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Mendelian Genetics
Jan 96,39
Use the following information to answer the next four questions
The human nose can distinguish up to 10,000 different odours. Airborne odour molecules are
trapped by ciliated olfactory receptors. These receptors initiate impulses to the olfactory bulbs,
which relay the signals to the temporal lobe of the cerebrum for interpretation.
Recently, researchers Linda Buck and Richard Axel of Columbia University have been able to
identify some odour-receptor proteins on the nasal cells of rats. These receptor proteins, embedded
in the cell membranes of ciliated neurons, are thought to bind with specific odour molecules,
initiating a nerve impulse. The researchers have also found the genes responsible for the
production of more than 100 types of odour-receptor proteins.
1.
Olfactory receptors are categorized as
A.
B.
C.
D.
mechanical receptors that are attached to the dendrites of neurons
mechanical receptors that are attached to the axons of neurons
chemoreceptors that form a synapse with the dendrites of neurons
chemoreceptors that form a synapse with the axons of neurons
Jan 96,40
2. Airborne molecules are thought to bind to odour-receptor proteins. These odour-receptor proteins
initiate the smelling process by
A.
B.
C.
D.
breaking down a neurotransmitter within the membrane of the neuron
making the membrane of the neuron more permeable to Na + ions
changing the threshold of the membrane of the neuron
causing the neuron to be polarized
Jan 96,41
3. The ability to smell some odours and not others is an inherited trait. This trait is an example of
A.
B.
C.
D.
incomplete dominance
crossing over
a phenotype
a genotype
Jan 96,42
4. If a gene responsible for the production of an odour-receptor protein underwent a mutation, one result
might be
A.
B.
C.
D.
a decrease in the ability to smell a specific odour
an increase in the ability to smell a variety of other odours
a decrease in protein synthesis within all olfactory neurons
an increase in threshold levels of stimulation for receptor neurons
Jan 96,43
Use the following information to answer the next two questions
Some researchers are investigating the hypothesis that feeding cow’s milk to children less than
nine months old could play a role in the development of type I (juvenile) diabetes mellitus in those
genetically prone to the disease. A child who has inherited certain genes synthesizes antibodies in
response to a particular fragment of a protein found in cow’s milk. These antibodies attack the
milk-protein fragment and some of the child’s cells. After a child reaches the age of nine months,
the immune system develops sufficiently to distinguish between the milk-protein fragment and the
child’s cells.
5.
If a child who is genetically prone to juvenile diabetes has consumed cow’s milk, it is likely that the
child’s blood contains antibodies that attack the
A.
B.
C.
D.
sensory cells of the hypothalamus
glycogen storage cells of the liver
secretory cells of the thyroid
islet cells of the pancreas
Jan 96,44
6. A correct inference is that the antibodies produced in response to cow’s milk are
A.
B.
C.
D.
nucleic acids containing a sequence of amino acids
nucleic acids containing a sequence of nucleotides
proteins containing a sequence of amino acids
proteins containing a sequence of nucleotides
Jan 96,45
Use the following information to answer the next three questions
The Romanovs, the Russian royal family, were shot dead during the Russian Revolution in 1918.
The family included Czar Nicholas II and his wife Alexandra; their daughters, Olga, Maria,
Tatiana, and Anastasia; and their son Alexis, who was a hemophiliac.
In 1992, scientists in Russia started work on the identification of skeletons thought to be the
remains of the Czar, his wife, and three of their daughters. The researchers crushed and partially
dissolved some bones to extract DNA.
To determine which skeletons belonged to the Romanovs, the researchers analyzed samples of
mitochondrial DNA. Mitochondrial DNA is passed on only from mother to child. Some parts of
mitochondrial DNA are very stable and mutate only once every 6,000 years. The researchers also
obtained a sample of mitochondrial DNA from Prince Philip of England. His maternal
grandmother was Alexandra’s sister.
7.
Which observation correctly accounts for the fact that mitochondrial DNA is passed on only from
mother to child?
A.
B.
C.
D.
eggs contain mitochondria and sperm do not
mitochondrial DNA is transcribed from a template on the XY chromosome set
mitochondrial DNA is transcribed from a template on the XX chromosome set
an egg contributes both its nucleus and its cytoplasm to a zygote, while a sperm
contributes only its nucleus
Jan 96,46
8. From which people would samples of mitochondrial DNA be nearly identical
A.
B.
C.
D.
