Chapter 12, Mendel's Experiments and Heredity Video Solutions, Biology for AP Courses | Numerade (2024)

Mendel performed hybridizations by transferring pollen to the female ova from what part of the male plant?
a. anther
b. pistil
c. stigma
d. seed

Which is one of the seven characteristics that Mendel observed in pea plants?
a. flower size
b. leaf shape
c. seed texture
d. stem color

Imagine you are performing a cross involving garden pea plants. What F1 offspring would you expect if you cross true-breeding parents with green seeds and yellow seeds? Yellow seed color is dominant over green.
a. 100 percent yellow-green seeds
b. 100 percent yellow seeds
c. 50 percent yellow, 50 percent green seeds
d. 25 percent green, 75 percent yellow seeds

Consider a cross to investigate the pea pod texture trait, involving constricted or inflated pods. Mendel found that the traits behave according to a dominant/recessive pattern in which inflated pods were dominant. If you performed this cross and obtained 650 inflated-pod plants in the F2 generation bred from true-breading stock, approximately how many constricted-pod plants would you expect to have?
a. 600
b. 165
c. 217
d. 468

The observable traits expressed by an organism are described as its
a. alleles
b. genotype
c. phenotype
d. zygote

A recessive trait will be observed in individuals that are what for that trait?
a. diploid
b. heterozygous
c. hom*ozygous or heterozygous
d. hom*ozygous

If black and white true-breeding mice are mated and the result is all gray offspring, what inheritance pattern would this be indicative of?
a. codominance
b. dominance
c. incomplete dominance
d. multiple alleles

The ABO blood groups in humans are controlled by theIA, IB, and I alleles. The IA allele encodes the A blood group antigen, IB encodes B, and I encodes O. Both A and B are dominant to O. If a heterozygous blood type A parent (iAi) and a heterozygous blood type B parent (iBi) mate, one quarter of their offspring will have AB blood type (IAIB) in which both antigens are expressed equally. Therefore, the ABO blood groups are an example of _______.
a. codominance and incomplete dominance
b. incomplete dominance only
c. multiple alleles and incomplete dominance
d. multiple alleles and codominance

In a mating between two individuals that are heterozygous for a recessive lethal allele that is expressed in utero, what genotypic ratio (hom*ozygous dominant : heterozygous : hom*ozygous recessive) would you expect to observe in the off-spring?
a. 1 : 2 : 1
b. 3 : 1 : 1
c. 1 : 2 : 0
d. 0 : 2 : 1

The forked line and probability methods make use of what probability rule?
a. monohybrid rule
b. product rule
c. sum rule
d. test cross

In pea plants, smooth seeds (S) are dominant to wrinkled seeds (s). In a genetic cross of two plants that are heterozygous for the seed shape trait, the Punnett square is shown. What is the missing genotype?
a. SS
b. Ss
c. sS
d. ss

If the inheritance of two traits fully obeys Mendelian laws of inheritance, where may you assume that the genes are located?
a. on any autosomal chromosome or chromosomes
b. on Y chromosomes
c. on the same chromosome
d. on separate chromosomes

How many different offspring genotypes are expected in a trihybrid cross between parents heterozygous for all three traits? How many phenotypes are expected if the traits behave in a dominant and recessive pattern?
a. 64 genotypes; 16 phenotypes
b. 16 genotypes; 64 phenotypes
c. 8 genotypes; 27 phenotypes
d. 27 genotypes; 8 phenotypes

Four-o’ clock flowers may be red, pink or white. In the crossing of a true-breeding red and true-breeding white plants, all the offspring are pink. Use a Punnett square to determine the correct genotype of the offspring if the red parent has genotype RR and the white parent has genotype rr.
a. RR and Rr
b. Rr and rr
c. Rr only
d. RR only

Which cellular process underlies Mendel’s law of independent assortment?
a. Chromosomes align randomly during meiosis.
b. Chromosomes can exchange genetic material during crossover.
c. Gametes contain half the number of chromosomes of somatic cells.
d. Daughter cells are genetically identical to parentcells after mitosis.

