Category: Ecology

Biomes of the World

%CODE1% Biomes of the World

 

TSOVCDARPRLVTTS
QSLUEIEBEGALUDR
XTESSMTTIVDENBT
NAERUEAOVOIMDHK
ZRUSOWRJINTORER
TMNTHFCOHBRIARE
SOLSNHNAVCEBCBC
CPERFKQIRITAFIU
NRSAVANNANNYCVD
FENIRAMFDRIMUOO
GRASSLANDZAVORR
SUOUDICEDGNJOEP
OHZGLJYXIBYXFRW
UCTCTNYAQBAJSUE
XNBSOOTDEKPTNBX

Find each term and then number and define it on the back of this sheet.

 

ABIOTICBIOMEBIOTIC
CARNIVORECONSUMERDECIDUOUS
DESERTFRESHWATERGRASSLAND
HERBIVOREINTERTIDALMARINE
OMNIVOREPRODUCERRAINFOREST
SAVANNATAIGATUNDRA

 

 

 

 

 

Genetics

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Genetics
 

Modified True/False
Indicate whether the statement is true or false.  If false, change the identified word or phrase to make the statement true.
 1.
Mendel discovered predictable patterns in the inheritance of traits. _________________________


 2.
Garden peas are difficult to grow because they mature slowly. _________________________


 3.
When Mendel cross-pollinated two varieties from the P generation that exhibited contrasting traits, he called the offspring the second filial, or F2, generation. _________________________


 4.
The contrasting forms of each character studied by Mendel appeared in a 3:1 ratio in the F2 generation. _________________________


 5.
A dominant allele masks the effect of a recessive allele. _________________________


 6.
The allele for a recessive trait is usually represented by a capital letter. _________________________


 7.
Heterozygous individuals have two of the same alleles for a particular gene. _________________________


 8.
The inheritance of sex-linked traits can be studied by making a pedigree of several generations of a family. _________________________


nar001-1.jpg
 9.
Refer to the illustration above. The father listed in the pedigree is most likely heterozygous for the trait. _________________________


 10.
Refer to the illustration above. Child #3 probably has the dominant phenotype. _________________________


 11.
Albinism is caused by a recessive allele. _________________________


 12.
The phenotype that results from an inherited pair of alleles depends on the instructions in the genes only. _________________________


 13.
Traits for different characters that are usually inherited together are said to be combined. _________________________


 

Multiple Choice
Identify the choice that best completes the statement or answers the question.
 14.
The passing of traits from parents to offspring is called
a.
genetics.
c.
development.
b.
heredity.
d.
maturation.
 15.
The difference between Mendel’s experiments in the area of heredity and those done by earlier researchers was that
a.
earlier researchers did not have microscopes.
b.
earlier researchers used detailed and numerical procedures.
c.
Mendel expressed the results of his experiments in terms of numbers.
d.
Mendel used pea plants with both purple and white flowers.
 16.
The scientific study of heredity is called
a.
meiosis.
c.
genetics.
b.
crossing-over.
d.
pollination.
 17.
The “father” of genetics was
a.
A. Knight.
c.
Gregor Mendel.
b.
Hans Krebs.
d.
Charles Darwin.
 18.
Garden peas are good subjects for studying heredity because they
a.
are difficult to grow.
c.
produce few offspring.
b.
mature quickly.
d.
have few traits.
 19.
Mendel obtained his P generation by allowing the plants to
a.
self-pollinate.
c.
assort independently.
b.
cross-pollinate.
d.
segregate.
 20.
Step 1 of Mendel’s garden pea experiment, allowing each variety of garden pea to self-pollinate for several generations, produced the
a.
F1 generation.
c.
P generation.
b.
F2 generation.
d.
P1 generation.
 21.
F2 : F1 ::
a.
P : F1
c.
F1 : P
b.
F1 : F2
d.
P : F2
 22.
An allele that is always expressed whenever it is present is called
a.
dominant.
c.
recessive.
b.
phenotypic.
d.
superior.
 23.
The discovery of chromosomes provided a link between the first law of heredity that stemmed from Mendel’s work and
a.
pollination.
c.
mitosis.
b.
inheritance.
d.
meiosis.
 24.
The law of segregation states that
a.
alleles of a gene separate from each other during meiosis.
b.
different alleles of a gene can never be found in the same organism.
c.
each gene of an organism ends up in a different gamete.
d.
each gene is found on a different molecule of DNA.
 25.
Mendel’s law of segregation states that
a.
pairs of alleles are dependent on one another when separation occurs during gamete formation.
b.
pairs of alleles separate independently of one another after gamete formation.
c.
each pair of alleles remains together when gametes are formed.
d.
the two alleles for a trait segregate independently when gametes are formed.
 26.
The phenotype of an organism
a.
is used to represent its genetic composition.
b.
is the physical appearance of a trait.
c.
occurs only in dominant individuals.
d.
cannot be seen.
 27.
If an individual possesses two recessive alleles for the same trait, the individual is said to be
a.
homozygous for the trait.
c.
heterozygous for the trait.
b.
haploid for the trait.
d.
mutated.
 28.
When an individual heterozygous for a trait is crossed with an individual homozygous recessive for the trait, the offspring produced will
a.
all have the same genotype.
b.
show two different phenotypes.
c.
show three different phenotypes.
d.
all have the same phenotype.
 29.
Tallness (T) is dominant to shortness (t) in pea plants. Which of the following represents a genotype of a pea plant that is heterozygous for tallness?
a.
T
c.
Tt
b.
TT
d.
tt
 30.
homozygous : heterozygous ::
a.
heterozygous : Bb
c.
probability : chance
b.
gg : Gg
d.
factor : gene
 31.
Mendel’s finding that the inheritance of one trait had no effect on the inheritance of another became known as the
a.
law of dominance.
b.
law of universal inheritance.
c.
law of separate convenience.
d.
law of independent assortment.

In humans, having freckles (F) is dominant to not having freckles (f). The inheritance of these traits can be studied using a Punnett square similar to the one shown below.

nar002-1.jpg
 32.
Refer to the illustration above. The child represented in box 1 in the Punnett square would
a.
be homozygous for freckles.
b.
have an extra freckles chromosome.
c.
be heterozygous for freckles.
d.
not have freckles.
 33.
Refer to the illustration above. The parents shown in the Punnett square could have children with a phenotype ratio of
a.
1:2:1.
c.
3:1.
b.
4:0.
d.
2:2.
 34.
Refer to the illustration above. Which box in the Punnett square represents a child who does not have freckles?
a.
box 1
c.
box 3
b.
box 2
d.
box 4
 35.
Refer to the illustration above. The child in box 3 of the Punnett square has the genotype
a.
FF.
c.
ff.
b.
Ff.
d.
FfFf.