Prince Philip, Czar Nicholas II, and Alexandra
Alexis, Prince Philip and Czar Nicholas II
Czar Nicholas II, Alexandra, and Alexis
Alexandra, Alexis, and Prince Philip
Jan 96,8
Numerical Response
1. Czar Nicholas II was not a hemophiliac. His wife, Alexandra, was a carrier for hemophilia. What was
the probability that any of their daughters was a carrier for hemophilia?
(Record your answer as a value from 0 to 1, rounded to two significant digits, in the numericalresponse section of the answer sheet.)
Answer: ___________
Jan 96,47
Use the following information to answer the next two questions
William Jia, a researcher at the University of British Collumbia, thinks a genetically altered
Herpes virus may one day be used to destroy gliomas, a type of brain cancer that attacks the
frontal lobes of the cerebrum.
Jia managed to genetically engineer a Herpes virus that replicates only in rapidly dividing cancer
cells and thus destroys only those cells. Jia handicaps the Herpes virus by deleting the genes that
code for an enzyme needed for viral replication. This prevents the virus from replicating in
normal, non-dividing brain cells, but enable it to continue replicating in cancerous brain cells. As
the virus replicates, the rapidly dividing cancer cells are killed.
9.
The deletion of genes from the Herpes virus likely involves the use of
A.
B.
C.
D.
biopsies and chorionic villus sampling
karyotyping and pedigree analysis
amniocentesis and chromosome mapping
restriction enzymes and ligases
Jan 96,48
10. After the destruction of the cells in a brain tumour, the altered Herpes virus will likely
A.
B.
C.
D.
change into a form capable of producing the enzymes necessary for the reproductive
cycle of the virus
spread to other areas of the body and attack other types of cancer cells that may be
present
enter a dormant stage because enzymes required for replication are absent
spread to a nearby region of the brain and begin normal replication
Jun 96,27
Use the following information to answer the next question.
When conducting his research into the genetics of pea plants, Gregor Mendel crossed truebreeding, tall pea plants with true-breeding, short pea plants.
The offspring of this P1 generation, the F1 generation, were interbred and produced the F2
generation. The height of each plant in the F1 and the F2 generations was recorded.
11. In pea plants, if tall is dominant over short, which graph best represents the expected distribution of
offspring in the F2 generation?
Jun 96,2
Use the following information to answer the next question
Mendel also found that in peas, yellow seeds (Y) are dominant over green seeds (y), and round
seed shape (R) is dominant over wrinkled seed shape (r). the alleles for these traits are located on
independent chromosomes.
Numerical Response
2. When a YyRr parent and a YYrr parent are crossed, what percentage of their offspring are expected to
have yellow seed-coat colour and wrinkled seed shape?
(Record your answer in the numerical-response section of the answer sheet.)
Answer: _____________ %
Jun 96,28
Use the following information to answer the next question
In his work with the fruit fly (Drosophila), Thomas Morgan bred normal red-eyed female flies
with white-eyed males. He bred enough flies to have acceptable statistical significance and
reliability. The F1 generation were all red-eyed. All the females of the F2 generation had red eyes,
but one-half of the males of the F2 generation had white eyes.
12. A conclusion that may be drawn from Morgan’s work is that the gene for white eyes is located on the
A.
B.
C.
D.
Y chromosome and is recessive
Y chromosome and is dominant
X chromosome and is recessive
X chromosome and is dominant
Jun 96,29
Use the following information to answer the next question
In Drosophila, the genes for eye colour (pr), wing shape (vg), and body colour (eb) are all found
on the same chromosome. The following crossover frequencies for these genes were determined
by experimentation.
Genes
pr and vg
pr and eb
vg and eb
Crossover Frequency
12.5%
6.0%
18.5%
13. Which is the correct sequence of the genes pr, vg, and eb on the chromosome?
A.
B.
C.
D.
pr–vg–eb
vg–eb–pr
eb–pr–vg
eb–vg–pr
Jun 96,31
Use the following information to answer the next two questions
A Pedigree Showing the Incidence of Opalescent Dentin in a Family
This pedigree shows the incidence of opalescent dentin in a family. People with this trait have insufficient
enamel on their teeth. The dentin inside the teeth shows through, giving the teeth a mottled, pearl-like
appearance.
14. The pedigree indicates that the trait for opalescent dentin is controlled by a gene that is
A.
B.
C.
D.
X - linked
recessive
dominant
codominant
Jun 96,3
Numerical Response
3. If individual III 1 and individual III 2 had another child, what is the probability that this child would
be a female with opalescent dentin?
(Record your answer as a value from 0 to 1, rounded to two decimal places, in the numericalresponse section of the answer sheet.)
