Plant Chapter 14 Concept Check

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Quizzes Created: 4 | Total Attempts: 858

Questions: 11 | Attempts: 292 | Updated: Mar 20, 2023

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Questions and Answers

1.

A pea plant heterozygous for inflated pods (Ii) is crossed with a plant homozygous for constricted pods (ii). Draw a Punnett square for this cross. Assume pollen comes from the ii plant. A cross of Ii x ii would yield offspring with a genotypic ratio of : and a phenotypic ratio of inflated : constricted
2.

Pea plants heterozygous for flower position and stem length (AaTt) are allowed to self-pollinate, and 400 of the resulting seeds are planted. Draw a Punnett square for this cross. How many offspring would be predicted to have terminal flowers and be dwarf? According to the law of , plants are predicted to be aatt, or recessive for both characters. The actual result is likely to differ slightly from this value.

Explanation
According to the law of independent assortment, genes for different traits segregate independently during the formation of gametes. In this case, the plants are heterozygous for flower position (Aa) and stem length (Tt). When these plants self-pollinate, there are four possible combinations of alleles in the offspring: AT, At, aT, and at. Among these combinations, the offspring with the genotype aatt would have terminal flowers and be dwarf. Since there are four possible combinations and each has an equal chance of occurring, 25% of the offspring (or 25 out of 100) would be predicted to have terminal flowers and be dwarf.

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3.

List the different gametes that could be made by a pea plant heterozygous for seed color, seed shape, and pod shape (YyRrIi). How large a Punnett square would you need to predict the offspring of a self-pollination of this “trihybrid”? The plant could make eight different gametes ( , , , , , , , ) To fit all the possible gametes in a self-pollination, a Punnet square would need rows and columns. It would have spaces for the possible unions of gametes in the offspring.

Explanation
A pea plant heterozygous for seed color, seed shape, and pod shape (YyRrIi) can produce eight different gametes, which are YRI, YRi, YrI, Yri, yRI, rRi, yrI, and yri. To predict the offspring of a self-pollination of this "trihybrid," a Punnett square would need 8 rows and 8 columns. This Punnett square would have spaces for the 64 possible unions of gametes in the offspring.

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4.

For any gene with a dominant allele C and recessive allele c, what proportions of the offspring from a CC x Cc cross are expected to be homozygous dominant, homozygous recessive, and heterozygous? homozygous dominant (CC), homozygous recessive (cc), and heterozygous (Cc)

Explanation
In a cross between CC and Cc, the dominant allele C will always be expressed in the offspring. Therefore, half of the offspring will be homozygous dominant (CC) and the other half will be heterozygous (Cc). Since the recessive allele c is masked by the dominant allele, there will be no homozygous recessive offspring (cc) in this cross.

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5.

An organism with the genotype BbDD is mated to one with the genotype BBDd. Assuming independent assortment of these two genes, write the genotypes of all possible offspring from this cross and use the rules of probability to calculate the chance of each genotype occurring. BBDD; BbDD; BBDd; BbDd

Explanation
The given genotypes of all possible offspring from the cross are BBDD, BbDD, BBDd, and BbDd. Since the genes assort independently, the chance of each genotype occurring is equal. Therefore, the probability of each genotype occurring is 1/4 or 25%.

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6.

Three characters (flower color, seed color, and pod shape) are considered in a cross between two pea plants (PpYyIi x ppYyii). What fraction of the offspring would be predicted to be homozygous recessive for at least two of the three characters? The genotypes and frequencies to fulfill this condition are

Explanation
The fraction of the offspring predicted to be homozygous recessive for at least two of the three characters can be calculated by adding up the frequencies of the genotypes that fulfill this condition. In this case, the genotypes that fulfill the condition are ppyyIi, ppYyii, Ppyyii, ppYYii, and ppyyii. The frequencies of these genotypes are 1/16, 2/16, 1/16, 1/16, and 1/16 respectively. Adding up these frequencies gives a total of 6/16, which simplifies to 3/8. Therefore, the fraction of the offspring predicted to be homozygous recessive for at least two of the three characters is 3/8.

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7.