In rabbits, black fur (B) is dominant to brown fur (b). Consider the following cross between two rabbits.

nar003-1.jpg
 36.
Refer to the illustration above. The device shown, which is used to determine the probable outcome of genetic crosses, is called a
a.
Mendelian box.
c.
genetic graph.
b.
Punnett square.
d.
phenotypic paradox.
 37.
Refer to the illustration above. Both of the parents in the cross are
a.
black.
c.
homozygous dominant.
b.
brown.
d.
homozygous recessive.
 38.
Refer to the illustration above. The phenotype of the offspring indicated by box 3 would be
a.
brown.
c.
a mixture of brown and black.
b.
black.
d.
white.
 39.
Refer to the illustration above. The genotypic ratio of the F1 generation would be
a.
1:1.
c.
1:3.
b.
3:1.
d.
1:2:1.
 40.
What is the expected genotypic ratio resulting from a homozygous dominant ´ heterozygous monohybrid cross?
a.
1:0
c.
1:2:1
b.
1:1
d.
1:3:1
 41.
What is the expected genotypic ratio resulting from a heterozygous ´ heterozygous monohybrid cross?
a.
1:2:1
c.
1:2
b.
1:3:1
d.
1:0
 42.
What is the expected phenotypic ratio resulting from a homozygous dominant ´ heterozygous monohybrid cross?
a.
1:3:1
c.
2:1
b.
1:2:1
d.
1:0
 43.
The unknown genotype of an individual with a dominant phenotype can be determined using a
a.
ratio.
c.
probability formula.
b.
dihybrid cross.
d.
test cross.
 44.
What is the probability that the offspring of a homozygous dominant individual and a homozygous recessive individual will exhibit the dominant phenotype?
a.
0.25
c.
0.66
b.
0.5
d.
1.0
 45.
Probability is calculated by dividing the number of one kind of possible outcome by the
a.
number of other kinds of outcomes.
b.
total number of all possible outcomes.
c.
number of genes being considered.
d.
total number of offspring produced.
 46.
If a characteristic is sex-linked, the gene for it is found on
a.
a sex chromosome.
c.
a linked chromosome.
b.
an autosome.
d.
an allele.
 47.
Since the allele for colorblindness is located on the X chromosome, colorblindness
a.
cannot be inherited.
c.
is sex-linked.
b.
occurs only in adults.
d.
occurs only in females.
 48.
A diagram in which several generations of a family and the occurrence of certain genetic characteristics are shown is called a
a.
Punnett square.
c.
pedigree.
b.
monohybrid cross.
d.
family karyotype.
 49.
In humans, eye color and height are controlled by
a.
simple dominance.
c.
polygenic inheritance.
b.
multiple alleles.
d.
incomplete dominance.
 50.
Which of the following traits is controlled by multiple alleles in humans?
a.
eye color
c.
colorblindness
b.
blood type
d.
albinism
 51.
What would be the blood type of a person who inherited an A allele from one parent and an O allele from the other?
a.
type A
c.
type AB
b.
type B
d.
type O
 52.
Which of the following is not an outcome of the environment modifying a phenotype?
a.
the changing of the color of an animal’s fur as the temperature changes
b.
the increased intelligence of a person who attended school for many years
c.
the very short stature of a kind of tree that grows at a high altitude in comparison with the same kind of tree growing at a lower altitude
d.
the pink-flowered snapdragons that result from crosses between red-flowered and white-flowered snapdragon plants
 53.
Genes that are close together on a single chromosome are considered to be
a.
alleles.
c.
independent.
b.
homozygous.
d.
linked.

 

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True/False
Indicate whether the sentence or statement is true or false.
1.
Hemophilia is a recessive, sex-linked disorder.
2.
An autosomal trait will occur with equal frequency in both males and females.
3.
Sex-linkd traits appear more often in females than in males.
4.
An individual who expresses a genetic disorder is called a carrier.
5.
A male can produce sperm that contains either an X or a Y chromosome.
6.
Mutations are always harmful.
7.
A pedigree is a family record that shows how a trait is inherited over several generations.
 

Multiple Choice
Identify the letter of the choice that best completes the statement or answers the question.
8.
Hemophilia is a genetic disorder that is
a.
sex-linked.
c.
fairly common.
b.
sex-influenced.
d.
more common in women than in men.
9.
Parents of a color blind female could have the genotypes
a.
XCXc and XCY
c.
XCXc and XcY
b.
XcXc and XCY
d.
XCXC and XcY
10.
People who are heterozygous for sickle cell anemia are
a.
partially resistant to the effect of ultraviolet radiation.
b.
totally resistant to the effect of ultraviolet radiation.
c.
partially resistant to malaria.
d.
totally resistant to malaria.
11.
Both sickle cell anemia and hemophilia
a.
are caused by genes coding for defective proteins.
b.
are seen in homozygous dominant individuals.
c.
provide resistance to malaria infections.
d.
are extremely common throughout the world.
12.
This genetic disease occurs most frequently in the black population.
a.
cystic fibrosis
c.
sickle-cell anemia
b.
Tay-Sachs disease
d.
hemophilia
13.
The trait for skin color in humans is cause by
a.
crossing-over
c.
nondisjunction
b.
multiple genes
d.
mutations
14.
A hereditary disease in which the red blood cells have an abnormal shape is
a.
galactosemia
c.
polydactyly
b.
Tay-Sachs disease
d.
sickle-cell anemia
15.
Sex-linked traits appear more often in males than in females because
a.
males are produced in greater numbers
b.
females with a sex-linked trait will die
c.
males have only one X chromosome
d.
sex-linked traits are carried on the Y chromosome
16.
A diagram in which several generations of a family and the occurrence of certain genetic characteristics are shown is called a
a.
Punnett square.
c.
pedigree.
b.
monohybrid cross.
d.
family karyotype.
17.
What is the percentage of chance that two people who carry a sickle-cell gene will have a child having the disease?
a.
0
c.
50
b.
25
d.
100
18.
Which genetic trait is most influenced by the environment?
a.
blood type
c.
baldness
b.
eye color
d.
weight
19.
Very dark-skinned people have alleles at all gene positions for skin color that code for
a.
red-green colorblindness.
c.
the production of Rh antigens.
b.
albinism.
d.
the production of melanin.
20.
Sex-linked traits such as color blindness
a.
are never carried by females
c.
occur more frequently in males
b.
occur more frequently in females
d.
never occur in males
21.
What would be the blood type of a person who inherited an A allele from one parent and an O allele from the other?
a.
type A
c.
type AB
b.
type B
d.
type O.
22.
Genetic counseling is a process that
a.
helps identify parents at risk for having children with genetic defects.
b.
assists parents in deciding whether or not to have children.
c.
uses a family pedigree.
d.
All of the above
23.
If both parents carry the recessive allele that causes cystic fibrosis, the chance that their child will develop the disease is
a.
one in two.
c.
two in five.
b.
one in four.
d.
100 percent.
24.
People with Down syndrome have
a.
45 chromosomes.
c.
47 chromosomes.
b.
46 chromosomes.
d.
no X chromosomes.
25.
If nondisjunction occurs,
a.
there will be too many gametes produced.
b.
no gametes will be produced.
c.
a gamete will receive too many or too few homologues of a chromosome.
d.
mitosis cannot take place.
26.
A human female inherits
a.
one copy of every gene located on each of the X chromosomes.
b.
twice as many sex chromosomes as a human male inherits.
c.
one copy of every gene located on the Y chromosome.
d.
all of the same genes that a human male inherits.
27.
A pedigree CANNOT be used to
a.
determine whether a trait is inherited.
b.
show how a trait is passed from one generation to the next.
c.
determine whether an allele is dominant or recessive.
d.
none of the above
28.
If a man with blood type A and a woman with blood type B produce an offspring, what might be the offspring’s blood type?
a.
AB or O
b.
A, B, or O
c.
A, B, AB, or O
d.
AB only
29.
Sickle cell disease is caused by a
a.
change in one DNA base.
b.
change in the size of a chromosome.
c.
change in two genes.
d.
change in the number of chromosomes in a cell.
30.
Many sex-linked genes are located on
a.
the autosomes.
b.
the X chromosome only.
c.
the Y chromosome only.
d.
both the X chromosome and the Y chromosome.

