A P biology Chapter 14 Review

Multiple Choice
Identify the choice that best completes the statement or answers the question.

____	1.	Pea plants were particularly well suited for use in Mendel's breeding experiments for all of the following reasons except that
a.	peas show easily observed variations in a number of characters, such as pea shape and flower color.	
b.	it is possible to control matings between different pea plants.	
c.	it is possible to obtain large numbers of progeny from any given cross.	
d.	peas have an unusually long generation time.	
e.	many of the observable characters that vary in pea plants are controlled by single genes.	

____	2.	What is the difference between a monohybrid cross and a dihybrid cross?
a.	A monohybrid cross involves a single parent, whereas a dihybrid cross involves two parents.	
b.	A monohybrid cross produces a single progeny, whereas a dihybrid cross produces two progeny.	
c.	A dihybrid cross involves organisms that are heterozygous for two characters and a monohybrid only one.	
d.	A monohybrid cross is performed for one generation, whereas a dihybrid cross is performed for two generations.	
e.	A monohybrid cross results in a 9:3:3:1 ratio whereas a dihybrid cross gives a 3:1 ratio.	

____	3.	A cross between homozygous purple-flowered and homozygous white-flowered pea plants results in offspring with purple flowers. This demonstrates
a.	the blending model of genetics.	
b.	true-breeding.	
c.	dominance.	
d.	a dihybrid cross.	
e.	the mistakes made by Mendel.	

____	4.	The  offspring of Mendel's classic pea cross always looked like one of the two parental varieties because
a.	one phenotype was completely dominant over another.	
b.	each allele affected phenotypic expression.	
c.	the traits blended together during fertilization.	
d.	no genes interacted to produce the parental phenotype.	
e.	different genes interacted to produce the parental phenotype.	

____	5.	What was the most significant conclusion that Gregor Mendel drew from his experiments with pea plants?
a.	There is considerable genetic variation in garden peas.	
b.	Traits are inherited in discrete units, and are not the results of "blending."	
c.	Recessive genes occur more frequently in the  than do dominant ones.	
d.	Genes are composed of DNA.	
e.	An organism that is homozygous for many recessive traits is at a disadvantage.	

____	6.	How many unique gametes could be produced through independent assortment by an individual with the genotype AaBbCCDdEE?
a.	4	
b.	8	
c.	16	
d.	32	
e.	64	

____	7.	Two plants are crossed, resulting in offspring with a 3:1 ratio for a particular trait. This suggests
a.	that the parents were true-breeding for contrasting traits.	
b.	incomplete dominance.	
c.	that a blending of traits has occurred.	
d.	that the parents were both heterozygous.	
e.	that each offspring has the same alleles.	

____	8.	Two characters that appear in a 9:3:3:1 ratio in the  generation should have which of the following properties?
a.	Each of the traits is controlled by single genes.	
b.	The genes controlling the characters obey the law of independent assortment.	
c.	Each of the genes controlling the characters has two alleles.	
d.	Four genes are involved.	
e.	Sixteen different phenotypes are possible.	

____	9.	A sexually reproducing animal has two unlinked genes, one for head shape (H) and one for tail length (T). Its genotype is HhTt. Which of the following genotypes is possible in a gamete from this organism?
a.	HT	
b.	Hh	
c.	HhTt	
d.	T	
e.	tt	

____	10.	It was important that Mendel examined not just the  generation in his breeding experiments, but the  generation as well, because
a.	he obtained very few  progeny, making statistical analysis difficult.	
b.	parental traits that were not observed in the  reappeared in the .	
c.	analysis of the  progeny would have allowed him to discover the law of segregation, but not the law of independent assortment.	
d.	the dominant phenotypes were visible in the  generation, but not in the .	
e.	many of the  progeny died.	

____	11.	When crossing an organism that is homozygous recessive for a single trait with a heterozygote, what is the chance of producing an offspring with the homozygous recessive phenotype?
a.	0%	
b.	25%	
c.	50%	
d.	75%	
e.	100%	

Use Figure 14.1 and the following description to answer the questions below.

In a particular plant, leaf color is controlled by gene locus D. Plants with at least one allele D have dark green leaves, and plants with the homozygous recessive dd genotype have light green leaves. A true-breeding dark-leaved plant is crossed with a light-leaved one, and the  offspring is allowed to self-pollinate. The predicted outcome of the  is diagrammed in the Punnett square shown in Figure 14.1, where 1, 2, 3, and 4 represent the genotypes corresponding to each box within the square.

