Chapter 4: Extensions of Mendelian Genetics Flashcards
True or False?
In a cross between two strains that are true breeding for purple and white flowers, the F2 phenotypic ratio would be ¼ purple, ¼ white, and ½ lavender if the flower color phenotype exhibits incomplete dominance.
True
The heterozygous offspring exhibit a phenotype that is intermediate between the parents.
Which of the following statements about an individual with the Bombay phenotype is correct?
The individual lacks the enzyme required to produce the H substance.
Individuals with the Bombay phenotype are homozygous hh and do not make the enzyme required to produce the H substance. Thus, they cannot modify the H substance to produce the antigens on which type A and B blood phenotypes are based.
What is the expected phenotypic ratio of a cross between a disc-shaped squash that is heterozygous at both loci and a long squash?
4⁄16 disc, 8⁄16 sphere, 4⁄16 long
The genotypic ratio for this cross is 4⁄16 AB, 4⁄16 Ab, 4⁄16 aB, and 4⁄16 ab, making the phenotypic ratio 4⁄16 disc, 8⁄16 sphere, 4⁄16 long.
How many different phenotypes are possible in a one gene/three allele system that displays codominance to each other?
6
With three alleles, a1, a2, a3, each unique combination of two alleles results in a distinct phenotype. Possible combinations are: a1/a1 a1/a2 a1/a3 a2/a2 a2/a3 a3/a3
Which of the following is the most likely example of codominance?
A pure-breeding tall plant is crossed to a pure-breeding short plant. All their progeny are of medium height. A pure-breeding plant with red flowers is crossed to a pure-breeding plant with white flowers. All their progeny have pink flowers. A plant with red flowers is crossed to a plant with white flowers. Half their progeny have red flowers, the other half have white flowers. A pure-breeding plant with red flowers is crossed to a pure-breeding plant with white flowers. All their progeny have flowers with some red patches and some white patches.
A pure-breeding plant with red flowers is crossed to a pure-breeding plant with white flowers. All their progeny have flowers with some red patches and some white patches.
Heterozygotes express both alleles.
The color dilution gene in horses is an example of incomplete dominance.
True
In mice, agouti fur is a dominant trait. A mouse with solid color fur is the recessive phenotype (A = agouti; a = solid color).
A separate gene, which is not linked to the agouti gene, can result in either a dominant black pigment or a recessive brown pigment (B = black; b = brown).
A litter of mice from the mating of two agouti black parents includes offspring with the following fur colors: solid color, black solid color, brown agouti black agouti brown
What would be the expected frequency of agouti brown offspring in the litter?
3/16
Because the two traits are determined by unlinked genes, they assort independently. As a result, you need to use the multiplication rule to calculate the probability of agouti brown offspring (A_ bb) from AaBb parents. The probability of A_offspring is 3/4, and the probability of bb offspring is 1/4. The combined probability is therefore 3/4 x 1/4 = 3/16.
In addition to A and a, the agouti gene has a third allele, AY.
The AY allele is dominant to both A and a.
The homozygous genotype (AYAY ) results in lethality before birth.
The heterozygous genotypes (AYA or AYa) result in yellow fur color, regardless of which alleles are present for the B/b gene. (This effect exhibited by the AY allele is known as epistasis–when the expression of one gene masks the expression of a second gene.)
In a mating of mice with the genotypes AYaBb x AYaBb , what is the probability that a Live-Born offspring will have yellow fur?
2/3
Because the presence of the AY allele is epistatic to (masks expression of) the B/b gene, the B/b gene does not need to be taken into consideration in this problem. For the AYa x AYa cross, 1/4 of the offspring would have the AYAY genotype, which is lethal before birth. For the live-born offspring, 2/3 would be AYa, and thus have yellow fur.
In the same mouse species, a third unlinked gene (gene C/c) also has an epistatic effect on fur color. The presence of the dominant allele C (for color), allows the A/a and B/b genes to be expressed normally. The presence of two recessive alleles (cc), on the other hand, prevents any pigment from being formed, resulting in an albino (white) mouse.