 

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Genetics Practice Test Bi

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Chapter 9 Test, Form A
 

Multiple Choice
Identify the letter of the choice that best completes the statement or answers the question.
1.
The “father” of genetics was
a.
T. A. Knight.
c.
Gregor Mendel.
b.
Hans Krebs.
d.
None of the above
2.
Mendel obtained his P generation by allowing the plants to
a.
self-pollinate.
c.
assort independently.
b.
cross-pollinate.
d.
segregate.
3.
What is the probability that the offspring of a homozygous dominant individual and a homozygous recessive individual will exhibit the dominant phenotype?
a.
0.25
c.
0.66
b.
0.5
d.
1.0
4.
True-breeding pea plants always
a.
are pollinated by hand.
b.
produce offspring with either form of a trait.
c.
produce offspring with only one form of a trait.
d.
are heterozygous.
5.
The first filial (F1) generation is the result of
a.
cross-pollination among parents and the next generation.
b.
crosses between individuals of the parental generation.
c.
crosses between the offspring of a parental cross.
d.
self-fertilization between parental stock.
6.
Which of the following is the designation for Mendel’s original pure strains of plants?
a.
P
c.
F1
b.
P1
d.
F2
7.
A genetic trait that appears in every generation of offspring is called
a.
dominant.
c.
recessive.
b.
phenotypic.
d.
superior.
8.
To describe how traits can disappear and reappear in a certain pattern from generation to generation, Mendel proposed
a.
the law of independent assortment.
b.
the law of segregation.
c.
the law of genotypes.
d.
that the F2 generation will only produce purple flowers.
9.
When Mendel crossed pea plants with two contrasting traits, such as flower color and plant height,
a.
these experiments led to his law of segregation.
b.
he found that the inheritance of one trait did not influence the inheritance of the other trait.
c.
he found that the inheritance of one trait influenced the inheritance of the other trait.
d.
these experiments were considered failures because the importance of his work was not recognized.
10.
If an individual has two recessive alleles for the same trait, the individual is said to be
a.
homozygous for the trait.
c.
heterozygous for the trait.
b.
haploid for the trait.
d.
mutated.
11.
An individual heterozygous for a trait and an individual homozygous recessive for the trait are crossed and produce many offspring that are
a.
all the same genotype.
c.
of three different phenotypes.
b.
of two different phenotypes.
d.
all the same phenotype.
In humans, having freckles (F) is dominant to not having freckles (f). The inheritance of these traits can be studied using a Punnett square similar to the one shown below.
      evhm2gen-b_files/i0100000.jpg
12.
Refer to the illustration above. The genotype represented in box “1” in the Punnett square would
a.
be homozygous for freckles.
b.
have an extra freckles chromosome.
c.
be heterozygous for freckles.
d.
have freckles chromosomes.
13.
Refer to the illustration above. The genotype in box “3” of the Punnett square is
a.
FF.
c.
ff.
b.
Ff.
d.
None of the above
14.
How many different phenotypes can be produced by a pair of codominant alleles?
a.
1
c.
3
b.
2
d.
4
evhm2gen-b_files/i0130000.jpg
15.
Refer to the illustration above. The phenotype represented by the cell labeled “1” is
a.
green, inflated.
c.
yellow, inflated.
b.
green, constricted.
d.
yellow, constricted.
16.
2,000 yellow seeds : 8,000 total seeds ::
a.
1 : 6
c.
1 : 3
b.
1 : 8
d.
1 : 4
In rabbits, black fur (B) is dominant to brown fur (b). Consider the following cross between two rabbits.
      evhm2gen-b_files/i0160000.jpg
17.
Refer to the illustration above. Both of the parents in the cross are
a.
black.
c.
homozygous dominant.
b.
brown.
d.
homozygous recessive.
18.
Refer to the illustration above. The phenotype of the offspring indicated by box “3” would be
a.
brown.
c.
a mixture of brown and black.
b.
black.
d.
The phenotype cannot be determined.
19.
What is the expected phenotypic ratio resulting from a homozygous dominant ´ heterozygous monohybrid cross?
a.
1:3:1
c.
2:1
b.
1:2:1
d.
1:0
      evhm2gen-b_files/i0240000.jpg
20.
Refer to the illustration above. The phenotype represented by the cell labeled “1” is
a.
round, yellow.
c.
wrinkled, yellow.
b.
round, green.
d.
wrinkled, green.
21.
An organism that has inherited two of the same alleles of a gene from its parents is called
a.
hereditary.
c.
homozygous.
b.
heterozygous.
d.
a mutation.
22.
In pea plants, yellow seeds are dominant over green seeds. What would be the expected genotype ratio in a cross between a plant with green seeds and a plant that is heterozygous for seed color?
a.
1:3
c.
4:1
b.
1:2:1
d.
1:1
23.
codominance : both traits are displayed ::
a.
probability : crosses
b.
heterozygous : alleles are the same
c.
homozygous : alleles are the same
d.
Punnett square : chromosomes combine
24.
The difference between a monohybrid cross and a dihybrid cross is that
a.
monohybrid crosses involve traits for which only one allele exists, while dihybrid traits involve two alleles.
b.
monohybrid crosses involve self-pollination, while dihybrid crosses involve cross-pollination.
c.
monohybrid crosses involve one gene; dihybrid crosses involve two genes.
d.
dihybrid crosses require two Punnett squares; monohybrid crosses need only one.
25.
What fraction of the offspring resulting from a heterozygous ´ heterozygous dihybrid cross are heterozygous for both traits?
a.
9/16
c.
3/16
b.
1/4
d.
1/16

 

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AP Heredity Study Guide

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ap heredity study guide

 

 

