotypes as follows:
In males:
mahogany & white à dominant (CMCM, CMCR)
red & white à recessive (CRCR)
In females:
red & white à dominant (CRCR, CMCR)
mahogany & white à (CMCM)
a. If a red & white male is crossed to a mahogany & white female, what phenotypic and genotypic proportions are expected in the F1 and F2 generations?
b. If a mahogany & white cow gives birth to a red & white calf, what sex is the calf?
1. Illustrate the following crosses among fruit flies and determine genotype and phenotypes of the resulting offspring:
a. true-breeding red-eyed female x white-eyed male
b. heterozygous red-eyed female x white-eyed male
c. white-eyed female x red-eyed male
2. Red- green colorblindness is caused by an X-linked recessive allele. A colorblind man marries a woman with normal vision whose father was colorblind.
a. What is the probability that their daughter will be colorblind?
b. What is the probability that they will have a colorblind son?
3. A man with hemophilia has a daughter of normal phenotype. She marries a man who is normal for the trait.
a. What is the probability that a daughter of this mating will be a hemophiliac?
b. That a son will be a hemophiliac?
4. Given that hypophosphatemia is an X-linked dominant trait. Predict the results of the mating between:
a. An unaffected father and an affected mother
b. An affected father and an unaffected mother
Sex-limited inheritance
5. In domestic fowl, the differences in plumage between males and females is sex-influenced. If the genotypes and phenotypes for plumage are as listed below:
Genotype Female Male
HH hen-feathered hen-feathered
Hh hen-feathered hen-feathered
hh hen-feathered rooster-feathered
Predict the F1 and F2 results of crossing a male that is rooster-feathered and a true-breeding hen-feathered female.
6. In the clover butterfly, all males are yellow, but females may be yellow if they are of the homozygous recessive phenotype (yy) or white if they possess the dominant allele (Y_). What phenotypic proportions are expected from a cross between heterozygous parents?
Sex-influenced inheritance
7. Spotting in cattle is controlled by two alleles where CM is the allele for mahogany and white, while CR is the allele for red and white. This trait is sex-influenced and the different sexes exhibit the two phenotypes as follows:
In males:
mahogany & white à dominant (CMCM, CMCR)
red & white à recessive (CRCR)
In females:
red & white à dominant (CRCR, CMCR)
mahogany & white à (CMCM)
a. If a red & white male is crossed to a mahogany & white female, what phenotypic and genotypic proportions are expected in the F1 and F2 generations?
b. If a mahogany & white cow gives birth to a red & white calf, what sex is the calf?
Saturday, August 4, 2007
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
Since the inheritance of flower color in snapdragons is governed by incomplete inheritance, let’s first designate the alleles and the genotypes for the different plants: Red = RR, White = WW and Pink = RW. You may also opt to use r for white, so that the white-flowered snapdragons will be rr and the pink-flowered ones Rr. In either case, it’s important to remember that the heterozygote exhibits a phenotype that is distinct from the dominant and the recessive traits.
RR x WW --> RW; all the offspring will be have pink flowers
RW x RW --> 1RR:2RW:1WW; 1 red:2 pink:1 white
RW x WW --> 1RW:1WW; 1 pink:1 white
RW x RR --> 1RR:1RW; 1 pink:1 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.
There are three alleles for the ABO blood type: IA, IB and i. IA encodes for the production of antigen A on the surface of RBCs, IB for the production of antigen B and i, for the absence of these antigens. An individual that has type AB blood has both the alleles for the production of antigens A (IA) and B (IB), thus the father in this problem has the genotype IAIB. On the other hand, the mother, has blood type O which means that she does not have either antigens A or B, resulting from a genotype of ii.
So this cross written out will be: IAIB x ii
The father can produce two types of gametes, IA and IB; while the mother can only produce one type, i. The allele combinations that are possible in the resulting offspring are IAi and IBi, so the offspring can have either type A or type B 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 gray rooster and a black hen?
The pattern of inheritance described in this problem is most likely incomplete dominance. The phenotypic ratio of the offspring approximates 1:2:1 (6 black-feathered:15 gray-feathered:1 white-feathered), which is characteristic of this type of inheritance. In addition, of course, there is a third distinct phenotype (gray) that is exhibited by the heterozygotes, which is a blend of black and white phenotypes exhibited by the homozygotes.
A cross between a gray (BW) rooster and a black (BB) hen would result in half of the offspring possessing gray feathers (BW) while the other half would have black feathers (BB).
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?
The blood types of each parent are given, so we already know that the mother, having type A blood, can have a genotype of either IAIA or IAi; while the father, having type B blood, can be either IBIB or IBi. But since their child has type O blood, then we can conclude that both mother and father each have the recessive allele i, in order to have such a child. And since we now know what their phenotypes are: IAi for the mother and IBi for the father, we can predict the offspring that may result from this cross: the genotypic ratio would then be 1 IAIB:1 IAIB:1 IBi:1 ii.
a. Red x White
b. Pink x Pink
c. Pink x White
d. Pink x Red
Since the inheritance of flower color in snapdragons is governed by incomplete inheritance, let’s first designate the alleles and the genotypes for the different plants: Red = RR, White = WW and Pink = RW. You may also opt to use r for white, so that the white-flowered snapdragons will be rr and the pink-flowered ones Rr. In either case, it’s important to remember that the heterozygote exhibits a phenotype that is distinct from the dominant and the recessive traits.
RR x WW --> RW; all the offspring will be have pink flowers
RW x RW --> 1RR:2RW:1WW; 1 red:2 pink:1 white
RW x WW --> 1RW:1WW; 1 pink:1 white
RW x RR --> 1RR:1RW; 1 pink:1 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.
There are three alleles for the ABO blood type: IA, IB and i. IA encodes for the production of antigen A on the surface of RBCs, IB for the production of antigen B and i, for the absence of these antigens. An individual that has type AB blood has both the alleles for the production of antigens A (IA) and B (IB), thus the father in this problem has the genotype IAIB. On the other hand, the mother, has blood type O which means that she does not have either antigens A or B, resulting from a genotype of ii.
So this cross written out will be: IAIB x ii
The father can produce two types of gametes, IA and IB; while the mother can only produce one type, i. The allele combinations that are possible in the resulting offspring are IAi and IBi, so the offspring can have either type A or type B 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 gray rooster and a black hen?
The pattern of inheritance described in this problem is most likely incomplete dominance. The phenotypic ratio of the offspring approximates 1:2:1 (6 black-feathered:15 gray-feathered:1 white-feathered), which is characteristic of this type of inheritance. In addition, of course, there is a third distinct phenotype (gray) that is exhibited by the heterozygotes, which is a blend of black and white phenotypes exhibited by the homozygotes.
A cross between a gray (BW) rooster and a black (BB) hen would result in half of the offspring possessing gray feathers (BW) while the other half would have black feathers (BB).
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?
The blood types of each parent are given, so we already know that the mother, having type A blood, can have a genotype of either IAIA or IAi; while the father, having type B blood, can be either IBIB or IBi. But since their child has type O blood, then we can conclude that both mother and father each have the recessive allele i, in order to have such a child. And since we now know what their phenotypes are: IAi for the mother and IBi for the father, we can predict the offspring that may result from this cross: the genotypic ratio would then be 1 IAIB:1 IAIB:1 IBi:1 ii.
Wednesday, July 25, 2007
Non-Mendelian Inheritance
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...
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...
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.
Subscribe to:
Comments (Atom)
