Exam 4

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Question #1

An albino corn snake is crossed with a normal colored corn snake. The offspring are all normal colored. When these first generation snakes are crossed among each other, they produce 34 normal snakes and 11 albino snakes: How many were heterozygous?

A 18

B 11

C 23

Question #2

A 18

B 11

C 23

Question #3

A aa and AA

B aa and Aa

Question #4

A Normal

B Albino

Question #5

Galactosemia is a recessive human disease that is treatable by restricting lactose and glucose in the diet. Susan and her partner Rick are heterozygous for this disease. Susan is pregnant with twins. If she has identical twins – – what is the probability that both of the twins will be girls who have galactosemia?

A 1/8

B 1/32

C 1/64

Question #6

Galactosemia is a recessive human disease that is treatable by restricting lactose and glucose in the diet. Susan and her partner Rick are heterozygous for this disease. Susan is pregnant with twins. If she has fraternal twins (nonidentical) – what is the probability that both of the twins will be girls who have galactosemia?

A 1/32

B 1/64

C 1/8

Question #7

In the following cases of disputed paternity, determine the probable father : Mother is type O and bears non-identical twins, one type A and one type B. Father #1 is type A; father #2 is type B.

A Father 1

B Both

Question #8

In the following cases of disputed paternity, determine the probable father: Mother is type B, child is type AB. Father #1 is A; father #2 is B.

A Father 1

B Father 2

Question #9

In the following cases of disputed paternity, determine the probable father: Mother is type B, child is type O. Father #1 is A; father #2 is AB

A Father 2

B Father 1

Question #10

Assume that D, E, F, G, H, and I are autosomal genes on different chromosomes. From the mating DdeeFfGGHhIi x DdEEFFGgHhii – calculate the probabilities of the following: 3. What is the probability that one of the offspring will have the genotype DDEEFfGGhhii?

A 0

B 1/64

C 1/32

Question #11

Assume that D, E, F, G, H, and I are autosomal genes on different chromosomes. From the mating DdeeFfGGHhIi x DdEEFFGgHhii – calculate the probabilities of the following: 2. What is the probability that one of the offspring will be heterozygous for each allele?

A 1/32

B 0

C 1/64

Question #12

Assume that D, E, F, G, H, and I are autosomal genes on different chromosomes. From the mating DdeeFfGGHhIi x DdEEFFGgHhii – calculate the probabilities of the following: 1. What is the probability that one of the offspring will have the genotype DdEeFFGghhIi?

A 1/64

B 1/32

C 0

Question #13

From the cross TTYyRr x TtYyrr, calculate the proportion of offspring that would match the following phenotypes. 5. tall plants with green, wrinkled seeds

A 1/8

B 0

C 3/8

Question #14

From the cross TTYyRr x TtYyrr, calculate the proportion of offspring that would match the following phenotypes. 4. tall plants with yellow, wrinkled seeds

A 3/8

B 1/8

C 0

Question #15

From the cross TTYyRr x TtYyrr, calculate the proportion of offspring that would match the following phenotypes. 3. dwarf plants with round, green seeds

A 0

B 1/8

C 3/8

Question #16

From the cross TTYyRr x TtYyrr, calculate the proportion of offspring that would match the following phenotypes. 2. tall plants with round, green seeds

A 1/8

B 0

C 3/8

Question #17

From the cross TTYyRr x TtYyrr, calculate the proportion of offspring that would match the following phenotypes. 1. tall plants with round, yellow seeds

A 3/8

B 0

C 1/8

Question #18

A new phenotype of mouse has been discovered – short-tails. You have been given a short-tailed mouse and do not know anything about its genotype. You decide to cross this short-tailed mouse with a long-tailed mouse. Multiple crosses of this short-tailed mouse to long-tailed mice produces 27 long-tailed mice and 25 short-tailed mice. When you cross these short-tailed progeny with each other, you get 31 short-tailed mice and 11 long-tailed mice. What is the genotype of the long-tailed mice used in this breeding experiment?

A AA

B aa

Question #19

A new phenotype of mouse has been discovered – short-tails. You have been given a short-tailed mouse and do not know anything about its genotype. You decide to cross this short-tailed mouse with a long-tailed mouse. Multiple crosses of this short-tailed mouse to long-tailed mice produces 27 long-tailed mice and 25 short-tailed mice. When you cross these short-tailed progeny with each other, you get 31 short-tailed mice and 11 long-tailed mice.
What was the genotype of the original short-tailed mouse?