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 completely 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.
A plant with purple flowers is allowed to self-pollinate. Generation after generation, it produces purple flowers. This is an example of
a.
hybridization.
b.
incomplete dominance.
c.
true-breeding.
d.
the law of segregation.
e.
polygenetics.
 3.
Which of the following statements about Mendel’s breeding experiments is correct?
a.
None of the parental (P) plants were true-breeding.
b.
All of the F2 progeny showed a phenotype that was intermediate between the two parental (P) phenotypes.
c.
Half of the F1 progeny had the same phenotype as one of the parental (P) plants, and the other half had the same phenotype as the other parent.
d.
All of the F1 progeny resembled one of the parental (P) plants, but only some of the F2 progeny did.
e.
none of the above
 4.
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 monohybrid cross involves organisms that are heterozygous for a single character, whereas a dihybrid cross involves organisms that are heterozygous for two characters.
d.
A monohybrid cross is performed only once, whereas a dihybrid cross is performed twice.
e.
A monohybrid cross results in a 9:3:3:1 ratio whereas a dihybrid cross gives a 3:1 ratio.
 5.
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.
 6.
The F1 offspring of Mendel’s classic pea cross always looked like one of the two parental varieties because
a.
one allele 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.
 7.
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 F1 than do dominant ones.
d.
Genes are composed of DNA.
e.
An organism that is homozygous for many recessive traits is at a disadvantage.
 8.
Which of the following is (are) true for alleles?
a.
They can be identical or different for any given gene in a somatic cell.
b.
They can be dominant or recessive.
c.
They can represent alternative forms of a gene.
d.
Only A and B are correct.
e.
A, B, and C are correct.
 9.
What is genetic cross between an individual showing a dominant phenotype (but of unknown genotype) and a homozygous recessive individual called?
a.
a self-cross
b.
a testcross
c.
a hybrid cross
d.
an F1 cross
e.
a dihybrid cross
 10.
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
 11.
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.
 12.
Two characters that appear in a 9:3:3:1 ratio in the F2 generation should have which of the following properties?
a.
Each of the characters is controlled by a single gene.
b.
The genes controlling the characters obey the law of independent assortment.
c.
Each of the genes controlling the characters has two alleles.
d.
Only A and C are correct.
e.
A, B, and C are correct.
 13.
A 9:3:3:1 phenotypic ratio is characteristic of which of the following?
a.
a monohybrid cross
b.
a dihybrid cross
c.
a trihybrid cross
d.
linked genes
e.
both A and D
 14.
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
 15.
It was important that Mendel examined not just the F1 generation in his breeding experiments, but the F2 generation as well, because
a.
he obtained very few F1 progeny, making statistical analysis difficult.
b.
parental traits that were not observed in the F1 reappeared in the F2, suggesting that the traits did not truly disappear in the F1.
c.
analysis of the F1 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 F2 generation, but not in the F1.
e.
all of the above
 16.
When crossing a homozygous recessive with a heterozygote, what is the chance of getting an offspring with the homozygous recessive phenotype?
a.
0%
b.
25%
c.
50%
d.
75%
e.
100%

Use the diagram and description below to answer the following question.

In a particular plant, leaf color is controlled by gene D. Plants with the dominant 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 F1 offspring is allowed to self-pollinate. The predicted outcome of this cross is diagrammed in the Punnett square shown below, where 1, 2, 3, and 4 represent the genotypes corresponding to each box within the square.

nar001-1.jpg

 17.
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
 18.
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
 19.
Which of the plants will be true-breeding?
a.
1 and 4
b.
2 and 3
c.
1-4
d.
1 only
e.
none
 20.
P = purple, pp = white. The offspring of a cross between two heterozygous purple-flowering plants (Pp Pp) results in
a.
all purple-flowered plants.
b.
purple-flowered plants and white-flowered plants.
c.
two types of white-flowered plants: PP and Pp.
d.
all white-flowered plants.
e.
all pink-flowered plants.
 21.
Mendel accounted for the observation that traits which had disappeared in the F1 generation reappeared in the F2 generation by proposing that
a.
new mutations were frequently generated in the F2 progeny, “reinventing” traits that had been lost in the F1.
b.
the mechanism controlling the appearance of traits was different between the F1 and the F2 plants.
c.
traits can be dominant or recessive, and the recessive traits were obscured by the dominant ones in the F1.
d.
the traits were lost in the F1 due to blending of the parental traits.
e.
members of the F1 generation had only one allele for each character, but members of the F2 had two alleles for each character.
 22.
What are Punnett squares used for?
a.
predicting the result of genetic crosses between organisms of known genotypes
b.
determining the DNA sequence of a given gene
c.
identifying the gene locus where allelic variations are possible
d.
testing for the presence of the recessive allele
e.
more than one of the above
 23.
Which of the following is false, regarding the law of segregation?
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 F2 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.
 24.
The fact that all seven of the pea plant traits studied by Mendel obeyed the principle of independent assortment means that
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.
 25.
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
 26.
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.
1/2
b.
1/4
c.
0
d.
1
e.
1/6
 27.
A couple has three children, all of whom have brown eyes and blond hair. Both parents are homozygous for brown eyes (BB) but one is a blond (rr) and the other is a redhead (Rr). What is the probability that their next child will be a brown-eyed redhead?
a.
1/16
b.
1/8
c.
1/4
d.
1/2
e.
1
 28.
Two true-breeding stocks of pea plants are crossed. One parent has red, axial flowers and the other has white, terminal flowers; all F1 individuals have red, axial flowers. If 1,000 F2 offspring resulted from the cross, approximately how many of them would you expect to have red, terminal flowers? (Assume independent assortment).
a.
65
b.
190
c.
250
d.
565
e.
750
 29.
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
 30.
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
 31.
A 1:2:1 phenotypic ratio in the F2 generation of a monohybrid cross is a sign of
a.
complete dominance.
b.
multiple alleles.
c.
incomplete dominance.
d.
polygenic inheritance.
e.
pleiotropy.

Refer to the result below to answer the following questions.

A tall plant is crossed with a short plant, and the progeny are all intermediate in size between the two parental plants.

 32.
This could be an example of
a.
incomplete dominance.
b.
polygenic inheritance.
c.
complete dominance.
d.
A and B
e.
B and C
 33.
If the intermediate F1 progeny were allowed to self-pollinate, and the F2 progeny were also intermediate in size, but following a normal distribution, this would suggest
a.
incomplete dominance.
b.
polygenic inheritance.
c.
complete dominance.
d.
a strong environmental influence.
e.
codominance.
 34.
If the intermediate F1 progeny were allowed to self-pollinate, and 25% of the F2 progeny were tall, 50% were intermediate in size, and 25% were short, this would suggest
a.
incomplete dominance.
b.
polygenic inheritance.
c.
complete dominance.
d.
pleiotropy.
e.
multifactorial inheritance.
 35.
In snapdragons, heterozygotes 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%
 36.
Tallness (T) 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 F1 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
 37.
Skin color in a 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
 38.
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.
 39.
The relationship between genes S and N is an example of
a.
incomplete dominance.
b.
epistasis.
c.
complete dominance.
d.
pleiotropy.
e.
codominance.
 40.
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.
 41.
If doubly heterozygous SsNn cactuses were allowed to self-pollinate, the F2 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 below to answer the following questions.

Feather color in budgies is determined by two different genes Y and B. YYBB, YyBB, or YYBb is green; yyBB or yyBb is blue; YYbb or Yybb is yellow; and yybb is white.

 42.
A blue budgie is crossed with a white budgie. Which of the following results is not possible?
a.
green offspring
b.
yellow offspring
c.
blue offspring
d.
A and B
e.
A, B, and C
 43.
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
 44.

Three babies were mixed up in a hospital. After consideration of the data below, which of the following represent the correct baby and parent combinations?

Couple #
Blood Groups		I
A and A		II
A and B		III
B and O
Baby #
Blood Groups		1
B		2
O		3
AB
a.
I-3, II-1, III-2
b.
I-1, II-3, III-2
c.
I-2, II-3, III-1
d.
I-2, II-1, III-3
e.
I-3, II-2, III-1

Use the following information to answer the questions below.

A woman who has blood type A, has a daughter who is type O positive and a son who is type B negative. Rh positive is a simple dominant trait over Rh negative.

