Question 1

In the genetic cross AaBbCcDdEE *AaBBCcDdEe, what fraction of the offspring will be heterozygous for all of these genes (AaBbCcDdEe)? Assume all genes are unlinked.
A) 1/32
B) 1/16
C) 3/32
D) 1/4
E) 5/32

Accepted Answer

Each gene pair has a 1/2 chance of being heterozygous in the offspring. Since there are five gene pairs, the probability is (1/2)^5 = 1/32.

Question 2

What is another word for "family tree"?
A) Pleiotropy
B) Punnett square
C) Pedigree
D) Heritable
E) Heterosis

Accepted Answer

A pedigree is a diagram that shows the genetic relationships among family members and is commonly referred to as a family tree. A, B, D, and E are all terms related to genetics but are not synonymous with family tree.

Question 3

In garden peas, the allele for tall plants is dominant over the allele for short plants.A true-breeding tall plant is crossed with a short plant, and one of their offspring is test crossed.Out of 60 offspring resulting from the test cross, _______ should be short.
A) 0
B) about 15
C) about 30
D) about 45
E) about 60

Accepted Answer

Since the allele for tall plants is dominant over the allele for short plants, the true-breeding tall plant must be homozygous dominant (TT) while the short plant must be homozygous recessive (tt). The offspring resulting from their cross would be heterozygous (Tt) and all tall. However, when one of these tall offspring is test crossed (crossed with a homozygous recessive individual), the possible genotypes of its gametes are T or t. This means that 50% of the offspring resulting from the test cross should inherit the recessive allele and be short (tt genotype). Since there were 60 offspring in the test cross, about half of them (30) should be short. Therefore, the answer is C) about 30.

Question 4

A true-breeding strain of tall soybean plants is crossed with a true-breeding strain of short soybean plants.All the offspring are short.These offspring are then crossed with each other to create an F₂ generation.If the height of soybean plants is due to a single gene, then of 200 F₂plants, _______ should be tall. A) 25 B) 50 C) 75 D) 100 E) 150

Accepted Answer

If the height of soybean plants is due to a single gene, it means that tallness is dominant (represented by uppercase "B" allele) over shortness (represented by lowercase "b" allele). From the initial cross, we can infer that the tall parent was homozygous dominant (BB) and the short parent was homozygous recessive (bb). When they were crossed, all the offspring were heterozygous (Bb) for the height gene. The Bb offspring were then crossed with each other, resulting in a Punnett square with the following genotypes: BB, Bb, Bb, and bb in a 1:2:1 ratio. Only the BB offspring will be tall, so out of 200 plants, 50 (25%) of them should be tall.

Question 5

Separation of the alleles of a single gene into different gametes is called
A) segregation.
B) synapsis.
C) independent assortment.
D) heterozygous separation.
E) recombination.

Accepted Answer

Segregation is the process by which the two different alleles of a gene separate and end up in different gametes during meiosis. This leads to the production of haploid gametes with only one allele of each gene.

Question 6

Classical albinism results from a recessive allele.What is the expected ratio for the progeny when a fully pigmented male with an albino father has children with a fully pigmented female with no history of albinism in her family?
A) ¾ pigmented; ¼ albino
B) ¾ albino; ¼ pigmented
C) ½ pigmented; ½ albino
D) All pigmented
E) All albino

Accepted Answer

The fully pigmented male with an albino father is a carrier (heterozygous) for the albinism allele. The fully pigmented female with no history of albinism is homozygous dominant. Therefore, all offspring will be pigmented, as they will inherit at least one dominant allele from the mother.

Question 7

When we multiply the probabilities of two genetic events to obtain the probability that both events will occur together, we
A) assume that the events involve particulate inheritance.
B) assume that the events are independent.
C) assume that we have performed a dihybrid cross.
D) assume that the second event depends on the result of the first event.
E) make no assumptions about the events.

Accepted Answer

Multiplying probabilities to obtain the probability of two events occurring together assumes that the events are independent. This means that the outcome of one event does not influence the outcome of the other event. It is a fundamental assumption of Mendelian genetics and the principle of segregation and independent assortment.

Question 8

You are working with an organism of the genotype AAbbCcDDeeFFGgHHiiJJKKLl.How many different gametes with respect to these genes could this individual produce, assuming no linkage among the different genes?
A) 4
B) 6
C) 8
D) 10
E) 12

Accepted Answer

The individual can produce 8 different gametes. This is calculated by identifying the heterozygous gene pairs (Cc, Gg, Ll) which are 3 in total. Since each heterozygous pair can produce two types of gametes (one for each allele), the total number of different gametes is 2^3 = 8, as each heterozygous pair contributes to the diversity independently.