Figure 14.1

____	12.	Which of the boxes marked 1-4 correspond to plants with dark leaves?
a.	1 only	
b.	1 and 2	
c.	2 and 3	
d.	4 only	
e.	1, 2, and 3	

____	13.	Which of the boxes correspond to plants with a heterozygous genotype?
a.	1	
b.	1 and 2	
c.	1, 2, and 3	
d.	2 and 3	
e.	2, 3, and 4	

____	14.	Which of the plants will be true-breeding?
a.	1 and 4	
b.	2 and 3	
c.	1—4	
d.	1 only	
e.	None	

____	15.	Mendel accounted for the observation that traits which had disappeared in the  generation reappeared in the  generation by proposing that
a.	new mutations were frequently generated in the  progeny, "reinventing" traits that had been lost in the .	
b.	the mechanism controlling the appearance of traits was different between the  and the  plants.	
c.	traits can be dominant or recessive, and the recessive traits were obscured by the dominant ones in the .	
d.	the traits were lost in the  due to blending of the parental traits.	
e.	members of the  generation had only one allele for each character, but members of the  had two alleles for each character.	

____	16.	Which of the following about the law of segregation is false?
a.	It states that each of two alleles for a given trait segregate into different gametes.	
b.	It can be explained by the segregation of homologous chromosomes during meiosis.	
c.	It can account for the 3:1 ratio seen in the  generation of Mendel's crosses.	
d.	It can be used to predict the likelihood of transmission of certain genetic diseases within families.	
e.	It is a method that can be used to determine the number of chromosomes in a plant.	

____	17.	The fact that all seven of the pea plant traits studied by Mendel obeyed the principle of independent assortment most probably indicates which of the following?
a.	None of the traits obeyed the law of segregation.	
b.	The diploid number of chromosomes in the pea plants was 7.	
c.	All of the genes controlling the traits were located on the same chromosome.	
d.	All of the genes controlling the traits behaved as if they were on different chromosomes.	
e.	The formation of gametes in plants occurs by mitosis only.	

____	18.	Mendel was able to draw his ideas of segregation and independent assortment because of the influence of which of the following?
a.	His reading and discussion of Darwin's Origin of Species	
b.	The understanding of particulate inheritance he learned from renowned scientists of his time	
c.	His discussions of heredity with his colleagues at major universities	
d.	His reading of the scientific literature current in the field	
e.	His experiments with the breeding of plants such as peas	

____	19.	Mendel's observation of the segregation of alleles in gamete formation has its basis in which of the following phases of cell division?
a.	Prophase I of meiosis	
b.	Prophase II of meiosis	
c.	Metaphase I of meiosis	
d.	Anaphase I of meiosis	
e.	Anaphase of mitosis	

____	20.	Mendel's second law of independent assortment has its basis in which of the following events of meiosis I?
a.	Synapsis of homologous chromosomes	
b.	Crossing over	
c.	Alignment of tetrads at the equator	
d.	Separation of homologs at anaphase	
e.	Separation of cells at telophase	

____	21.	Black fur in mice (B) is dominant to brown fur (b). Short tails (T) are dominant to long tails (t). What fraction of the progeny of the cross BbTt ´ BBtt will have black fur and long tails?
a.	1/16	
b.	3/16	
c.	3/8	
d.	1/2	
e.	9/16	

____	22.	In certain plants, tall is dominant to short. If a heterozygous plant is crossed with a homozygous tall plant, what is the probability that the offspring will be short?
a.	1	
b.	1/2	
c.	1/4	
d.	1/6	
e.	0	

____	23.	Two true-breeding stocks of pea plants are crossed. One parent has red, axial flowers and the other has white, terminal flowers; all  individuals have red, axial flowers. The genes for flower color and location assort independently. If 1,000  offspring resulted from the cross, approximately how many of them would you expect to have red, terminal flowers?
a.	65	
b.	190	
c.	250	
d.	565	
e.	750	

____	24.	In a cross AaBbCc ´ AaBbCc, what is the probability of producing the genotype AABBCC?
a.	1/4	
b.	1/8	
c.	1/16	
d.	1/32	
e.	1/64	

____	25.	Given the parents AABBCc ´ AabbCc, assume simple dominance and independent assortment. What proportion of the progeny will be expected to phenotypically resemble the first parent?
a.	1/4	
b.	1/8	
c.	3/4	
d.	3/8	
e.	1	

Use the following information to answer the questions below.

Labrador retrievers are black, brown, or yellow. In a cross of a black female with a brown male, results can be either all black puppies, 1/2 black to 1/2 brown puppies, or 3/4 black to 1/4 yellow puppies.