While studying meiosis, you observe that gametes receive one copy of each pair of hom*ologous chromosomes and one copy of the sex chromosomes. This observation is the physical explanation of Mendel’s law of ______.
a. dominance
b. independent assortment
c. random distribution of traits
d. segregation

In some primroses, the petal color blue is dominant. A cross between a true-breed blue primrose and a white primrose yields progeny with white petals. A second gene at another locus prevented the expression of the dominant coat color. This is an example of ______.
a. codominance
b. hemizygosity
c. incomplete dominance
d. epistasis

Purple flowers (P) are dominant over red flowers (p) and long pollen grains are dominant over round pollen grains. When purple flowers and long pollen grain plants were crossed with plants with white flowers and round pollen grains, all the F1 plants showed purple flowers and long pollen grains. The F1 plants were crossed and the results are in the table. What conclusions about the physical relationship between the traits can be drawn from the experiment?
a. The traits are probably linked.
b. The traits follow the law of independent assortment
c. The traits are located on different chromosomes
d. There was epistasis.

When the expression of one gene pair masks or modifies the expression of another, the genes show _______.
a. codominance
b. epistasis
c. incomplete dominance
d. partial linkage

Describe one reason why the garden pea was an excellent model system for studying inheritance.
a. The garden pea has flowers that close tightly to promote cross-fertilization.
b. The garden pea has flowers that close tightly to prevent cross-fertilization.
c. The garden pea does not mature in one season and is a perennial plant.
d. Male and female reproductive parts attain maturity at different times, promoting selffertilization.

How would you perform a reciprocal cross to test stem height in the garden pea?
a. First cross is performed by transferring the pollen of a heterozygous tall plant to the stigmaof a true breeding dwarf plant. Second cross is
performed by transferring the pollen of a heterozygous dwarf plant to the stigma of a true breeding tall plant.
b. First cross is performed by transferring the pollen of a true breeding tall plant to the stigma of a true breeding dwarf plant. Second cross is performed by transferring the pollen of a true breeding dwarf plant to the stigma of a true breeding tall plant.
c. First cross is performed by transferring the pollen of a true breeding tall plant to the stigma of a heterozygous dwarf plant. Second cross is performed by transferring the pollen of a heterozygous dwarf plant to the stigma of a true breeding tall plant.
d. First cross is performed by transferring the pollen of a heterozygous tall plant to the stigma of a heterozygous dwarf plant. Second cross is performed by transferring the pollen of a heterozygous tall plant to the stigma of a heterozygous dwarf plant

Flower position in pea plants is determined by a gene with axial and terminal alleles. Given that axial is dominant to terminal, list all of the possible F1 and F2 genotypes and phenotypes from a cross involving parents that are hom*ozygous for each trait. Express genotypes with conventional genetic abbreviations.
a. F1: All AA-axial; F2: AA-Axial and aa-terminal.
b. F1: All aa-terminal; F2: AA-Axial and Aaterminal.
c. F1: AA-axial and Aa-terminal; F2: All AA-axial.
d. F1: All Aa-axial; F2: AA-Axial, Aa-Axial, and aa-terminal.

Use a Punnett square to predict the offspring in a cross between a dwarf pea plant (hom*ozygous recessive) and a tall pea plant (heterozygous). What is the phenotypic ratio of the offspring?
a. 1 Tall : 1 dwarf
b. 1 tall : 2 dwarf
c. 3 tall : 1 dwarf
d. 1 dwarf : 4 tall

Can a human male be a carrier of red-green color blindness?
a. Yes, males can be the carriers of red-green color blindness, as color blindness is autosomal dominant.
b. No, males cannot be the carriers of red-green color blindness, as color blindness is X-linked.
c. No, males cannot be the carriers of red-green color blindness, as color blindness is Y-linked.
d. Yes, males can be the carriers of red-green color blindness, as color blindness is autosomal recessive.