A aA

B Aa

Question #20

The definition of a multifactorial trait is one in which more than one set of alleles is involved, in which the environment plays no role.

A TRUE

B FALSE

Question #21

You breed a black, long-haired rabbit with a white, short-haired rabbit. All of the offspring have long, black hair. If the genes for hair color and length are closely linked on the same chromosome, what would you expect to see in the F2 generation?

A All long-haired black

B 3 black, long-haired: 9 black, short-haired: 9 white, long-haired: 1 white, short-haired

C 15 black, long-haired: 1 black, short-haired: 1 white, long-haired: 13 white, short-haired

D 3 black, long-haired: 1 white, short-haired

E 9 black, long-haired: 3 black, short-haired: 3 white, long-haired: 1 white, short-hairedite, short-haired

Question #22

You are breeding different varieties of roses in your garden. When you cross a true-breeding yellow “Texas Beauty” rose with a true-breeding “Ruby” red rose, you get all red roses. But when you cross a “Texas Beauty” yellow with the yellow variety called “Jealousy,” you get a 9:3:4 ratio of phenotypes. What can you conclude from these results?

A There is a mutation in the gene for yellow pigment in the “Jealousy” variety.

B There are epistatic interactions between at least two genes for rose flower pigmentation.

C The environment, and not genes, influences pigment in these rose varieties.

D The red allele is dominant to the yellow allele.

E There is no recessive allele for the red pigment gene.

F There is a mutation in the gene for yellow pigment in the “Jealousy” variety.

Question #23

Two babies are mixed up in the hospital nursery. The blood types of Couple 1 are A and O and the blood types of Couple 2 are AB and B. Baby Joe has blood type O and Baby Jane has blood type B. Who are the parents of Baby Joe and Baby Jane?

B Couple 1, Baby Joe; Couple 2, Baby Jane

C Couple 2, Baby Joe; Couple 1 could not be the parents of either baby

D Couple 1, Baby Joe; Couple 2 could not be the parents of either baby

E Couple 1, Baby Jane; Couple 2, Baby Joe

F From these combinations of blood types, it is impossible to determine

Question #24

Which of the following organelle(s) has/have a genome separate from the genome in the cell nucleus?

A chloroplasts

B mitochondria

C both the mitochondria and chloroplasts

D nucleolus

E both the mitochondria and the nucleolus

Question #25

The ABO system of blood types is an example of:

A epistasis

B incomplete dominane and co-dominance

C multiple-alleles and co-dominance

D multiple-alleles

E co-dominance

Question #26

A man who carries an X-linked allele will pass it on to

A half of his sons.

B all of his children.

C all of his daughters.

D all of his sons.

E half of his daughters.

Question #27

Why must the life cycle of sexually reproducing species alternate between haploid and diploid stages?

A Meiosis must occur at some point in the life cycle to prevent a doubling of chromosomes in each generation.

B Meiosis must occur at some point in the life cycle to prevent a reduction of chromosomes in each generation.

C The diploid stage allows for genetic recombination.

D Mitosis must occur at some point in the life cycle to prevent a reduction of chromosomes in each generation.

E Haploid cells cannot grow independently of diploid cells.

Question #28

How many unique gametes could be produced through independent assortment by an individual with the genotype AaBbCCDdEEFf?

A 16

B 64

C 4

D 8

Question #29

The inheritance pattern where two or more genes do not assort independently because they are close together on the same chromosome is called

A genomic imprinting.

B dominance.

C linkage.

D epistasis.

E Mendelian inheritance.

Question #30

Polyploidy in plants

A is very common.

B allows for adaptation to diverse environments.

C All of these statements are true regarding polyploidy in plants.

D allows humans to create plants with unusual flowers.

E may alter the levels of gene expression.

Question #31

If you want to know the probability that tossing two coins will result in a particular outcome (two heads, for example), you would use the

A punnett rule

B sum rule.

C product rule.

D Bergman’s rule.

E genetics rule.

Question #32

A woman has six sons. The chance that her next child will be a daughter is

A 1

B 1/6

C 1/2

D 0

Question #33

What is a dominant allele?

A An allele that will only have an affect on phenotype in a haploid organism; otherwise its presence will be masked

B An allele that is very prevalent in a population

C An allele that totally beats up on a recessive allele

D In a heterozygous individual, the allele that determines the phenotype

E An allele that has no noticeable affect on an organism’s phenotype

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