____	26.	These results indicate which of the following?
a.	Brown is dominant to black.	
b.	Black is dominant to brown and to yellow.	
c.	Yellow is dominant to black.	
d.	There is incomplete dominance.	
e.	Epistasis is involved.	

____	27.	How many genes must be responsible for these coat colors in Labrador retrievers?
a.	1	
b.	2	
c.	3	
d.	4	

____	28.	In one type cross of black ´ black, the results were as follows:
9/16 black
4/16 yellow
3/16 brown
The genotype aabb must result in which of the following?
a.	Black	
b.	Brown	
c.	Yellow	
d.	A lethal result	

Use the following information to answer the questions below.

Radish flowers may be red, purple, or white. A cross between a red-flowered plant and a white-flowered plant yields all-purple offspring. The part of the radish we eat may be oval or long, with long being the dominant characteristic.

____	29.	If true-breeding red long radishes are crossed with true breeding white oval radishes, the  will be expected to be which of the following?
a.	Red and long	
b.	Red and oval	
c.	White and long	
d.	Purple and long	
e.	Purple and oval	

____	30.	In the  generation of the above cross, which of the following phenotypic ratios would be expected?
a.	9:3:3:1	
b.	9:4:3	
c.	1:1:1:1	
d.	1:1:1:1:1:1	
e.	6:3:3:2:1:1	

____	31.	Drosophila (fruit flies) usually have long wings (+) but mutations in two different genes can result in bent wings (bt) or vestigial wings (vg). If a homozygous bent wing fly is mated with a homozygous vestigial wing fly, which of the following offspring would you expect?
a.	All +bt +vg heterozygotes	
b.	1/2 bent and 1/2 vestigial flies	
c.	All homozygous + flies	
d.	3/4 bent to 1/4 vestigial ratio	
e.	1/2 bent and vestigial to 1/2 normal	

____	32.	The flower color trait in radishes is an example of which of the following?
a.	A multiple allelic system	
b.	Sex linkage	
c.	Codominance	
d.	Incomplete dominance	
e.	Epistasis	

____	33.	A 1:2:1 phenotypic ratio in the  generation of a monohybrid cross is a sign of
a.	complete dominance.	
b.	multiple alleles.	
c.	incomplete dominance.	
d.	polygenic inheritance.	
e.	pleiotropy.	

____	34.	In snapdragons, heterozygotes for one of the genes have pink flowers, whereas homozygotes have red or white flowers. When plants with red flowers are crossed with plants with white flowers, what proportion of the offspring will have pink flowers?
a.	0%	
b.	25%	
c.	50%	
d.	75%	
e.	100%	

____	35.	Tallness (T) in snapdragons is dominant to dwarfness (t), while red (R) flower color is dominant to white (r). The heterozygous condition results in pink (Rr) flower color. A dwarf, red snapdragon is crossed with a plant homozygous for tallness and white flowers. What are the genotype and phenotype of the  individuals?
a.	ttRr–dwarf and pink	
b.	ttrr–dwarf and white	
c.	TtRr–tall and red	
d.	TtRr–tall and pink	
e.	TTRR–tall and red	

____	36.	Skin color in a certain species of fish is inherited via a single gene with four different alleles. How many different types of gametes would be possible in this system?
a.	1	
b.	2	
c.	4	
d.	8	
e.	16	

____	37.	In cattle, roan coat color (mixed red and white hairs) occurs in the heterozygous (Rr) offspring of red (RR) and white (rr) homozygotes. Which of the following crosses would produce offspring in the ratio of 1 red : 2 roan : 1 white?
a.	red ´ white	
b.	roan ´ roan	
c.	white ´ roan	
d.	red ´ roan	
e.	The answer cannot be determined from the information provided.	

Refer to the following to answer the questions below.

Gene S controls the sharpness of spines in a type of cactus. Cactuses with the dominant allele, S, have sharp spines, whereas homozygous recessive ss cactuses have dull spines. At the same time, a second gene, N, determines whether cactuses have spines. Homozygous recessive nn cactuses have no spines at all.

____	38.	The relationship between genes S and N is an example of
a.	incomplete dominance.	
b.	epistasis.	
c.	complete dominance.	
d.	pleiotropy.	
e.	codominance.	

____	39.	A cross between a true-breeding sharp-spined cactus and a spineless cactus would produce
a.	all sharp-spined progeny.	
b.	50% sharp-spined, 50% dull-spined progeny.	
c.	25% sharp-spined, 50% dull-spined, 25% spineless progeny	
d.	all spineless progeny.	
e.	It is impossible to determine the phenotypes of the progeny.	