 45.
Which of the following is a possible genotype for the son?
a.
IBIB
b.
IBIA
c.
ii
d.
IBi
e.
IAIA
 46.
Which of the following is a possible genotype for the mother?
a.
IAIA
b.
IBIB
c.
ii
d.
IAi
e.
IAIB
 47.
Which of the following is a possible phenotype for the father?
a.
A
b.
O
c.
B
d.
AB
e.
impossible to determine
 48.
Which of the following is the probable genotype for the mother?
a.
IAIARR
b.
IAIARr
c.
IAirr
d.
IAiRr
e.
IAiRR
 49.
Which of the following is a possible phenotype of the father?
a.
A negative
b.
O negative
c.
B positive
d.
A positive
e.
O positive
 50.
What is the chromosomal system for determining sex in mammals?
a.
A
b.
B
c.
C
d.
D
e.
E
 51.
What is the chromosomal system for sex determination in grasshoppers and certain other insects?
a.
A
b.
B
c.
C
d.
D
e.
E
 52.
What is the chromosomal system for sex determination in birds?
a.
A
b.
B
c.
C
d.
D
e.
E
 53.
What is the chromosomal system of sex determination in most species of ants and bees?
a.
A
b.
B
c.
C
d.
D
e.
E

Use the terms listed below to answer the following questions. Each term may be used once, more than once, or not at all.

A. incomplete dominance
B. multiple alleles
C. pleiotropy
D. epistasis
 54.
the ability of a single gene to have multiple phenotypic effects
a.
A
b.
B
c.
C
d.
D
 55.
the ABO blood group system
a.
A
b.
B
c.
C
d.
D
 56.
the phenotype of the heterozygote differs from the phenotypes of both homozygotes
a.
A
b.
B
c.
C
d.
D
 57.
cystic fibrosis affects the lungs, the pancreas, the digestive system, and other organs, resulting in symptoms ranging from breathing difficulties to recurrent infections
a.
A
b.
B
c.
C
d.
D
 58.
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
 59.
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
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.

Use the information below to answer the following questions.

A woman and her spouse both show the normal phenotype for pigmentation, but both had one parent who was an albino. Albinism is an autosomal recessive trait.

 60.
What is the probability that their first child will be an albino?
a.
0
b.
1/4
c.
1/2
d.
3/4
e.
1
 61.
If their first two children have normal pigmentation, what is the probability that their third child will be an albino?
a.
0
b.
1/4
c.
1/2
d.
3/4
e.
1
 62.
Huntington’s disease is caused by a dominant allele. If one of your parents has the disease, what is the probability that you, too, will have the disease?
a.
1
b.
3/4
c.
1/2
d.
1/4
e.
0
 63.
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 pedigree chart below is for a family, some of whose members exhibit the recessive trait, wooly hair. Affected individuals are indicated by an open square or circle. Use the chart to answer the following questions.

nar007-1.jpg

 64.
What is the genotype of individual B-5?
a.
WW
b.
Ww
c.
ww
d.
WW or ww
e.
ww or Ww
 65.
What is the likelihood that the progeny of D-3 and D-4 will have wooly hair?
a.
0%
b.
25%
c.
50%
d.
75%
e.
100%
 66.
What is the probability that individual C-1 is Ww?
a.
3/4
b.
1/4
c.
2/4
d.
2/3
e.
1
 67.
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
 68.
When a disease is said to have a multifactorial basis, it means that
a.
many factors, both genetic and environmental, 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.
 69.
Which of the following terms is least related to the others?
a.
pedigree
b.
karyotype
c.
amniocentesis
d.
chorionic villus sampling
e.
epistasis

Use the answers below to answer the following questions. Each answer may be used once, more than once, or not at all.