Use the probability method to calculate the genotypes and genotypic proportions of a cross between AABBCc and Aabbcc parents.
a. Possible genotypes are AABBcc, AaBbCc, AaBbcc and the ratio 1 : 2 : 1 .
b. Possible genotypes are AABbcc, AaBbCc, AaBbcc and the ratio 1 : 3 : 1 .
c. Possible genotypes are AABbCc, AABbcc, AaBbCc, AaBbcc and the ratio 1 : 1 : 1 : 1 .
d. Possible genotypes are AABbcc, AaBbCC, AaBbcc and the ratio 1 : 1 : 1 .

How does the segregation of traits result in different combinations of gametes at the end of meiosis?
a. The chromosomes randomly align during metaphase I at the equator, and separation of hom*ologous chromosomes occurs during anaphase I. Similarly separation of sister chromatids occurs at anaphase II of meiosis II. At the end of meiosis II, four different gametic combinations are produced, each containing a haploid set of chromosomes.
b. The chromosomes randomly align during anaphase I at the equator. Separation of bivalent chromosomes occur during metaphase I of meiosis I. Similarly, separation of sister chromatids occurs at metaphase II of meiosis II. At the end of meiosis II, four different gametic combinations are produced, each containing a haploid set of chromosomes.
c. The chromosomes randomly align during prophase I at the equator, and separation of sister chromatids occurs during metaphase I of meiosis I. Similarly separation of bivalent chromosomes occur at metaphase II of meiosis II. At the end of meiosis II, four different gametic combinations are produced, each containing a diploid set of chromosomes.
d. The chromosomes randomly align during prophase I at the equator, and separation of bivalent chromosomes occur during anaphase I of meiosis I. Similarly, separation of hom*ologous chromosomes occurs at metaphase II of meiosis II. At the end of meiosis II, four different gametic combinations are produced, each containing a diploid set of chromosomes.

In Section 12.3, “Laws of Inheritance,” an example of epistasis was given for summer squash. Cross white WwYy heterozygotes to demonstrate the phenotypic ratio of 12 white : 3 yellow : 1 green that was given in the text.
a. 12 offspring are white, as the W gene is epistatic to the Y gene. Three offspring are yellow, because w is not epistatic. Green offspring is obtained when the recessive form of both genes (wwyy) are present.
b. 12 offspring are white as W gene is hypostatic to Y gene. Three offspring are yellow because Y is epistatic to w. Green offspring is obtained when the dominant form of both the genes (WWYY) is present.
c. 12 offspring are white as W gene is dominant. Three offspring are yellow because Y is dominant and w is recessive. Green offspring is obtained when the recessive form of both the genes (wwyy) is present, showing codominance.
d. 12 offspring are white as W is epistatic to Y gene. Three offspring are yellow because Y is hypostatic to w. Green offspring is obtained when the recessive form of both the genes (wwyy) are present, showing codominance.

The trait for widow’s peak can be considered a monoallelic dominant trait in humans. If a man with a widow’s peak and a woman with a straight hairline have a child together, what is the probability that the child will inherit the widow’s peak if you know that the father’s mother had a straight hairline?
a. 0.25
b. 0.5
c. 0.75
d. 1

Don’t like Brussels sprouts? Blame your genes. The chemical PTC (phenylthiocarbamide), which is nearly identical to a compound found in the cabbage family, tastes very bitter for some people. Others cannot detect a taste. The ability to taste PTC is incompletely dominant and is controlled by a gene on chromosome 7. A woman who finds Brussel sprouts mildly distasteful (in other words, who can taste PTC weakly) has a child with a man who hates Brussel sprouts (in other words, who can taste PTC strongly). What is the probability that their son likes Brussel sprouts (in other words, cannot taste PTC)?
a. 0
b. 0.25
c. 0.5
d. 1

Tay-Sachs disease is an autosomal recessive disorder that causes severe problems in neurons. Children who receive two copies of the gene rarely live beyond the age of five. There is no cure for the disease. During a genetic screening, a couple is told that both partners carry there cessive gene. What kind of issue must the couple confront?
a. scientific
b. financial
c. ethical
d. educational

A couple has three daughters. What is the probability that the next child they have will be a daughter?
a. 0%
b. 25%
c. 50%
d. 100%