Question 9

In Drosophila, the genes for body coloration and eye size are on different chromosomes.Normal-colored bodies are dominant to ebony-colored (very dark) bodies, and normal-sized eyes are dominant to being eyeless.Line A is true breeding for normal body and normal eye, whereas line B is true breeding for ebony bodies and being eyeless.From a dihybrid cross between lines A and B, 400 flies are scored.How many F₂ flies are expected to have ebony-colored bodies and normal eyes? A) 9 B) 25 C) 75 D) 225 E) 300

Accepted Answer

Given that the traits are on different chromosomes and follow Mendelian inheritance, a dihybrid cross between homozygous parents (one with dominant traits and the other with recessive traits) will result in a 9:3:3:1 phenotypic ratio in the F2 generation. The question asks for the number of offspring with ebony-colored bodies and normal eyes, which corresponds to the 3/16 portion of the 9:3:3:1 ratio. Therefore, out of 400 flies, 3/16 * 400 = 75 flies are expected to have ebony-colored bodies and normal eyes.

Question 10

Suppose that in a hypothetical population of lizards, the probability of obtaining a male is 0.6.Assume that the sexes of each individual are independent.What is the probability of obtaining exactly two males in a clutch of three? (Note that there are three ways this can occur, depending on the order in which the probabilities are calculated.)
A) 0.144
B) 0.216
C) 0.376
D) 0.432
E) 0.648

Accepted Answer

The answer of Suppose that in a hypothetical population of...

Question 11

Suppose that tomatoes with the RR genotype are red and those with rr are green.Assuming that this reflects conventional genetics shorthand of upper- and lowercase letters, which situation is true?
A) R is dominant to r, and Rr individuals are green.
B) R is dominant to r, and Rr individuals are red.
C) r is dominant to R, and Rr individuals are green.
D) r is dominant to R, and Rr individuals are red.
E) r is dominant to R, and Rr individuals are yellow.

Accepted Answer

In conventional genetics shorthand, uppercase letters represent dominant alleles and lowercase letters represent recessive alleles. Therefore, R must be dominant to r. The homozygous dominant genotype would be RR and the homozygous recessive genotype would be rr. The heterozygous genotype would be Rr. Because R is dominant, individuals with at least one R allele (i.e. RR or Rr) will have the red phenotype. Thus, option B is correct as it explains that R is dominant to r and individuals with the heterozygous Rr genotype will be red.

Question 12

The most likely explanation for the observation that round is dominant to wrinkled in Mendel's peas is
A) the round allele encodes a functional enzyme, the wrinkled allele encodes a defective enzyme, and the round trait could arise with only one functional gene for the enzyme.
B) the wrinkled allele encodes a functional enzyme, the round allele encodes a defective enzyme, and the wrinkled trait could arise with only one functional gene for the enzyme.
C) the wrinkled allele encodes a functional enzyme, the round allele encodes a defective enzyme, and the round trait could arise with only one functional gene for the enzyme.
D) the round allele poisons the wrinkled allele.
E) the round allele has an uppercase letter.

Accepted Answer

The expression of the round phenotype is dominant to the wrinkled phenotype, suggesting that only one functional copy of the gene is necessary to produce the round trait. This is most likely due to the fact that the enzyme encoded by the round allele is functional, while the enzyme encoded by the wrinkled allele is defective.

Question 13

A test cross is used to determine whether an individual with
A) a dominant phenotype is heterozygous or homozygous.
B) a dominant genotype is heterozygous or homozygous.
C) a recessive phenotype is heterozygous or homozygous.
D) a recessive genotype is heterozygous or homozygous.
E) None of the above

Accepted Answer

A test cross involves mating the individual with the dominant phenotype (but unknown genotype) with a homozygous recessive individual. The phenotype of the offspring can reveal whether the dominant individual is heterozygous or homozygous for the dominant allele. If all the offspring show the dominant phenotype, then the dominant individual is likely homozygous. If some offspring show the recessive phenotype, then the dominant individual is likely heterozygous.

Question 14

Which value cannot be a probability?
A) -0.14
B) 0.11
C) 0.7
D) 0.99
E) 1.0

Accepted Answer

A probability cannot be negative, so -0.14 cannot be a probability.

Question 15

Consider the F₂ progeny of a dihybrid cross.For every double recessive homozygote observed, you should expect _______ double dominant homozygote(s). A) 1/9 B) 1/3 C) 1 D) 3 E) 9

Accepted Answer

In a dihybrid cross, the ratio of double dominant homozygotes to double recessive homozygotes is 1:1, assuming simple Mendelian inheritance with independent assortment.