A. Huntington’s disease
B. Tay-Sachs disease
C. phenylketonuria
D. cystic fibrosis
E. sickle-cell disease
 70.
Substitution of the “wrong” amino acid in the hemoglobin protein results in this disorder.
a.
A
b.
B
c.
C
d.
D
e.
E
 71.
Individuals with this disorder are unable to metabolize certain lipids, affecting proper brain development. Affected individuals die in early childhood.
a.
A
b.
B
c.
C
d.
D
e.
E
 72.
This is caused by a dominant single gene defect and generally does not appear until the individual is 35-45 years of age.
a.
A
b.
B
c.
C
d.
D
e.
E
 73.
Effects of this recessive disorder can be completely overcome by regulating the diet of the affected individual.
a.
A
b.
B
c.
C
d.
D
e.
E
 74.
This results from a defect in membrane proteins that normally function in chloride ion transport.
a.
A
b.
B
c.
C
d.
D
e.
E
 75.
Which of the following techniques involves the preparation of a karyotype?
a.
amniocentesis
b.
chorionic villus sampling
c.
fetoscopy
d.
A and B only
e.
A, B, and C
Media Activity Questions
 76.
All the offspring of a cross between a black-eyed MendAlien and an orange-eyed MendAlien have black eyes. This means that the allele for black eyes is ____ the allele for orange eyes.
a.
codominant to
b.
recessive to
c.
more aggressive than
d.
dominant to
e.
better than
 77.
What is the expected phenotypic ratio of a cross between two orange-eyed MendAliens?
a.
3 black-eyed:1 orange-eyed
b.
0 black-eyed:1 orange-eyed
c.
1 black-eyed:3 orange-eyed
d.
1 black-eyed:0 orange-eyed
e.
1 black-eyed:1 orange-eyed
 78.
Andalusian chickens with the genotype CBCB are black, those with the genotype CBCW are gray. What is the relationship between the CB and the CW alleles?
a.
CB is dominant to CW.
b.
CB is recessive to CW.
c.
CW is dominant to CB.
d.
The relationship is one of incomplete dominance.
e.
CB and CW are codominant.
 79.
Black eyes are dominant to orange eyes, and green skin is dominant to white skin. Sam, a MendAlien with black eyes and green skin, has a parent with orange eyes and white skin. Carole is MendAlien with orange eyes and white skin. If Sam and Carole were to mate, the predicted ratio of their offspring would be:
a.
1 black eyes, green skin: 1 black eyes, white skin: 1 orange eyes, green skin: 1 orange eyes, white skin
b.
3 black eyes, green skin: 3 black eyes, white skin: 9 orange eyes, green skin: 1 orange eyes, white skin
c.
1 black eyes, green skin: 3 black eyes, white skin: 3 orange eyes, green skin: 9 orange eyes, white skin
d.
9 black eyes, green skin: 3 black eyes, white skin: 3 orange eyes, green skin: 1 orange eyes, white skin
e.
There is insufficient information to determine Sam’s genotype.
 80.
All the offspring of a cross between a red-flowered plant and a white-flowered plant have pink flowers. This means that the allele for red flowers is ____ to the allele for white flowers.
a.
dominant
b.
codominant
c.
pleiotropic
d.
incompletely dominant
e.
recessive
 81.
You conduct a dihybrid cross and then testcross the F1 generation. A ____ ratio would make you suspect that the genes are linked.
a.
3:1
b.
1:2:1
c.
1:1:1:1
d.
7:7:1:1
e.
9:3:3:1
 82.
The recombination frequency between gene A and gene B is 8.4%, the recombination frequency between gene A and gene C is 6.8%, and the recombination frequency between gene B and gene C is 15.2%. Which is the correct arrangement of these genes?
a.
ABC
b.
ACB
c.
BCA
d.
CAB
e.
CBA
 83.
Hypophosphatemia (vitamin D-resistant rickets) is inherited as a X-linked dominant disorder. An unaffected woman mates with a male with hypophosphatemia. What is the expected phenotypic ratio of their offspring?
a.
1 normal daughter: 1 daughter with hypophosphatemia
b.
1 normal daughter : 1 son with hypophosphatemia
c.
1 daughter with hypophosphatemia : 1 normal son
d.
2 normal daughters : 1 normal son : 1 son with hypophosphatemia
e.
3 normal daughters : 1 son with hypophosphatemia
 84.
The sex chromosome complements of both normal human and normal MendAlien males is
a.
XO.
b.
XX.
c.
XY.
d.
YY.
e.
YO.
 85.
Mutant tetraploid plants
a.
are usually sickly.
b.
are able to interbreed with their parents.
c.
have an odd number of chromosomes.
d.
are unable to breed with a diploid plant.
e.
are unable to self-fertilize.
 86.
Chromosomes and genes share all of the following characteristics except that
a.
they are both present in pairs in all diploid cells.
b.
they both undergo segregation during meiosis.
c.
their copy numbers in the cell decrease after meiosis, and increase during fertilization.
d.
they are both copied during the S phase of the cell cycle.
e.
they both pair up with their homologues during prophase of mitosis.
 87.
The improvement of microscopy techniques in the late 1800s set the stage for the emergence of modern genetics because
a.
it revealed new and unanticipated features of Mendel’s pea plant varieties.
b.
it allowed biologists to study meiosis and mitosis, revealing the parallels between the behaviors of genes and chromosomes.
c.
it allowed scientists to see the DNA present within chromosomes.
d.
it led to the discovery of mitochondria.
e.
All of the above are true.
 88.
When Thomas Hunt Morgan crossed his red-eyed F1 generation flies to each other, the F2 generation included both red- and white-eyed flies. Remarkably, all the white-eyed flies were male. What was the explanation for this result?
a.
The involved gene was on the X chromosome.
b.
The involved gene was on the Y chromosome.
c.
The involved gene was on an autosome.
d.
Other male-specific factors influence eye color in flies.
e.
Other female-specific factors influence eye color in flies.
 89.
Which of the following statements is (are) true?
a.
The closer two genes are on a chromosome, the higher the probability that a crossover will occur between them.
b.
The observed frequency of recombination of two genes that are far apart from each other has a maximum value of 50%.
c.
Two of the traits that Mendel studied-seed color and flower color-are linked on the same chromosome.
d.
Only B and C are correct.
e.
A, B, and C are correct.
 90.
How would one explain a testcross involving F1 dihybrid flies in which more parental-type offspring than recombinant-type offspring are produced?
a.
The two genes are linked.
b.
The two genes are unlinked.
c.
Recombination did not occur in the cell during meiosis.
d.
The testcross was improperly performed.
e.
Both of the characters are controlled by more than one gene.
 91.
New combinations of linked genes are due to which of the following?
a.
nondisjunction
b.
crossing over
c.
independent assortment
d.
mixing of sperm and egg
e.
both A and C
 92.
What does a frequency of recombination of 50% indicate?
a.
The two genes likely are located on different chromosomes.
b.
All of the offspring have combinations of traits that match one of the two parents.
c.
The genes are located on sex chromosomes.
d.
Abnormal meiosis has occurred.
e.
Independent assortment is hindered.

The following questions refer to the data and figures below.

CROSS I. Purebred lines of wild-type fruit flies (gray body and normal wings) are mated to flies with black bodies and vestigial wings.

nar010-1.jpg

F1 offspring all have a normal phenotype.

nar010-2.jpg

CROSS II. F1 flies are crossed with flies recessive for both traits (a testcross).

Resulting OffspringNormalPercentage
Gray body; normal wings
575
25.1
Black body; vestigial wings
571
24.9
Black body; normal wings
577
25.2
Gray body; vestigial wings
568
24.8

KEY:
A. CROSS I results give evidence supporting the statement.
B. CROSS I results give evidence against the statement.
C. CROSS II results give evidence supporting the statement.
D. CROSS II results give evidence against the statement.
E. Neither CROSS I nor CROSS II results support the statement.

 93.
Vestigial wings are a recessive trait.
a.
A
b.
B
c.
C
d.
D
e.
E
 94.
The genes for body color and wing shape are linked.
a.
A
b.
B
c.
C
d.
D
e.
E
 95.
An F1 cross should produce flies that will fall into a Mendelian 9:3:3:1 ratio.
a.
A
b.
B
c.
C
d.
D
e.
E
 96.
There are 25 centimorgans (map units) between the genes for body color and wing shape.
a.
A
b.
B
c.
C
d.
D
e.
E
 97.
A 0.1% frequency of recombination is observed
a.
only in sex chromosomes.
b.
only on genetic maps of viral chromosomes.
c.
on unlinked chromosomes.
d.
in any two genes on different chromosomes.
e.
in genes located very close to one another on the same chromosome.
 98.

The following is a map of four genes on a chromosome:

mc098-1.jpg
Between which two genes would you expect the highest frequency of recombination?

a.
A and W
b.
W and E
c.
E and G
d.
A and E
e.
A and G
 99.
The reason that linked genes are inherited together is that
a.
they are located on the same chromosome.
b.
the number of genes in a cell is greater than the number of chromosomes.
c.
chromosomes are unbreakable.
d.
alleles are paired.
e.
genes align that way during metaphase I.
 100.
What is the mechanism for the production of genetic recombinants?
a.
X inactivation
b.
methylation of cytosine
c.
crossing over and independent assortment
d.
nondisjunction
e.
deletions and duplications during meiosis
 101.
There is good evidence for linkage when
a.
two genes occur together in the same gamete.
b.
a gene is associated with a specific phenotype.
c.
two genes work together to control a specific characteristic.
d.
genes do not segregate independently during meiosis.
e.
two characteristics are caused by a single gene.