Answer: __________
Jun 96,32
Use the following information to answer the next two questions
15. Which row provides correct information about the genotypes for individuals I 1, II 1, and II 4?
Row
A
B
C
D
Individual I 1
IAi
IAIA
IAi
IAIA
Individual II 1
IAIA or IAi
IAIA or IAi
IAIA or IAi or IAIB or IBi
IAIA or IAi or IAIB or IBi
Individual II 4
ii
IAi
IAi
ii
Jun 96,33
16. Which row correctly shows the probability of individuals II 2 and III 3 having type O blood?
Row
A
B
C
D
Probability of individual
II 2 having type O blood
0.00
0.00
0.25
0.25
Probability of individual
III 3 having type O blood
0.00
0.25
0.00
0.25
Jun 96,34
17. Protanopia, or red colour blindness, is a recessive X-linked characteristic. A woman with normal
colour vision who carries the allele for red colour blindness and a man who is a red colour-blind are
expecting a child. Which statement about the child is likely correct?
A.
B.
C.
D.
If the child is a boy, there is a 100% chance that he will be colour blind
If the child is a girl, thre is a 100% chance that she will be colour-blind
There is a 25% chance that the child will be colour-blind
There is a 50% chance that the child will be colour-blind
Jun 96,35
Use the following information to answer the next question
The ability to taste the chemical phenylthiocarbamide (PTC) is controlled by the dominant allele T.
Individuals who have the allele T find that PTC tastes bitter. Homozygous recessive individuals find
that PTC has no taste. In a sample of 320 people, 218 could taste PTC and 102 could not taste PTC.
18. Which row provides the probable percentage of this population with each genotype indicated?
Row
A
B
C
D
TT
15 %
44 %
19 %
23 %
Genotype
Tt
29 %
37 %
49 %
45 %
tt
56 %
19 %
32 %
32 %
Jun 96,36
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A plant breeder was trying to produce a true-breeding variety of radishes with an oval shape.
However, each time plants with oval radishes were self-fertilized, only half of the offspring were
oval. The rest of the plants produced either round or long radishes.
19. What would be the expected phenotype ratio if plants with round radishes were crossed with plants
with oval radishes?
A.
B.
C.
D.
All round
All oval
1 round : 1 oval
1 round : 1 long
Jun 96,4
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In guinea pigs, black coat (B) is dominant over white coat (b) an straight hair (S) is dominant over curly hair (s). Each
pair of alleles assorts independently.
Two guinea pigs were crossed. The gametes produced by the parents are shown in the Punnett Square.
♂
BS
♀
Bs
bS
bs
Note: ♂
male
♀ female
Bs
bs
Numerical Response
4. What was the phenotypic ratio of the offspring ( including all eight possible genotypes) produced by
the parents?
(Record your four-digit answer in the numerical-response section of the answer sheet.)
Answer:
__________
Black-straight
__________
black-curly
__________
white-straight
__________
white-curly
Jun 96,5
Use the following information to answer the next question
In some cattle, coat colour and pattern is controlled by an autosomal gene locus that may contain
one of four alleles. The dominance hierarchy for these alleles is
S (Dutch belt)  sh (Hereford)  sc (Solid)  s (Holstein)
Numerical Response
5. A Dutch belt bull carrying the allele for Hereford was crossed with a pure-breeding Holstein cow.
What is the probability that the offspring would be a Hereford?
(Record your answer as a value from 0 to 1, rounded to two decimal places, in the numericalresponse section of the answer sheet.)
Answer: ___________
Jun 96,6
Numerical Response
6. Snapdragons may be tall or short, and they may be red, white, or pink. The allele for short is recessive
and the alleles for colour display incomplete dominance. A plant, heterozygous for both traits, was
self-pollinated and produced 200 offspring. How many of these 200 offspring would be expected to be
both tall and pink/
(Record your answer in the numerical-response section of the answer sheet.)
Answer: ______________ plants
Jun 97,36
Use the following information to answer the next three questions
Hypohidrotic ectodermal dysplasia is a rare genetic disease that cause problems with most embryonic
ectoderm development, but does not usually affect the nervous system. Individuals are born apparently
healthy but lack sweat glands.
Note: Carriers have not been identified and all marriage partners are known to be unrelated.
From Discover
20. The inheritance pattern of this disease is most likely
A.
B.
C.
D.
X-linked recessive
X-linked dominant
autosomal recessive
autosomal dominant
Jun 97,37
21. Which structure or organ would not develop properly in a person with hypohidrotic ectodermal
dysplasia?