____	40.	If doubly heterozygous SsNn cactuses were allowed to self-pollinate, the  would segregate in which of the following ratios?
a.	3 sharp-spined : 1 spineless	
b.	1 sharp-spined : 2 dull-spined : 1 spineless	
c.	1 sharp spined : 1 dull-spined : 1 spineless	
d.	1 sharp-spined : 1 dull-spined	
e.	9 sharp-spined : 3 dull-spined : 4 spineless	

Use the information given here to answer the following questions.

Feather color in budgies is determined by two different genes Y and B, one for pigment on the outside and one for the inside of the feather. YYBB, YyBB, or YYBb is green; yyBB or yyBb is blue; YYbb or Yybb is yellow; and yybb is white.

____	41.	A blue budgie is crossed with a white budgie. Which of the following results is not possible?
a.	Green offspring only	
b.	Yellow offspring only	
c.	Blue offspring only	
d.	Green and yellow offspring	
e.	a 9:3:3:1 ratio	

____	42.	Two blue budgies were crossed. Over the years, they produced 22 offspring, 5 of which were white. What are the most likely genotypes for the two blue budgies?
a.	yyBB and yyBB	
b.	yyBB and yyBb	
c.	yyBb and yyBb	
d.	yyBB and yybb	
e.	yyBb and yybb	

Use the following information to answer the questions below.

A woman who has blood type A positive has a daughter who is type O positive and a son who is type B negative. Rh positive is a trait that shows simple dominance over Rh negative and is designated by the alleles R and r, respectively. A third gene for the MN blood group has codominant alleles M and N.

____	43.	Which of the following is a possible partial genotype for the son?
a.		
b.		
c.	ii	
d.		
e.		

____	44.	Which of the following is a possible genotype for the mother?
a.		
b.		
c.	ii	
d.		
e.		

____	45.	Which of the following is a possible phenotype for the father?
a.	A negative	
b.	O negative	
c.	B positive	
d.	AB negative	
e.	Impossible to determine	

____	46.	Which of the following is the probable genotype for the mother?
a.		
b.		
c.		
d.		
e.		

____	47.	If both children are of blood group MM, which of the following is possible?
a.	Each parent is either M or MN.	
b.	Each parent must be type M.	
c.	Both children are heterozygous for this gene.	
d.	Neither parent can have the N allele.	
e.	The MN blood group is recessive to the ABO blood group.	

____	48.	Which describes the ability of a single gene to have multiple phenotypic effects?
a.	Incomplete dominance	
b.	Multiple alleles	
c.	Pleiotropy	
d.	Epistasis	

____	49.	Which describes the ABO blood group system?
a.	Incomplete dominance	
b.	Multiple alleles	
c.	Pleiotropy	
d.	Epistasis	

____	50.	Which of the following terms best describes when the phenotype of the heterozygote differs from the phenotypes of both homozygotes?
a.	Incomplete dominance	
b.	Multiple alleles	
c.	Pleiotropy	
d.	Epistasis	

____	51.	Cystic fibrosis affects the lungs, the pancreas, the digestive system, and other organs, resulting in symptoms ranging from breathing difficulties to recurrent infections. Which of the following terms best describes this?
a.	Incomplete dominance	
b.	Multiple alleles	
c.	Pleiotropy	
d.	Epistasis	

____	52.	Which of the following is an example of polygenic inheritance?
a.	Pink flowers in snapdragons	
b.	The ABO blood groups in humans	
c.	Huntington's disease in humans	
d.	White and purple flower color in peas	
e.	Skin pigmentation in humans	

____	53.	Hydrangea plants of the same genotype are planted in a large flower garden. Some of the plants produce blue flowers and others pink flowers. This can be best explained by which of the following?
a.	Environmental factors such as soil pH	
b.	The allele for blue hydrangea being completely dominant	
c.	The alleles being codominant	
d.	The fact that a mutation has occurred	
e.	Acknowledging that multiple alleles are involved	

____	54.	Which of the following provides an example of epistasis?
a.	Recessive genotypes for each of two genes (aabb) results in an albino corn snake.	
b.	The allele b17 produces a dominant phenotype, although b1 through b16 do not.	
c.	In rabbits and many other mammals, one genotype (cc) prevents any fur color from developing.	
d.	In Drosophila (fruit flies), white eyes can be due to an X-linked gene or to a combination of other genes.	

____	55.	Most genes have many more than two alleles. However, which of the following is also true?
a.	At least one allele for a gene always produces a dominant phenotype.	
b.	Most of the alleles will never be found in a live-born organism.	
c.	All of the alleles but one will produce harmful effects if homozygous.	
d.	There may still be only two phenotypes for the trait.	
e.	More than two alleles in a genotype is lethal.	