Question 16

In Drosophila, the genes for body coloration and eye size are on different chromosomes.Normal-colored bodies are dominant to ebony-colored (very dark) bodies, and normal-sized eyes are dominant to being eyeless.Line A is true breeding for normal body and normal eye, whereas line B is true breeding for ebony bodies and being eyeless.From a dihybrid cross between lines A and B, 480 flies are scored.How many F₂ flies are expected to have both ebony body color and to be eyeless? A) 1 B) 30 C) 90 D) 120 E) 270

Accepted Answer

The answer of In Drosophila, the genes for body coloration...

Question 17

Suppose you are a genetics counselor who is working with a 21-year-old pregnant woman who has just discovered that her father has Huntington's disease, a rare dominant autosomal trait.This disease usually develops in middle age, so people carrying this trait do not find out they have this genetic disorder until midlife.What are the chances that the child she is carrying will develop the disease? (Assume that her husband's family has no history of the disease.)
A) 10%
B) 25%
C) 50%
D) 75%
E) 100%

Accepted Answer

The answer of Suppose you are a genetics counselor who...

Question 18

Suppose you are a genetics counselor who is working with a 21-year-old pregnant woman who has just discovered that her father has Huntington's chorea, a rare dominant autosomal trait.This disease usually develops in middle age, so people carrying this trait do not find out they have this genetic disorder until midlife.What is the chance that she has Huntington's chorea?
A) 10%
B) 25%
C) 50%
D) 75%
E) 100%

Accepted Answer

Huntington's chorea is a dominant autosomal trait, which means that an individual who inherits the gene from one parent will have the disease. Since the woman's father has Huntington's chorea, there is a 50% chance that he passed the gene to his daughter. However, genetic counseling is recommended for further testing and information.

Question 19

You are a lab assistant working with Drosophila.You suspect that vestigial-winged flies may have contaminated a vial of wild-type flies.You know that vestigial wings are recessive to the wild type.The best way to tell whether a male fly with normal wings taken from the questionable vial is homozygous wild type is to mate it to females that you know are _______.The male is homozygous wild type if _______.
A) homozygous for vestigial wings; all the offspring have wild-type wings
B) homozygous for vestigial wings; all the offspring have vestigial wings
C) heterozygous; all the offspring are wild type
D) homozygous for wild-type wings; all the offspring have wild-type wings
E) homozygous for vestigial wings; half the offspring have wild-type wings and half have vestigial wings

Accepted Answer

If the male fly is homozygous wild type and mated with females homozygous for vestigial wings, all offspring will have wild-type wings because the wild-type allele is dominant.

Question 20

Jane is a better tennis player than Jack.If the probability that Jane will beat Jack in any given game is 0.7, what is the probability that they will each win one game if they play two games? (Note that there are two ways they can arrive at this outcome: Jane wins the first game or Jack wins the first game.)
A) 0.21
B) 0.3
C) 0.42
D) 0.6
E) 1

Accepted Answer

There are four possible outcomes for the two games: JJ, JJ, JJ, and JJ.

The probability that they will each win one game if they play two games is the sum of the probabilities of the first and second outcomes:

P(JW, JW) + P(JL, LJ)

= P(JW)P(JL) + P(JL)P(JW)

= 0.7(1-0.7) + (1-0.7)0.7

= 0.42

Therefore, the answer is Choice C.

Quiz 12: Inheritance, Genes, and Chromosomes

In the genetic cross AaBbCcDdEE *AaBBCcDdEe, what fraction of the offspring will be heterozygous for all of these genes (AaBbCcDdEe) ? Assume all genes are unlinked.

What is another word for "family tree"?

In garden peas, the allele for tall plants is dominant over the allele for short plants.A true-breeding tall plant is crossed with a short plant, and one of their offspring is test crossed.Out of 60 offspring resulting from the test cross, _______ should be short.

Separation of the alleles of a single gene into different gametes is called

Classical albinism results from a recessive allele.What is the expected ratio for the progeny when a fully pigmented male with an albino father has children with a fully pigmented female with no history of albinism in her family?

When we multiply the probabilities of two genetic events to obtain the probability that both events will occur together, we

You are working with an organism of the genotype AAbbCcDDeeFFGgHHiiJJKKLl.How many different gametes with respect to these genes could this individual produce, assuming no linkage among the different genes?

Suppose that tomatoes with the RR genotype are red and those with rr are green.Assuming that this reflects conventional genetics shorthand of upper- and lowercase letters, which situation is true?

The most likely explanation for the observation that round is dominant to wrinkled in Mendel's peas is

A test cross is used to determine whether an individual with

Which value cannot be a probability?

Consider the F₂ progeny of a dihybrid cross.For every double recessive homozygote observed, you should expect _______ double dominant homozygote(s) .