A.
B.
C.
D.
Muscle
Heart
Liver
Skin
Jun 97,38
22. Which individuals from the pedigree are definitely carriers of the disease?
A.
B.
C.
D.
I-1 and II-5
I-2 and II-1
III-1 and III-4
III-2 and III-4
Jun 97,39
Use the following information to answer the next two questions
23. The row that correctly shows the expected ratio of offspring phenotypes of a cross between two dogs
heterozygous for height and tail curl is
Row
A.
B.
C.
D.
Normal Height
Curled Tail
0
1
1
9
Normal Height
Uncurled Tail
1
1
3
3
Dwarf
Curled Tail
1
1
3
3
Dwarf
Uncurled Tail
0
1
9
1
Jun 97,40
24. If a dog that is heterozygous for height ad tail curl is crossed with a dog that is recessive for both traits,
the four possible genotypes for the offspring are
A.
B.
C.
D.
HhCc, Hhcc, hhCc, hhcc
HhCc, HhCC, hhCc, hhcc
HhCc, Hhcc, hhCC, hhcc
HhCc, HHCc, hhCc, hhcc
Jun 97,41
Use the following information to answer the next four questions
Feather colour for Andalusian fowl is governed by incomplete dominance of a pair of alleles.
Fowl may have black, whit, or blue feathers. Blue-feathered birds are heterozygotes. In a
randomly mating population of 400 fowl, there were 49 white-feathered birds.
25. The allele frequencies p (black) and q (white), respectively, are
A.
B.
C.
D.
p = 0.3 and q =0.7
p = 0.7 and q = 0.3
p = 0.35 and q = 0.65
p = 0.65 and q = 0.35
Jun 97,4
Numerical Response
7. What is the frequency of the heterozygous genotype in this population of Andalusian fowl?
(Record your answer as a value from 0 to 1, rounded to two decimal places, in the numericalresponse section of the answer sheet.)
Answer: ___________
Jun 97,42
25. When a black-feathered hen is mated with a white-feathered rooster, what feather colour will the
offspring have?
A.
B.
C.
D.
All will have blue feathers
All will have black feathers
Some will have black and some will have white feathers
Some will black, some will have white, and some will have blue feathers
Jun 97,5
Numerical Response
8. In a cross between a blue-feathered rooster and a white-feathered hen, what percentage of the offspring
are expected to be white-feathered?
(Record your answer as a whole number percentage in the numerical-response section of the answer
sheet.)
Answer: _____________%
Jun 97,43
Use the following information to answer the next question
26. How many people in Generation III are homozygous for blood type?
A.
B.
C.
D.
1
2
3
4
Jun 97,6
Numerical Response
Use the following information to answer the next question
The characteristics plexus wings, scabrous eyes, speckled body, and brown eyes are influenced
by genes found on chromosome 2 of the common fruit fly, Drosophila melanogaster.
Crossover frequencies between the genes are given in the following table.
Characteristic (Identification number )
Speckled body (3) and scabrous eye (2)
Brown eyes (4) and plexus wings (1)
Plexus wings (1) and scabrous eyes (2)
Speckled body (3) and brown eyes (4)
Crossover frequency
41 %
4%
34 %
3%
-from An Introduction to Genetic Analysis
9.
Predict the order in which the genes occur on chromosome 2. Use the numbers following the
characteristics to code your answer
(Record your four-digit answer in the numerical-response section of the answer sheet.)
(There are two possible ways of recording the answer; either will be acceptable.)
Answer: ____________
Jan 98,30
Use the following information to answer the next three questions
27. Phenotypically, generation III offspring will be
A.
B.
C.
female, and each individual will be genotypically different
female, and each individual will be genotypically identical
50% male and 50% female, and will be genotypically different
D.
50% male and 50% female, and will be gentotypically identical
Jan 98,31
28. If a new disease-causing organism from which lizards have no protection reaches an island where a
uniform population of lizards reproduce only asexually, a likely outcome is that
A.
B.
C.
D.
rapid extinction of the lizard population will occur
the lizard population will begin to reproduce sexually
gene frequencies will change in the lizard population’s gene pool
unique individuals in the lizard population will not find a mating partner
Jan 98,2
Numerical Response
10. From the list below, identify the processes that correspond to A, B, C and D in the diagram on the
previous page
Normal Processes
1
Fertilization
2
Meiosis
3
Mitosis
(Record your four-digit answer in the numerical-response section of the answer sheet.)