____	56.	Huntington's disease is a dominant condition with late age of onset in humans. If one parent has the disease, what is the probability that his or her child will have the disease?
a.	1	
b.	3/4	
c.	1/2	
d.	1/4	
e.	0	

____	57.	A woman has six sons. The chance that her next child will be a daughter is
a.	1.	
b.	0.	
c.	1/2.	
d.	1/6.	
e.	5/6.	

The following questions refer to the pedigree chart in Figure 14.2 for a family, some of whose members exhibit the dominant trait, wooly hair. Affected individuals are indicated by an open square or circle.

Figure 14.2

____	58.	What is the genotype of individual II-5?
a.	WW	
b.	Ww	
c.	ww	
d.	WW or ww	
e.	ww or Ww	

____	59.	What is the likelihood that the progeny of IV-3 and IV-4 will have wooly hair?
a.	0%	
b.	25%	
c.	50%	
d.	75%	
e.	100%	

____	60.	What is the probability that individual III-1 is Ww?
a.	3/4	
b.	1/4	
c.	2/4	
d.	2/3	
e.	1	

____	61.	People with sickle-cell trait
a.	are heterozygous for the sickle-cell allele.	
b.	are usually healthy.	
c.	have increased resistance to malaria.	
d.	produce normal and abnormal hemoglobin.	
e.	All of the above	

____	62.	When a disease is said to have a multifactorial basis, it means that
a.	both genetic and environmental factors contribute to the disease.	
b.	it is caused by a gene with a large number of alleles.	
c.	it affects a large number of people.	
d.	it has many different symptoms.	
e.	it tends to skip a generation.	

____	63.	An ideal procedure for fetal testing in humans would have which of the following features?
a.	Lowest risk procedure that would provide the most reliable information	
b.	The procedure that can test for the greatest number of traits at once	
c.	A procedure that provides a 3D image of the fetus	
d.	The procedure that can be performed at the earliest time in the pregnancy	
e.	A procedure that could test for the carrier status of the fetus	

____	64.	A scientist discovers a DNA-based test for the allele of a particular gene. This and only this allele, if homozygous, produces an effect that results in death at or about the time of birth. Of the following, which is the best use of this discovery?
a.	To screen all newborns of an at-risk population	
b.	To design a test for identifying heterozygous carriers of the allele	
c.	To introduce a normal allele into deficient newborns	
d.	To follow the segregation of the allele during meiosis	
e.	To test school-age children for the disorder	

____	65.	An obstetrician knows that one of her patients is a pregnant woman whose fetus is at risk for a serious disorder that is detectable biochemically in fetal cells. The obstetrician would most reasonably offer which of the following procedures to her patient?
a.	CVS	
b.	Ultrasound imaging	
c.	Amniocentesis	
d.	Fetoscopy	
e.	X-ray	

____	66.	The frequency of heterozygosity for the sickle cell anemia allele is unusually high, presumably because this reduces the frequency of malaria. Such a relationship is related to which of the following?
a.	Mendel's law of independent assortment	
b.	Mendel's law of segregation	
c.	Darwin's explanation of natural selection	
d.	Darwin's observations of competition	
e.	The malarial parasite changing the allele	

____	67.	Cystic fibrosis (CF) is a Mendelian disorder in the human population that is inherited as a recessive. Two normal parents have two children with CF. The probability of their next child being normal for this characteristic is which of the following?
a.	0	
b.	1/2	
c.	1/4	
d.	3/4	
e.	1/8	

____	68.	Hutchinson-Gilford progeria is an exceedingly rare human genetic disorder in which there is very early senility, and death, usually of coronary artery disease, at an average age of approximately 13. Patients, who look very old even as children, do not live to reproduce. Which of the following represents the most likely assumption?
a.	All cases must occur in relatives; therefore, there must be only one mutant allele.	
b.	Successive generations of a family will continue to have more and more cases over time.	
c.	The disorder may be due to mutation in a single protein-coding gene.	
d.	Each patient will have had at least one affected family member in a previous generation.	
e.	The disease is autosomal dominant.	

____	69.	A pedigree analysis for a given disorder's occurrence in a family shows that, although both parents of an affected child are normal, each of the parents has had affected relatives with the same condition. The disorder is then which of the following?
a.	Recessive	
b.	Dominant	
c.	Incompletely dominant	
d.	Maternally inherited	
e.	A new mutation	

____	70.	One of two major forms of a human condition called neurofibromatosis (NF 1) is inherited as a dominant, although it may be either mildly to very severely expressed. If a young child is the first in her family to be diagnosed, which of the following is the best explanation?
a.	The mother carries the gene but does not express it at all.	
b.	One of the parents has very mild expression of the gene.	
c.	The condition skipped a generation in the family.	
d.	The child has a different allele of the gene than the parents.	