Incomplete dominance and epistasis are both terms that define genetic relationships. What is the most basic distinction between these terms? describes the relationship between two alleles of a , whereas relates o the genetic relationship between two

Explanation
Incomplete dominance describes the relationship between two alleles of a single gene, where neither allele is completely dominant over the other. In incomplete dominance, a heterozygous individual displays an intermediate phenotype that is a blend of the two homozygous phenotypes. On the other hand, epistasis relates to the genetic relationship between two genes. It occurs when the expression of one gene masks or modifies the expression of another gene. Epistasis can result in the suppression or alteration of a trait controlled by another gene.

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8.

If a man with type AB blood marries a woman with type O blood, what blood types would you expect in their children? of the children would be expected to have to have type A blood and type B blood.

Explanation
When a man with type AB blood marries a woman with type O blood, their children can inherit either type A or type B blood from the father, and they can only inherit type O blood from the mother. Therefore, it is expected that half of their children will have type A blood and the other half will have type B blood.

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9.

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 phenotypes would you expect in the offspring of a cross between a gray rooster and a black hen? The black and white alleles ar expressing , with gray in color. A cross between a gray rooster and a black hen should yield gray and blackoffspring.

Explanation
The simplest explanation for the inheritance of these colors in chickens is incomplete dominance. This means that neither the gray allele nor the black allele is completely dominant over the other, resulting in a blending of the two colors in the offspring. The phenotype of the heterozygotes, which have one gray allele and one black allele, is gray. Therefore, in a cross between a gray rooster and a black hen, we would expect to see gray and black offspring in equal numbers.

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10.

Beth and Tom each have a sibling with cystic fibrosis, but neither Beth nor Tom nor any of their parents have the disease. Calculate the probability that if this couple has a child, the child will have cystic fibrosis. What would be the probability if a test revealed that Tom is a carrier but Beth is not? (Since cystic fibrosis si caused by a , Beth and Tom's siblings who have CF must be ) Therefore, each parent must be a . Since neither Beth nor Tom has CF, this means they each have a chance of being a carrier. If they are both carriers, there is a chance that they will have a child with CF. If a test proves Tom is a carrier and Beth is not, there isa % chance that they produce a child with CF.

Explanation
Beth and Tom each have a sibling with cystic fibrosis, indicating that they are both carriers of the recessive allele for the disease. Since neither Beth nor Tom has cystic fibrosis, they must each have a 2/3 chance of being carriers. If they are both carriers, there is a 1/9 chance that they will have a child with cystic fibrosis. If a test reveals that Tom is a carrier but Beth is not, there is a 1/4 chance that they will produce a child with cystic fibrosis.

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11.

Joan was born with six toes on each foot, a dominant trait called polydactyly. Two of her five siblings and her mother, but not her father, also have extra digits. What is Joan’s genotype for the number-of-digits character? Joan's genotype is because the allele for polydactyly is to the allele for five digits per appendage. Because Joan's father does not have polydactyly, his genotype must be , therefore Joan, who does have the trait, must be .

Explanation
Joan's genotype is Dd because the allele for polydactyly is dominant to the allele for five digits per appendage. Because Joan's father does not have polydactyly, his genotype must be dd, therefore Joan, who does have the trait, must be heterozygous.

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Mar 20, 2023

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Mar 26, 2011

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Plant Chapter 14 Concept Check

For any gene with a dominant allele C and recessive allele c, what proportions of the offspring from a CC x Cc cross are expected to be homozygous dominant, homozygous recessive, and heterozygous? homozygous dominant (CC), homozygous recessive (cc), and heterozygous (Cc)

An organism with the genotype BbDD is mated to one with the genotype BBDd. Assuming independent assortment of these two genes, write the genotypes of all possible offspring from this cross and use the rules of probability to calculate the chance of each genotype occurring. BBDD; BbDD; BBDd; BbDd

Incomplete dominance and epistasis are both terms that define genetic relationships. What is the most basic distinction between these terms? describes the relationship between two alleles of a , whereas relates o the genetic relationship between two

If a man with type AB blood marries a woman with type O blood, what blood types would you expect in their children? of the children would be expected to have to have type A blood and type B blood.