Answer:
Process
______
A
______
B
______
C
______
D
Jan 98, 32
Use the following information to answer the next question
Many adult newts and salamanders have a remarkable ability to regenerate amputated limbs.
After amputation of a foot, a newt will regenerate the lost foot. However, if a newt has its foot
amputated and receives a particular dosage of vitamin A, the animal grows back a whole new
forelimb, not just the foot!
-from Pietsch
29. A logical interpretation that can be drawn from this information is that vitamin A may play a major
role in growth by
A.
B.
C.
D.
increasing the amount of mitosis
increasing the amount of meiosis
decreasing the amount of mitosis
decreasing the amount of meiosis
Jan 98,33
Use the following information to answer the next three questions
Nondisjunction in meiosis disrupts the chromosome number in the gametes that are produced.
Nondisjunction can occur in either the first or second division of meiosis and results in various
genetic disorders
30. One homologous pair of chromosomes in a human spermatogonium undergoes nondisjunction during
the first division of meiosis. After meiosis is completed, what number of chromosomes will the four
newly produced cells contain?
A.
B.
C.
D.
All four cells will have 23 chromosomes
Two cells will have 22 chromosomes, and two cells will have 24 chromosomes
One cell will have 22 chromosomes, and three cells will have 24 chromosomes
One cell will have 24 chromosomes, and three cells will have 22 chromosomes
Jan 98,34
31. Which expression of chromosome content represents somatic cells in people with trisomy disorders
such as Down syndrome?
A.
B.
C.
D.
n–1
n+1
2n – 1
2n + 1
Jan 98,35
Use the following additional information to answer the next question
Polyploids are organisms with three or more complete sets of chromosomes. If a diploid
organism is 2n, then a triploid is 3n, a tetraploid is 4n, and so on.
All major groups of seed plants have some polyploid members. Plant polyploids are larger
than plants with 2n chromosome number. Plant polyploids with even chromosome numbers
(e.g., 4n) can usually produce pollen and seeds. Plant polyploids with odd chromosome
numbers (e.g., 5n) are nearly always sterile.
32. If a diploid plant and a tetraploid plant, each capable of normal meiosis, were crossed, the chromosome
number in their offspring would be
A.
B.
C.
D.
2n
3n
4n
6n
Jan 98,36
Use the following information to answer the next question
33. This diagram of a cellular structure shows
A.
B.
C.
D.
one crossed-over pair of chromosomes
two homologous chromosomes
one tetrad of chromatids
two sister chromatids
Jan 98,37
Use the following information to answer the next question
The San Diego Zoo is preserving DNA from hundreds of species by freezing cell samples in
its Centre for the Reproduction of Endangered Species. The cell specimens are often from
individuals in wild populations that are chosen for distinctive characteristics.
-from Vedantam
34. To obtain all the representative DNA of an organism, it would be necessary to collect only
A.
B.
C.
D.
an egg
a sperm
a body cell
a cell from each type of body tissue
Jan 98,3
Numerical Response
11. Of the nitrogen-based molecules present in the DNA of sea urchins, 17.5% are cytosine molecules.
Calculate the percentage composition of thymine in sea urchin DNA
(Record your answer as a percentage rounded to one decimal place in the numerical-response
section of the answer sheet.)
Answer: _____________%
Jan 98,38
Use the following information to answer the next three questions
Gregor Mendel examined the inheritance of two traits in pea plants: seed coat texture and
colour. Seed coat texture can be represented as S-smooth and s-wrinkled, and seed coat
colour can be represented as Y-yellow and y-green. SSYY plants were crossed with ssyy
plants to yield F1 pea seeds that were all smooth and all yellow. By crossing plants grown
from these F1 seeds, Mandel obtained four different phenotypes of F2 seeds:
 smooth and green seeds
 wrinkled and green seeds
 smooth and yellow seeds
 wrinkled and yellow seeds
35. If the traits for seed coat texture and seed coat colour had been located close together on lthe same
chromosome, Mendel might not have conceptualized
A.
B.
C.
D.
gene pairs
dominance
the Law of Segregation
the Law of Independent Assortment
Jan 98,4
Numerical Response
12. The F2 seed phenotype ration that Mendel obtained upon crossing two heterozygous smooth and
yellow F1 individuals would have been ____________.
(Record your four-digit answer in the numerical-response section of the answer sheet.)