Short Answer

	71.	Two pea plants heterozygous for the characters of pod color and pod shape are crossed. Draw a Punnett square to determine the phenotypic ratios of the offspring.

	72.	In some plants, a true-breeding, red-flowered strain gives all pink flowers when crossed with a white-flowered strain:  (red) ´  (white) ®  (pink). If flower position (axial or terminal) is inherited as it is in peas (see Table 14.1 in your textbook), what will be the ratios of genotypes and phenotypes of the  generation resulting from the following cross: axial-red (true-breeding) ´ terminal-white? What will be the ratios in the  generation?

	73.	Flower position, stem length, and seed shape were three characters that Mendel studied. Each is controlled by an independently assorting gene and has dominant and recessive expression as follows:

Character 	Dominant 	Recessive	
Flower position 	Axial (A) 	Terminal (a)	
Stem length 	Tall (T) 	Dwarf (t)	
Seed shape 	Round (R) 	Wrinkled (r)	

If a plant that is heterozygous for all three characters is allowed to self-fertilize, what proportion of the offspring would you expect to be as follows? (Note: Use the rules of probability instead of a huge Punnett square.)
A) Homozygous for the three dominant traits
B) Homozygous for the three recessive traits
C) Heterozygous for all three characters
D) Homozygous for axial and tall, heterozygous for seed shape

	74.	A black guinea pig crossed with an albino guinea pig produces 12 black offspring. When the albino is crossed with a second black one, 7 blacks and 5 albinos are obtained. What is the best explanation for this genetic situation? Write genotypes for the parents, gametes, and offspring.

	75.	In sesame plants, the one-pod condition (P) is dominant to the three-pod condition (p), and normal leaf (L) is dominant to wrinkled leaf (l). Pod type and leaf type are inherited independently. Determine the genotypes for the two parents for all possible matings producing the following offspring:
A) 318 one-pod, normal leaf and 98 one-pod, wrinkled leaf
B) 323 three-pod, normal leaf and 106 three-pod, wrinkled leaf
C) 401 one-pod, normal leaf
D) 150 one-pod, normal leaf, 147 one-pod, wrinkled leaf, 51 three-pod, normal leaf, and 48 three-pod, wrinkled leaf
E) 223 one-pod, normal leaf, 72 one-pod, wrinkled leaf, 76 three-pod, normal leaf, and 27 three-pod, wrinkled leaf

	76.	A man with type A blood marries a woman with type B blood. Their child has type O blood. What are the genotypes of these individuals? What other genotypes, and in what frequencies, would you expect in offspring from this marriage?

	77.	Phenylketonuria (PKU) is an inherited disease caused by a recessive allele. If a woman and her husband, who are both carriers, have three children, what is the probability of each of the following?
A) All three children are of normal phenotype.
B) One or more of the three children have the disease.
C) All three children have the disease.
D) At least one child is phenotypically normal.
(Note: Remember that the probabilities of all possible outcomes always add up to 1.)

	78.	The genotype of  individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these four genes, what are the probabilities that  offspring will have the following genotypes?
A) aabbccdd
B) AaBbCcDd
C) AABBCCDD
D) AaBBccDd
E) AaBBCCdd

	79.	What is the probability that each of the following pairs of parents will produce the indicated offspring? (Assume independent assortment of all gene pairs.)
A) AABBCC ´ aabbcc ® AaBbCc
B) AABbCc ´ AaBbCc ® AAbbCC
C) AaBbCc ´ AaBbCc ® AaBbCc
D) aaBbCC ´ AABbcc ® AaBbCc

	80.	Karen and Steve each have a sibling with sickle-cell disease. Neither Karen nor Steve nor any of their parents have the disease, and none of them have been tested to reveal sickle-cell trait. Based on this incomplete information, calculate the probability that if this couple has a child, the child will have sickle-cell disease.

	81.	In 1981, a stray black cat with unusual rounded, curled-back ears was adopted by a family in California. Hundreds of descendants of the cat have since been born, and cat fanciers hope to develop the curl cat into a show breed. Suppose you owned the first curl cat and wanted to develop a true-breeding variety. How would you determine whether the curl allele is dominant or recessive? How would you obtain true-breeding curl cats? How could you be sure they are true-breeding?