What is the probability that a couple will have three daughters?
a. $\frac{1}{2}$
b. $\frac{1}{3}$
c. $\frac{1}{6}$
d. $\frac{1}{8}$

Petunias can be blue, red, or violet. When a blue flower is crossed with a red flower, all the resulting flowers are violet. When a violet flower is crossed with a red flower, about half of the flowers are violet and half are red. How do you characterize the color trait?
a. complete dominance
b. codominance
c. incomplete dominance
d. sex-linked

Petunias can be blue, red, or violet. When a blue flower is crossed with a red flower, all the resulting flowers are violet. Two violet petunias are crossed. Which is the most probable result of the cross?
a. 75% of the flowers are blue and 25% of the flowers are red.
b. 50% of the flowers are blue and 50% of the flowers are red.
c. 75% of the flowers are red and 25% are blue.
d. 25% of the flowers are blue, 50% of the flowers are violet, and 25% of the flowers are red.

Fruit flies (Drosophila melanogaster) with a wild-type phenotype have gray bodies and red eyes. Certain mutations can cause changes to these traits. Mutant flies may have a black body and/or cinnabar eyes. To study the genetics of these traits, a researcher crossed a truebreeding wild-typed male fly with a true-breeding female fly with a black body and cinnabar eyes. All of the F1 progeny displayed a wild type phenotype. Which of the following is correct about the traits observed?
a. Gray body and cinnabar eyes are dominant.
b. Eye color is sex-linked.
c. Body color is sex-linked.
d. Gray body and red eyes are dominant.

$\begin{array}{|c|c|}\hline \text { Body Color } & {\text { Eye Color }} & {\text { Number Predicted }} \\ \hline \text { Gray } & {\text { Red }} & {244} \\ \hline \text { Black } & {\text { Cinnabar }} & {244} \\ \hline \text { Gray } & {\text { Cinnabar }} & {244} \\ \hline \text { Black } & {\text { Red }} & {244} \\ \hline\end{array}$
Female flies from the F1 generation were crossed with true-breeding male flies with black bodies and cinnabar eyes. The table represents the predicted outcome and the data obtained from the cross. What was the assumption that lead to the predicted numbers?
a. The traits assort independently.
b. The traits are located on the X chromosome.
c. The traits are on the same chromosome.
d. The female flies were hom*ozygous for wild type alleles.

Cats can be black, yellow, or calico (black and yellow patches). Coat color is carried on the X chromosome. What type of inheritance is color coat in cats?
1. Complete dominance
2. Codominance
3. Incomplete dominance
4. Sex-linked
a. 2
b. 3
c. 2,4
d. 3,4

Cats can be black, yellow or calico (black and yellow patches). Coat color is carried on the X chromosome. A yellow cat is crossed with a black cat. Assume that the offspring are both male and female. What are the phenotypes of the offspring and in what proportions?
a. All the cats are yellow.
b. All the cats are black.
c. All the cats are calico.
d. There is not enough information to answer the question.

Two genes, A and B, are located adjacent to each other (linked) on the same chromosome. In the original cross (P0), one parent is hom*ozygous dominant for both traits (AB), whereas the other parent is recessive (ab).
A. Describe the distribution of genotypes and phenotypes in F1.
B. Describe the distribution of genotypes and phenotypes when F1 is crossed with the ab parent.
C. Describe the distribution of genotypes and phenotypes when F1 is crossed with the AB parent.
D. Explain the observed non-Mendelian results in terms of the violation of the laws governing Mendelian genetics.