Answer:
___________
smooth and
green
: ____________
wrinkled and
green
: __________
smooth and
yellow
:
___________
wrinkled and
yellow
Jan 98,39
Use the following information to answer the next question
Mendel selected two varieties of pea plants from seeds he had grown. One variety of peas
came from a field planted with smooth, yellow seeds. Another variety of peas came from a
field planted with wrinkled, green seeds. These two varieties of peas were crossed to produce
255 plants with smooth and green seeds
268 plants with wrinkled and green seeds
237 plants with smooth and yellow seeds
240 plants with wrinkled and yellow seeds
From the phenotype ratio of the offspring, Mendel deduced that the smooth and yellow parents
had the genotype YySs
36. This type of cross is referred to as a
A.
B.
C.
D.
test cross
monohybrid cross
homozygous cross
heterozygous cross
Jan 98,40
Use the following information to answer the next two questions
In Labrador retriever dogs, two alleles, B and b, determines whether the coat colour will be
black (B) and brown (b). Black coat colour is dominant. A second pair of alleles, E and e, are
on a separate chromosome from B and b. the homozygous recessive condition, ee, prevents the
expression of either allele B or b and produces a dog with a yellow-coloured coat. Some
examples of genotypes and phenotypes for Labrador retrievers are shown below.
Genotype
BBEe
bbEe
Bbee
Phenotype
black
brown
yellow
-from Davol
37. What is the probability of obtaining a black puppy from the following cross?
BbEe x BbEE
A.
B.
C.
D.
9/16
3/16
3/4
1/4
Jan 98,5
Numerical Response
13. Two dogs, each with the genotype BbEe, were crossed. What is the percentage probability that their
offspring would have yellow coat colour?
(Record your answer as a whole number percentage in the numerical-response section of the answer
sheet.)
Answer: __________
Jan 98,41
Use the following information to answer the next question
A recessive allele causes Drosophila to have white eyes instead of wild-type eyes. This eye
colour gene is known to be X-linked. In a cross between homozygous wild-type females
and white-eyed males, all F1 progeny have wild-type eyes.
38. What ratio of wild-type to white-eyed progeny can be expected in each sex if F1 females are crossed to
males of the same genotype as their father?
A.
B.
C.
D.
Males – 1:0, females – 1:0
Males – 1:1, females – 1:0
Males – 0:1, females – 1:1
Males – 1:1, females – 1:1
Jan 98,42
Use the following information to answer the next question
Crossover Frequencies for Some Genes on Drosophila Chromosome One
Genes
White eyes (w) and Facet eyes (f)
White eyes (w) and Echinus eyes (e)
White eyes (w) and Ruby eye (r)
Facet eyes (f) and Echinus eyes (e)
Facet eyes (f) and Ruby eyes (r)
39. The crossover frequency between genes e and r is
A.
B.
C.
D.
3.5 %
2.0 %
1.5 %
0.5 %
Crossover Frequency
1.5 %
4.0 %
6.0 %
2.5 %
4.5 %
Jan 98,43
Use the following information to answer the next two questions
40. During meiosis, which pair of genes have the best chance of being transferred together to a new cell?
A.
B.
C.
D.
Black body and purple eyes
Purple eyes and speck body
Dumpy wings and purple eyes
Dumpy wings and speck body
Jan 98,44
41. To determine whether this is an X chromosome or an autosome, a reasearcher would have to determine
whether these traits are
A.
B.
C.
D.
recessive
dominant
passed from male parents to their male offspring
passed from female parents to their male offspring
Jan 98,45
Use the following information to answer the next two questions
42. Which individual is a known homozygote for blood type?
A.
B.
C.
D.
I–1
I–2
II – 2
II – 3
Jan 98,46
43. Which of the following rows correctly identifies the genotypes of individuals III – 2 and III – 3?
Row
A.
B.
C.
D.
Individual III – 2
IBi or IBIB
IBI
IBi or IBIB
IBi
Individual III – 3
IAi
IAi or IAIA
IAi or IAIA
IAi
Jan 98,6
Numerical Response
Use the following information to answer the next two questions
In sheep, white wool is a dominant trait and black wool is a recessive trait. In a herd of 500
sheep, 20 sheep have black wool.
14.
In this herd, what is the frequency of the heterozygous genotype?
(Record you answer as a value from 0 to 1, rounded to two decimal places, in the numericalresponse section of the answer sheet.)
Answer: __________
Jan 98,7
Numerical Response
15. If two heterozygous sheep mated, what would be the probability of them having a white lamb?
(Record your answer as a value from 0 to 1, rounded to two decimal places, in the numericalresponse section of the answer sheet.)