	82.	Imagine that a newly discovered, recessively inherited disease is expressed only in individuals with type O blood, although the disease and blood group are independently inherited. A normal man with type A blood and a normal woman with type B blood have already had one child with the disease. The woman is now pregnant for a second time. What is the probability that the second child will also have the disease? Assume that both parents are heterozygous for the gene that causes the disease.

	83.	In tigers, a recessive allele causes an absence of fur pigmentation (a white tiger) and a cross-eyed condition. If two phenotypically normal tigers that are heterozygous at this locus are mated, what percentage of their offspring will be cross-eyed? What percentage will be white?

	84.	In corn plants, a dominant allele I inhibits kernel color, while the recessive allele i permits color when homozygous. At a different locus, the dominant allele P causes purple kernel color, while the homozygous recessive genotype pp causes red kernels. If plants heterozygous at both loci are crossed, what will be the phenotypic ratio of the offspring?

	85.	The pedigree below traces the inheritance of alkaptonuria, a biochemical disorder. Affected individuals, indicated here by the colored circles and squares, are unable to metabolize a substance called alkapton, which colors the urine and stains body tissues. Does alkaptonuria appear to be caused by a dominant allele or by a recessive allele? Fill in the genotypes of the individuals whose genotypes can be deduced. What genotypes are possible for each of the other individuals?

	86.	A man has six fingers on each hand and six toes on each foot. His wife and their daughter have the normal number of digits. Extra digits is a dominant trait. What fraction of this couple's children would be expected to have extra digits?

	87.	Imagine that you are a genetic counselor, and a couple planning to start a family comes to you for information. Charles was married once before, and he and his first wife had a child with cystic fibrosis. The brother of his current wife, Elaine, died of cystic fibrosis. What is the probability that Charles and Elaine will have a baby with cystic fibrosis? (Neither Charles nor Elaine has cystic fibrosis.)

	88.	In mice, black color (B) is dominant to white (b). At a different locus, a dominant allele (A) produces a band of yellow just below the tip of each hair in mice with black fur. This gives a frosted appearance known as agouti. Expression of the recessive allele (a) results in a solid coat color. If mice that are heterozygous at both loci are crossed, what is the expected phenotypic ratio of their offspring?