Gregor Mendel’s 1865 paper described experiments on the inheritance of seven characteristics of Pisum sativum shown in the first column in the table below. Many years later, based on his reported outcomes and analysis of the inheritance of a single characteristic, Mendel developed the concepts of genes, their alleles, and dominance. These concepts are defined in the second column of the table using conventional symbols for the dominant allele for each characteristic. Even later, the location of each of these genes on one of the seven chromosomes in P. sativum were determined, as shown in the third column.
A. Before the acceptance of what Mendel called “factors” as the discrete units of inheritance, the accepted model was that the traits of progeny were “blended” traits of the parents. Evaluate the evidence provided by Mendel’s experiments in disproving the blending theory of inheritance.
B. Mendel published experimental data and analysis for two experiments involving the inheritance of more than a single characteristic. He examined two-character inheritance of seed shape and seed color. He also reported three-character inheritance of seed shape, seed color, and flower color. Evaluate the evidence provided by the multiple-character experiments. Identify which of the following laws of inheritance depend upon these multiplecharacter experiments for support:
a. During gamete formation, the alleles for each gene segregate from each other so that each gamete carries only one allele for each gene.
b. Genes for different traits can segregate independently during the formation of gametes.
c. Some alleles are dominant, whereas others are recessive. An organism with at least one dominant allele will display the effect of the dominant allele.
d. All three laws can be inferred from the singlecharacter experiments.
C. As shown in the table above, some chromosomes contain the gene for more than one of the seven characteristics Mendel studied, for example, seed color and flowers. The table below shows, with filled cells above the dashed diagonal line, the combinations of characteristics for which Mendel reported results. In the cells below the dotted diagonal line, identify with an X each cell where deviations from the law or laws identified in part B might be expected.
D. Explain the reasons for the expected deviations for those combinations of characteristics identified in part C.
E. In one of the experiments reported by Mendel, deviations from the law identified in part B might be expected. Explain how the outcomes of this experiment were consistent with Mendel’s laws.

A dihybrid cross involves two traits. A cross of parental types AaBb and AaBb can be represented with a Punnett square: This representation clearly organizes all of the possible genotypes and reveals the 9:3:3:1 distribution of phenotypes and a 4×4 grid of 16 cells. Expressed as a fraction of the 16 possible genotypes of the offspring, the phenotypic ratio describes the probability of each phenotype among the offspring: 3 (AA, Aa, aA) × 3 (BB, bB, Bb)/16 = 9/16; 3 (AA, Aa, aA) × 1 (bb) /16 = 3/16; 1 (aa) × 3 (BB, bB, Bb) = 3/16; and 1 (aa) × 1 (bb) = 1/16.
A. Using the probability method, calculate the likelihood of these phenotypes from each dihybrid cross:
recessive in the gene with alleles A and a from the cross AaBb × aabb
dominant in both genes from the cross AaBb × aabb
recessive in both genes from the cross AaBb × aabb
recessive in either gene from the cross AaBb × aabb
A Punnett square representation of a trihybrid cross, such as the self-cross of AaBbCc, is more cumbersome because there are eight columns and rows (2×2×2 ways to choose parental genotypes) and 64 cells. A less tedious representation is to calculate the number of each type of genotype in the offspring directly by counting the unique permutations of the letters representing the alleles. For example, the probability of the cross AaBbCc × AaBbCc is 3 (AA, Aa, aA) × 3 (BB, Bb, bB) × 3 (CC, Cc, cC)/64 = 27/64.
B. Using the probability method, calculate the likelihood of these phenotypes from each trihybrid cross:
recessive in all traits from the cross AaBbCc × aabbcc
recessive in the gene with alleles C and c and dominant in the other two traits from the cross AaBbCc × AaBbCc
dominant in the gene with alleles A and a and recessive in the other two traits from the cross AaBbcc × AaBbCc
C. The probability method is an easy way to calculate the likelihood of each particular phenotype, but it doesn’t simultaneously display the probability of all possible phenotypes. The forked line representation described in the text allows the entire phenotypic distribution to be displayed. Using the forked line method, calculate the probabilities in a cross between AABBCc and Aabbcc parents:
all traits are recessive: aabbcc
traits are dominant at each loci, A?B?C?
traits are dominant at two genes and recessive at the third
traits are dominant at one gene and recessive at the other two

Construct a representation showing the connection between the process of meiosis and the transmission of six possible phenotypes from parents to F2 offspring. The phenotypes are labeled A, a, B, b and C, c. Expression of each phenotype is controlled by a separate Mendelian gene. Your representation should show the proportion of every possible combination of phenotypes (e.g., ABC, AbC, etc.) that will be present in the F2 offspring.