Answer: __________
Jun 98,27
Use the following information to answer the next two questions
In cattle, hornless or polled (P) is dominant over the horned (p) condition. This is an autosomal
trait. The semen of a polled bull is used to artificially inseminate three cows. Cow 1 (horned)
produces a horned calf, cow 2 (polled) produces a horned calf, and cow 3 (polled) produces a
polled calf
44. Which of the cattle must have a heterozygous genotype for this trait?
A.
B.
C.
D.
Cow 2 and calf 2
Cow 3 and calf 3
The polled bull and cow 1
The polled bull and cow 2
Jun 98,28
45. Which of the above cattle could have two possible genotype?
A.
B.
C.
D.
Cow 1
Cow 2
Cow 3
The polled bull
Jun 98,29
Use the following information to answer the next two questions
Four babies were born in a hospital on the same day. Due to a mix-up at the hospital, there was
some confusion as to the identity of the babies.
Parents 1
Parents 2
Parents 3
Parents 4
Mother
Type A
Type AB
Type AB
Type O
Father
Type O
Type B
Type O
Type B
46. Which of the above parents could have a baby with blood type O?
A.
B.
C.
D.
Parents 1 and 3
Parents 1 and 4
Parents 2 and 3
Parents 2 and 4
Jun 98,30
47. The predicted phenotypic ratio for the children of parents 3 is
A.
B.
C.
D.
1/2 type A and 1/2 type B
1/2 type AB and 1/2 type O
1/4 type A, 1/2 type O, and 1/4 type B
1/4 type A, 1/4 type B, 1/4 type AB, and 1/4 type O
Jun 98,31
Use the following information to answer the next two questions
Assume that there are two gene pairs involved in determining eye colour: one codes for
pigment in the front of the iris and the other codes for pigment in the back of the iris.
If the genotype is
then the eye colour is
AABB  black-brown
AABb  dark brown
Aabb  brown
AaBB  brown-green flecked
AaBb  light brown
Aabb  grey-blue
AaBB  green
aaBb
 dark blue
aabb
 light blue
-from Audesirk, 1996
48. A man has grey-blue eyes and a woman has green eyes. Which eye colour phenotypes would be
possible for children born to this man and woman?
A.
B.
C.
D.
Grey-blue and green
Dark blue and brown
Light brown and dark blue
Brown-green flecked and light blue
Jun 98,5
Numerical Response
16. If one parent has light brown eyes and the other has dark brown eyes, what is the probability that they
would have an offspring with grey-blue eyes.
(Record your answer as a percentage to three digits in the numerical-response section of the answer
sheet.)
Answer: __________ %
Jun 98,32
Use the following information to answer the next three questions
Scientists have identified a genetic condition that apparently makes some men prone to
impulsive, violent behaviour. A pedigree was drawn highlighting the violent members of a
particular family. It appeared, from the pedigree, that men who displayed this violent
behaviour inherited this condition form their mothers, not their fathers. Further evidence
showed that this was the mode of inheritance.
-from Richardson, 1993
49. The inheritance pattern described indicates that this condition is
A.
B.
C.
D.
X – linked
Y – linked
autosomal
codominant
Jun 98,33
50. Which of the following pedigrees is not possible for the above condition?
(Note: Carriers have not been identified.)
Jun 98,34
Use the additional information to answer the next question
Researchers found that the allele that caused violent behaviour was situated on a chromosome in
a region that codes for an enzyme responsible for breaking down neurotransmitters in the brain.
- from Richardson, 1993
51. The substance affected by the genetic defect could be
A.
B.
C.
D.
ACTH
acetycholine
cholinesterase
norepinephrine
Jun 98, 35
Use the following information to answer the next two questions
Two alleles (HB and HN) exist for the pattern baldness gene. The HB allele is dominant in men
but recessive in women. The HN (normal) allele is dominant in females bur recessive in males. A
single HB allele seems to cause pattern baldness only in the presence of the level of testosterone
normally found in adult males. Males will develop pattern baldness with the genotypes H BHN or
HBHB; whereas, for females to develop pattern baldness, they must inherit two HB alleles. It
appears that testosterone destroys or inhibits the production of an enzyme necessary for hair
growth in the hair follicle.
- from Hoffman and Lingna, 1995
52. The difference in expression of pattern baldness in women and men results from the pattern baldness
gene’s being
A.
B.
C.
D.
X – linked
Y – linked
independent of hormonal balance
influenced by the level of testosterone
Jun 98,36
53. Parents who do not display pattern baldness have a son who exhibited pattern baldness by the age of
thirty. If they also have a daughter, she has
A.
B.
C.