Refer to the figure below to answer the following questions.

nar011-1.jpg

 102.
In a series of mapping experiments, the recombination frequencies for four different linked genes of Drosophila were determined as shown in the figure. What is the order of these genes on a chromosome map?
a.
rb-cn-vg-b
b.
vg-b-rb-cn
c.
cn-rb-b-vg
d.
b-rb-cn-vg
e.
vg-cn-b-rb
 103.
Which of the following two genes are closest on a genetic map of Drosophila?
a.
b and vg
b.
vg and cn
c.
rb and cn
d.
cn and b
e.
b and rb
X, Y, and Z are three genes in Drosophila. The recombination frequencies for two of the three genes are shown below.

nar012-1.jpg
 104.
Genes X and Y could be
a.
located on different chromosomes.
b.
located very near to each other on the same chromosome.
c.
located far from each other on the same chromosome.
d.
both A and B
e.
both A and C
 105.
If the recombination frequency for Y and Z was found to be 50%, this would mean that
a.
genes X and Y are on the same chromosome.
b.
genes X and Y are on different chromosomes.
c.
genes Y and Z are on different chromosomes.
d.
both A and C.
e.
both B and C
 106.
Which of the following is true regarding linkage maps? They
a.
always have a total of 100 map units.
b.
can be used to pinpoint the precise physical position of a gene on a chromosome.
c.
are a genetic map based on recombination frequencies.
d.
require preparation of karyotypes.
e.
reflect the frequency of crossing over between X and Y chromosomes.
 107.
The frequency of crossing over between any two linked genes is
a.
higher if they are recessive.
b.
different between males and females.
c.
determined by their relative dominance.
d.
the same as if they were not linked.
e.
proportional to the distance between them.
 108.
Sturtevant provided genetic evidence for the existence of four pairs of chromosomes in Drosophila by showing that
a.
there are four major functional classes of genes in Drosophila.
b.
Drosophila genes cluster into four distinct groups of linked genes.
c.
the overall number of genes in Drosophila is a multiple of four.
d.
the entire Drosophila genome has approximately 400 map units.
e.
Drosophila genes have, on average, four different alleles.
 109.
Map units on a linkage map cannot be relied upon to calculate physical distances on a chromosome because
a.
the frequency of crossing over varies along the length of the chromosome.
b.
the relationship between recombination frequency and map units is different in every individual.
c.
physical distances between genes change during the course of the cell cycle.
d.
the gene order on the chromosomes is slightly different in every individual.
e.
all of the above
 110.
A map of a chromosome that includes the positions of genes relative to visible chromosomal features, such as stained bands, is called a
a.
linkage map.
b.
physical map.
c.
recombination map.
d.
cytogenetic map.
e.
banded map.
 111.
Males are more often affected by sex-linked traits than females because
a.
males are hemizygous for the X chromosome.
b.
male hormones such as testosterone often exacerbate the effects of mutations on the X chromosome.
c.
female hormones such as estrogen often compensate for the effects of mutations on the X.
d.
X chromosomes in males generally have more mutations than X chromosomes in females.
e.
mutations on the Y chromosome often exacerbate the effects of X-linked mutations.
 112.
SRY is
a.
a gene present on the Y chromosome that triggers male development.
b.
a gene present on the X chromosome that triggers female development.
c.
an autosomal gene that is required for the expression of genes on the Y chromosome.
d.
an autosomal gene that is required for the expression of genes on the X chromosome.
e.
required for development, and males or females lacking the gene do not survive past early childhood.
 113.
In cats, black fur color is caused by an X-linked allele; the other allele at this locus causes orange color. The heterozygote is tortoiseshell. What kinds of offspring would you expect from the cross of a black female and an orange male?
a.
tortoiseshell female; tortoiseshell male
b.
black female; orange male
c.
orange female; orange male
d.
tortoiseshell female; black male
e.
orange female; black male
 114.
Red-green color blindness is a sex-linked recessive trait in humans. Two people with normal color vision have a color-blind son. What are the genotypes of the parents?
a.
XcXc and XcY
b.
XcXc and XCY
c.
XCXC and XcY
d.
XCXC and XCY
e.
XCXc and XCY
 115.
In the following list, which term is least related to the others?
a.
Duchenne muscular dystrophy
b.
autosome
c.
sex-linked genes
d.
color blindness
e.
hemophilia
 116.
Cinnabar eyes is a sex-linked recessive characteristic in fruit flies. If a female having cinnabar eyes is crossed with a wild-type male, what percentage of the F1 males will have cinnabar eyes?
a.
0%
b.
25%
c.
50%
d.
75%
e.
100%
 117.
Most calico cats are female because
a.
a male inherits only one of the two X-linked genes controlling hair color.
b.
the males die during embryonic development.
c.
the Y chromosome has a gene blocking orange coloration.
d.
only females can have Barr bodies.
e.
multiple crossovers on the Y chromosome prevent orange pigment production.
 118.
A recessive allele on the X chromosome is responsible for red-green color blindness in humans. A woman with normal vision whose father is color-blind marries a color-blind male. What is the probability that a son of this couple will be color-blind?
a.
0
b.
1/4
c.
1/2
d.
3/4
e.
1
 119.
In birds, sex is determined by a ZW chromosome scheme. Males are ZZ and females are ZW. A lethal recessive allele that causes death of the embryo is sometimes present on the Z chromosome in pigeons. What would be the sex ratio in the offspring of a cross between a male that is heterozygous for the lethal allele and a normal female?
a.
2:1 male to female
b.
1:2 male to female
c.
1:1 male to female
d.
4:3 male to female
e.
3:1 male to female
 120.
A man who carries an X-linked allele will pass it on to
a.
all of his daughters.
b.
half of his daughters.
c.
all of his sons.
d.
half of his sons.
e.
all of his children.

Refer to the information below to answer the following questions.

An achondroplastic male dwarf with normal vision marries a color-blind woman of normal height. The man’s father was six-feet tall, and both the woman’s parents were of average height. Achondroplastic dwarfism is autosomal dominant, and red-green color blindness is X-linked recessive.

 121.
How many of their daughters might be expected to be color-blind dwarfs?
a.
all
b.
none
c.
half
d.
one out of four
e.
three out of four
 122.
How many of their sons would be color-blind and of normal height?
a.
all
b.
none
c.
half
d.
one out of four
e.
three out of four
 123.
They have a daughter who is a dwarf with normal color vision. What is the probability that she is heterozygous for both genes?
a.
0
b.
0.25
c.
0.50
d.
0.75
e.
1.00
 124.
Male calico cats could be the result of
a.
sex-linked inheritance.
b.
nondisjunction, leading to the male calico having two X chromosomes.
c.
incomplete dominance of multiple alleles.
d.
recessive alleles retaining their fundamental natures even when expressed.
e.
a reciprocal translocation.
 125.
A Barr body is normally found in the nucleus of which kind of human cell?
a.
unfertilized egg cells only
b.
sperm cells only
c.
somatic cells of a female only
d.
somatic cells of a male only
e.
both male and female somatic cells
 126.
Which of these syndromes afflicts mostly males?
a.
Turner syndrome
b.
Down syndrome
c.
Duchenne muscular dystrophy
d.
cri du chat syndrome
e.
chronic myelogenous leukemia
 127.
If a human interphase nucleus of a person contains three Barr bodies, it can be assumed that the person
a.
has hemophilia.
b.
is a male.
c.
has four X chromosomes.
d.
has Turner syndrome.
e.
has Down syndrome.
 128.
If nondisjunction occurs in meiosis II during gametogenesis, what will be the result at the completion of meiosis?
a.
All the gametes will be diploid.
b.
Two gametes will be n + 1, and two will be n – 1.
c.
One gamete will be n + 1, one will be n – 1, and two will be n.
d.
There will be three extra gametes.
e.
Two of the four gametes will be haploid, and two will be diploid.
 129.