A P biology Chapter 14 Review
Answer Section

MULTIPLE CHOICE

	1.	ANS:	D	PTS:	1

	2.	ANS:	C	PTS:	1

	3.	ANS:	C	PTS:	1

	4.	ANS:	A	PTS:	1

	5.	ANS:	B	PTS:	1

	6.	ANS:	B	PTS:	1

	7.	ANS:	D	PTS:	1

	8.	ANS:	B	PTS:	1

	9.	ANS:	A	PTS:	1

	10.	ANS:	B	PTS:	1

	11.	ANS:	C	PTS:	1

	12.	ANS:	E	PTS:	1

	13.	ANS:	D	PTS:	1

	14.	ANS:	A	PTS:	1

	15.	ANS:	C	PTS:	1

	16.	ANS:	E	PTS:	1

	17.	ANS:	D	PTS:	1

	18.	ANS:	E	PTS:	1

	19.	ANS:	D	PTS:	1

	20.	ANS:	C	PTS:	1

	21.	ANS:	D	PTS:	1

	22.	ANS:	B	PTS:	1

	23.	ANS:	B	PTS:	1

	24.	ANS:	E	PTS:	1

	25.	ANS:	C	PTS:	1

	26.	ANS:	E	PTS:	1

	27.	ANS:	B	PTS:	1

	28.	ANS:	C	PTS:	1

	29.	ANS:	D	PTS:	1

	30.	ANS:	E	PTS:	1

	31.	ANS:	A	PTS:	1

	32.	ANS:	D	PTS:	1

	33.	ANS:	C	PTS:	1

	34.	ANS:	E	PTS:	1

	35.	ANS:	D	PTS:	1

	36.	ANS:	C	PTS:	1

	37.	ANS:	B	PTS:	1

	38.	ANS:	B	PTS:	1

	39.	ANS:	A	PTS:	1

	40.	ANS:	E	PTS:	1

	41.	ANS:	D	PTS:	1

	42.	ANS:	C	PTS:	1

	43.	ANS:	D	PTS:	1

	44.	ANS:	D	PTS:	1

	45.	ANS:	C	PTS:	1

	46.	ANS:	D	PTS:	1

	47.	ANS:	A	PTS:	1

	48.	ANS:	C	PTS:	1

	49.	ANS:	B	PTS:	1

	50.	ANS:	A	PTS:	1

	51.	ANS:	C	PTS:	1

	52.	ANS:	E	PTS:	1

	53.	ANS:	A	PTS:	1

	54.	ANS:	C	PTS:	1

	55.	ANS:	D	PTS:	1

	56.	ANS:	C	PTS:	1

	57.	ANS:	C	PTS:	1

	58.	ANS:	C	PTS:	1

	59.	ANS:	C	PTS:	1

	60.	ANS:	E	PTS:	1

	61.	ANS:	E	PTS:	1

	62.	ANS:	A	PTS:	1

	63.	ANS:	A	PTS:	1

	64.	ANS:	B	PTS:	1

	65.	ANS:	C	PTS:	1

	66.	ANS:	C	PTS:	1

	67.	ANS:	C	PTS:	1

	68.	ANS:	C	PTS:	1

	69.	ANS:	A	PTS:	1

	70.	ANS:	B	PTS:	1

SHORT ANSWER

	71.	ANS:	

PTS:	1

	72.	ANS:	
Parental cross is . Genotype of  is , phenotype is all axial-pink. Genotypes of  are 4  : 2  : 2  : 2  : 2  : 1  : 1  : 1  : 1 . Phenotypes of  are 6 axial-pink : 3 axial-red : 3 axial-white : 2 terminal-pink : 1 terminal-white : 1 terminal-red.

PTS:	1

	73.	ANS:	
A) 1/64, B) 1/64, C) 1/8, D) 1/32

PTS:	1

	74.	ANS:	
Albino (b) is a recessive trait; black (B) is dominant. First cross: parents BB ´ bb; gametes B and b; offspring all Bb (black coat). Second cross: parents Bb ´ bb; gametes 1/2 B and 1/2 b (heterozygous parent) and b; offspring 1/2 Bb and 1/2 bb.

PTS:	1

	75.	ANS:	
A) PPLl ´ PPLl, PPLl ´ PpLl, or PPLl ´ ppLl. B) ppLl ´ ppLl. C) PPLL ´ any of the 9 possible genotypes or PPll ´ ppLL. D) PpLl ´ Ppll. E) PpLl ´ PpLl.

PTS:	1

	76.	ANS:	
Man ; woman ; child ii. Other genotypes for children are 1/4 , 1/4 , 1/4 .

PTS:	1

	77.	ANS:	
A) 3/4 ´ 3/4 ´ 3/4 = 27/64, B) 1 — 27/64 = 37/64, C) 1/4 ´ 1/4 ´ 1/4 = 1/64, D) 1 — 1/64=63/64

PTS:	1

	78.	ANS:	
A) 1/256, B) 1/16, C) 1/256, D) 1/64, E) 1/128

PTS:	1

	79.	ANS:	
A) 1, B) 1/32, C) 1/8, D) 1/2

PTS:	1

	80.	ANS:	
1/9

PTS:	1

	81.	ANS:	
Matings of the original mutant cat with true-breeding noncurl cats will produce both curl and noncurl  offspring if the curl allele is dominant, but only noncurl offspring if the curl allele is recessive. You would obtain some true-breeding offspring homozygous for the curl allele from matings between the  cats resulting from the original curl ´ noncurl crosses whether the curl trait is dominant or recessive. You know that cats are true-breeding when curl ´ curl matings produce only curl offspring. As it turns out, the allele that causes curled ears is dominant.

PTS:	1

	82.	ANS:	
1/16

PTS:	1

	83.	ANS:	
25% will be cross-eyed; all of the cross-eyed offspring will also be white.

PTS:	1

	84.	ANS:	
The dominant allele I is epistatic to the P/p locus, and thus the genotypic ratio for the  generation will be 9 I_P_ (colorless) : 3 I_pp (colorless) : 3 iiP_ (purple) : 1 iipp (red). Overall, the phenotypic ratio is 12 colorless : 3 purple : 1 red.

PTS:	1

	85.	ANS:	
Recessive. All affected individuals (Arlene, Tom, Wilma, and Carla) are homozygous recessive aa. George is Aa, since some of his children with Arlene are affected. Sam, Ann, Daniel, and Alan are each Aa, since they are all unaffected children with one affected parent. Michael also is Aa, since he has an affected child (Carla) with his heterozygous wife Ann. Sandra, Tina, and Christopher can each have the AA or Aa genotype.

PTS:	1

	86.	ANS:	
1/2

PTS:	1

	87.	ANS:	
1/6

PTS:	1

	88.	ANS:	
9 B_A_ (agouti) : 3 B_aa (black) : 3 bbA_ (white) : 1 bbaa (white). Overall, 9 agouti : 3 black : 4 white.

PTS:	1