Because the C/c gene is epistatic to both the A/a and B/b genes, any offspring with the cc allele combination will be albino. Otherwise, the A/a and B/b genes are expressed normally.
AaBbcc-Albino, AaBBCC-Agouti Black Aabbcc-Albono AAbbCc Agouti Brown aaBbCc-Solid color, Black AABBcc-Albino
piebaldism
Piebaldism is a rare autosomal dominant disorder of melanocyte development. Common characteristics include a congenital white forelock, scattered normal pigmented and hypopigmented macules and a triangular shaped depigmented patch on the forehead.
Although piebaldism may visually appear to be partial albinism, it is a fundamentally different condition. The vision problems associated with albinism are not usually present as eye pigmentation is normal. Piebaldism differs from albinism in that the affected cells maintain the ability to produce pigment but have that specific function turned off. In albinism the cells lack the ability to produce pigment altogether.
In the same mouse species, a 4th unlinked gene (gene P/p) also affects fur color.
For mice that are either homozygous dominant (PP) or heterozygous (Pp), the organism’s fur color is dictated by the other three genes (A/a, B/b, and C/c).
For mice that are homozygous recessive (pp), large patches of the organism’s fur are white. This condition is called piebaldism.
In a cross between 2 mice that are heterozygous for agouti, black, color, and piebaldism, what is the probability that offspring will have solid black fur along with large patches of white fur?
9/256
Because each gene segregates independently, you need to determine the probability of each genotype independently and then multiply the four probabilities together. The probability of offspring with solid color (aa) is 1/4; the probability of offspring with black fur (BBor Bb) is 3/4; the probability of colored fur (CCor Cc) is 3/4; and the probability of piebald, or white patches (pp), is 1/4. The combined probability is1/4 x 3/4 x 3/4 x 1/4 = 9/256.
White eye color is an X-linked trait in one line of fruit flies. White eyes is recessive to red eyes. If a red-eyed female and a white-eyed male are crossed, _______.
some of their male progeny may have white eyes
If the female is heterozygous, approximately half of the male progeny will have white eyes.
TRUE or FALSE?
For X-linked traits in Drosophila, the male phenotype is determined by the maternally inherited allele.
True
Males inherit only one X chromosome. That chromosome is contributed by the female parent.
Eye color in Drosophila is an X-linked trait. White eyes is recessive to red eyes. If a Drosophila male has white eyes, which of the following must also be true?
His mother had at least one white allele.
Because this male had white eyes, he must have inherited a white allele from his mother.
The inheritance of eye color in Drosophila is controlled by genes on each of the fly’s four chromosome pairs. One eye-color gene is on the fly’s X chromosome, so the trait is inherited in a sex-linked manner. For this sex-linked trait, the wild-type (brick red) allele is dominant over the mutant vermilion (bright red) allele.
A homozygous wild-type female fly is mated with a vermilion male fly.
Predict the eye colors of F1 and F2 generations. (Assume that the F1 flies interbreed to produce the F2 generation.)
Both F1 females and F1 males will have the wild-type eye color, but because the trait is sex-linked, F1 females will be heterozygous for the trait.
As a result, F2 males have a 1/2 chance of inheriting each allele, and thus of having that eye color. F2 females, on the other hand, will all have the wild-type eye color because they inherit the dominant allele on the X chromosome from their fathers (the F1 males).
In cases involving complete dominance, homozygous dominant individuals cannot be distinguished from heterozygous individuals by phenotype alone. Therefore, this phenotypic category contains two genotypes.
In cases involving incomplete dominance, each genotype has a distinct phenotype.
Like incomplete dominance, codominance produces a distinct phenotype for each of the three genotypes in a monohybrid cross.
Unlike incomplete dominance, however, codominance results in both alleles fully expressing their phenotype in the heterozygote.
Is the condition recessive or dominant?
Dominant conditions require that an affected individual have at least one affected parent.