The figure below represents the stained nucleus from a cheek epithelial cell of an individual whose genotype would probably be

mc129-1.jpg

a.
XX.
b.
XY.
c.
XYY.
d.
XXX.
e.
XXY.
 130.
If a pair of homologous chromosomes fails to separate during anaphase of meiosis I, what will be the chromosome number of the four resulting gametes with respect to the normal haploid number (n)?
a.
n + 1; n + 1; n – 1; n – 1
b.
n + 1; n – 1; n; n
c.
n + 1; n – 1; n – 1; n – 1
d.
n + 1; n + 1; n; n
e.
n – 1; n – 1; n; n
 131.
A cell that has 2n + 1 chromosomes is
a.
trisomic.
b.
monosomic.
c.
aneuploid.
d.
polyploid.
e.
both A and C
 132.
If a chromosome lacks certain genes, what has most likely occurred?
a.
disjunction
b.
an inversion
c.
a deletion
d.
a translocation
e.
a nonduplication
 133.
One possible result of chromosomal breakage is for a fragment to join a nonhomologous chromosome. This is called a (an)
a.
deletion.
b.
disjunction.
c.
inversion.
d.
translocation.
e.
duplication.
 134.
In the following list, which term is least related to the others?
a.
trisomic
b.
monosomic
c.
aneuploid
d.
triploid
e.
nondisjunction
 135.
A nonreciprocal crossover causes which of the following products?
a.
deletion
b.
duplication
c.
nondisjunction
d.
A and B
e.
B and C
 136.
One possible result of chromosomal breakage can be that a fragment reattaches to the original chromosome in a reverse orientation. This is called
a.
disjunction.
b.
translocation.
c.
deletion.
d.
inversion.
e.
aneuploidy.
 137.
A human individual is phenotypically female, but her interphase somatic nuclei do not show the presence of Barr bodies. Which of the following statements concerning her is probably true?
a.
She has Klinefelter syndrome.
b.
She has an extra X chromosome.
c.
She has Turner syndrome.
d.
She has the normal number of sex chromosomes.
e.
She has two Y chromosomes.
 138.

The karyotype shown below is associated with which of the following genetic disorders?

mc138-1.jpg

a.
Turner syndrome
b.
Down syndrome
c.
Klinefelter syndrome
d.
hemophilia
e.
male-pattern baldness
 139.
In humans, male-pattern baldness is controlled by a gene that occurs in two allelic forms. Allele Hn determines nonbaldness, and allele Hb determines pattern baldness. In males, because of the presence of testosterone, allele Hb is dominant over Hn. If a man and woman both with genotype HnHb have a son, what is the chance that he will eventually be bald?
a.
0%
b.
25%
c.
33%
d.
50%
e.
75%
 140.
Of the following human trisomies, the one that generally has the most severe impact on the health of the individual is
a.
trisomy 21.
b.
Klinefelter syndrome (XXY).
c.
trisomy X.
d.
XYY.
e.
All of the above have equal impact.
 141.
What do all human males inherit from their mother?
a.
mitochondrial DNA
b.
an X chromosome
c.
the SRY gene
d.
A and B only
e.
A, B, and C
 142.
Which of the following statements is true regarding genomic imprinting?
a.
It explains cases in which the gender of the parent from whom an allele is inherited affects the expression of that allele.
b.
It is greatest in females because of the larger maternal contribution of cytoplasm.
c.
It may explain the transmission of Duchenne muscular dystrophy.
d.
It involves an irreversible alteration in the DNA sequence of imprinted genes.
e.
All of the above are correct.
 143.

The pedigree in the figure below shows the transmission of a trait in a particular family. Based on this pattern of transmission, the trait is most likely

mc143-1.jpg

a.
mitochondrial.
b.
autosomal recessive.
c.
sex-linked dominant.
d.
sex-linked recessive.
e.
autosomal dominant.
 144.
Which of the following statements about mitochondria is false?
a.
Because of the role of the mitochondria in producing cellular energy, mitochondrial diseases often affect the muscles and nervous system.
b.
Because mitochondria are present in the cytoplasm, mitochondrial diseases are transmitted maternally.
c.
Like nuclear genes, mitochondrial genes usually follow Mendelian patterns of inheritance.
d.
Mitochondria contain circular DNA molecules that code for proteins and RNAs.
e.
Many mitochondrial genes encode proteins that play roles in the electron transport chain and ATP synthesis.
 

Short Answer
Self-Quiz Questions
 145.
A man with hemophilia (a recessive, sex-linked condition. has a daughter of normal phenotype) She marries a man who is normal for the trait. What is the probability that a daughter of this mating will be a hemophiliac? That a son will be a hemophiliac? If the couple has four sons, what is the probability that all four will be born with hemophilia?

 146.
Pseudohypertrophic muscular dystrophy is an inherited disorder that causes gradual deterioration of the muscles. It is seen almost exclusively in boys born to apparently normal parents and usually results in death in the early teens. Is this disorder caused by a dominant or a recessive allele? Is its inheritance sex-linked or autosomal? How do you know? Explain why this disorder is almost never seen in girls.

 147.
Red-green color blindness is caused by a sex-linked recessive allele. A color-blind man marries a woman with normal vision whose father was color-blind. What is the probability that they will have a color-blind daughter? What is the probability that their first son will be color-blind? (Note: The two questions are worded a bit differently).

 148.
A wild-type fruit fly (heterozygous for gray body color and normal wings) is mated with a black fly with vestigial wings. The offspring have the following phenotypic distribution: wild type, 778; black-vestigial, 785; black-normal, 158; gray-vestigial, 162. What is the recombination frequency between these genes for body color and wing size?

 149.
In another cross, a wild-type fruit fly (heterozygous for gray body color and red eyes) is mated with a black fruit fly with purple eyes. The offspring are as follows: wild type, 721; black-purple, 751; gray-purple, 49; black-red, 45. What is the recombination frequency between these genes for body color and eye color? Using information from problem 4, what fruit flies (genotypes and phenotypes) would you mate to determine the sequence of the body-color, wing-size, and eye-color genes on the chromosome?

 150.
What pattern of inheritance would lead a geneticist to suspect that an inherited disorder of cell metabolism is due to a defective mitochondrial gene?

 151.
Women born with an extra X chromosome (XXX) are healthy and phenotypically indistinguishable from normal XX women. What is a likely explanation for this finding? How could you test this explanation?

 152.
Determine the sequence of genes along a chromosome based on the following recombination frequencies: A-B, 8 map units; A-C, 28 map units; A-D, 25 map units; B-C, 20 map units; B-D, 33 map units.

 153.
Assume that genes A and B are linked and are 50 map units apart. An animal heterozygous at both loci is crossed with one that is homozygous recessive at both loci. What percentage of the offspring will show phenotypes resulting from crossovers? If you did not know that genes A and B were linked, how would you interpret the results of this cross?

 154.
A space probe discovers a planet inhabited by creatures who reproduce with the same hereditary patterns seen in humans. Three phenotypic characters are height (T = tall, t = dwarf), head appendages (A = antennae, a = no antennae), and nose morphology (S = upturned snout, s = downturned snout). Since the creatures are not “intelligent,” Earth scientists are able to do some controlled breeding experiments, using various heterozygotes in testcrosses. For tall heterozygotes with antennae, the offspring are: tall-antennae, 46; dwarf-antennae, 7; dwarf-no antennae, 42; tall-no antennae, 5. For heterozygotes with antennae and an upturned snout, the offspring are: antennae-upturned snout, 47; antennae-downturned snout, 2; no antennae-downturned snout, 48; no antennae-upturned snout, 3. Calculate the recombination frequencies for both experiments.