An affected offspring of two unaffected parents indicates a recessive condition.
Is the condition autosomal or X-linked?
An affected female offspring of two unaffected parents indicates an autosomal recessive condition. If the condition were X-linked, the male parent would be affected.
To distinguish between rare X-linked and autosomal recessive conditions, choose the simpler mode for the observed pattern—that is, the mode that requires fewer unrelated individuals to carry a rare allele. In this example, the autosomal recessive mode would require two unrelated horses to carry a rare recessive allele for condition B, whereas the X-linked recessive mode requires only one. Therefore, the X-linked recessive mode is a better choice.
A pedigree provides a pictorial representation of the genetic relationships in a family and shows each person’s sex and phenotype with respect to the trait in question. In this family, Leah is the only family member (on both sides) that has galactosemia.
Based on the inheritance pattern, which mode of inheritance must be the cause of galactosemia?
autosomal recessive
Because neither Jane nor John has the same condition as their daughter, and there is no evidence of sex-linkage, galactosemia must be an autosomal recessive trait.
If Jane and John want to have another child, they plan to see a genetic counselor to find out when it would be best to test for galactosemia. A newborn with galactosemia must be put on a lactose- and galactose-free diet as soon as possible after birth. Even on this diet, affected individuals may still suffer from learning disabilities, ovarian failure (in young women), late-onset cataracts, and early death.
Which of the following tests would be most useful for Jane and John to have?
newborn screening (either assaying for the GALT enzyme or measuring excess galactose in the newborn’s blood)
Chromosome analysis will not identify this abnormality because it is caused by a mutation in a single gene.
Newborn screening by either of these methods would be effective in identifying the condition in a newborn. It would provide definitive information early enough to begin dietary restrictions, if needed.
The study of modern genetics began with Thomas Hunt Morgan in the early 1900s. Morgan’s experiments with the fruit fly Drosophila were the first to demonstrate that genes are located on chromosomes and are the basis of heredity.
One of Morgan’s first challenges was to find fruit flies with mutant phenotypes. He discovered a single male fly with white eyes, instead of the usual (wild-type) red eye color. Through his genetic experiments with the white-eyed fly, Morgan deduced that a fruit fly’s eye color was somehow linked to its sex.
Reciprocal crosses
When Gregor Mendel conducted his genetic experiments with pea plants, he observed that a trait’s inheritance pattern was the same regardless of whether the trait was inherited from the maternal or paternal parent. Mendel made these observations by carrying out reciprocal crosses: For example, he first crossed a female plant homozygous for yellow seeds with a male plant homozygous for green seeds, and then crossed a female plant homozygous for green seeds with a male plant homozygous for yellow seeds.
Unlike Mendel, however, Morgan obtained very different results when he carried out reciprocal crosses involving eye color in his fruit flies. The diagram below shows Morgan’s reciprocal cross: He first crossed a homozygous red-eyed female with a white-eyed male, and then crossed a homozygous white-eyed female with a red-eyed male.
When Morgan crossed a homozygous red-eyed female with a white-eyed male, all of the offspring had red eyes. However, in the reciprocal cross (homozygous white-eyed female with a red-eyed male), all of the females had red eyes while all of the males had white eyes.
When a homozygous red-eyed female was crossed with the white-eyed male (w+w+ × wY), the resulting F1 females were w+w and the F1 males were w+ Y. Crossing the F1 males and F1 females would yield these results:
All the F2 females would have red eyes, although some would be homozygous (w+w+ ) and others would be heterozygous (w+w).
Half the F2 males would have red eyes (w+ Y), and half would have white eyes (wY).
Use these rules when assigning genotypes for X-linked recessive conditions:
Males have only one allele for every X-linked gene. Affected males have the recessive allele, and non-affected males have the dominant (wild-type) allele.
Unaffected females with affected sons are heterozygous (carriers).
Unaffected females with affected fathers are heterozygous.
All sons of affected females will be affected.
All daughters of affected females will be heterozygous