D.
no chance of going bald, as the HBHB genotype is not possible
no chance of going bald, as the HBHN genotype is not possible
a 1 in 3 chance of going bald, as the HBHN genotype is possible
a 1 in 2 chance of going bald, as the HBHB genotype is possible
Jun 98,6
Use the following information to answer the next question
In Drosophila (fruit flies), the genes for pink eyes, rough eyes, curled wings and hairless bristles are
located on chromosome 3.
Genes
 pink eyes and hairless bristles
 hairless bristles and curled wings
 rough eyes and curled wings
 pink eyes and rough eyes
 rough eyes and hairless bristles
Crossover Frequency
21.5
19.5
41.1
43.1
21.6
Legend
Pink eyes – 1
Rough eyes – 2
Curled wings – 3
Hairless bristles - 4
-from Griffiths, 1993
Numerical Response
17. Use the legend to indicate the order of these genes along a chromosome.
(Record your four-digit answer in the numerical-response section of the answer sheet.)
Answer: _____ _____ _____ _____
(There are two possible ways of recording the answer; either will be acceptable.)
Jun 99,32
Use the following information to answer the next two questions
Melanin pigments range in colour from yellow to reddish –brown to black. The amount and the
colour of melanin in the skin account for differences in human skin coloration.
Albinism is a genetic disorder that results in unpigmented skin and other tissues. About 1 in
20,000 humans has albinism. In humans, it can be caused by an autosomal recessive allele (a). Its
dominant allele (A) results in normal pigmentation.
54. The fact that exposure to sunlight increases melanin production in many humans and produces a tan
demonstrates that
A.
B.
C.
D.
some people have mutations that prevent melanin production
the expression of some genes is influenced by the environment
the environment causes mutations that increase the chance of survival
the environment causes mutations that have no effect on the chance of survival
Jun 99,33
55. In the type of albinism described above, because melanin production is controlled by an autosomal
gene, it is expected that
A.
B.
C.
D.
males will develop albinism as they mature
males will inherit albinism from their mothers
albinism will occur more frequently among males than females
albinism will occur with equal frequencies among males and females
Jun 99,5
Use the following information to answer the next question
Piebald spotting is a rate human disorder. Although this disorder occurs in all races, piebald spotting is most
obvious in people with dark skin. A dominant allele appears to interfere with the migration of pigmentproducing cells; thus, patches of skin and hair lack pigment, allowing “spots” to form.
Pedigree Chart for Piebald Spotting
Numerical Response
18. What is the probability that any offspring produced by individuals II-5 and II-6 would have piebald
spotting?
(Record your answer as a value form 0 to 1 rounded to two decimal places in the numericalresponse section on the answer sheet.)
Answer: __________
Jun 99,34
Use the following information to answer the next question
In garden peas, the allele for tall plant height (T) is dominant over the allele for short plant
height (t), and the allele for axial flower position (A) is dominant over the allele for terminal
flower position (a). The alleles for plant height and flower position assort independently.
56. A plant heterozygous for both traits was crossed with a plant homozygous recessive for both traits.
What percentage of the offspring produced would be expected to display at least one of the dominant
traits?
A.
B.
C.
D.
25%
50%
75%
100%
Jun 99,6
Use the following information to answer the next question
“Alligator men” or “fish women” were exhibited for their physical abnormalities in fairs or
circuses earlier this century. These people probably suffered from X-linked ichthyosis, which
produces symmetric dark scales on the body. The disease occurs in 1 in 6,000 males and much
more rarely in females. Ichthyosis is likely a recessive disorder.
-from Cummings, 1994
Numerical Response
19. If an “alligator man” were to marry a woman homozygous for the normal condition, what is the
percentage probability that their children would have ichthyosis?
(Record your answer as a whole number percentage in the numerical-response section on the answer
sheet.)
Answer: __________ %
Jun 99,35
Use the following information to answer the next question
F1 Blood Type Cross
IA IB
IA i
IA IB
IA i
57. The genotypes of the parents to whom this Punnett square applies are
A.
B.
C.
D.
heterozygous Band homozygous A
heterozygous O and homozygous A
homozygous B and heterozygous A
heterozygous Band heterozygous A
Jun 99,36
Use the following information to answer the next question
A, B, M, N, O, and Rh Blood Typing
The alleles for A (IA) and B (IB) are codominant, and both are dominant to O (i).
The alleles for M and N are codominant
The allele for Rh+ is dominant to the allele for Rh-.
Blood groups can be used to determine relationships for a variety for legal and medical
purposes. The following is a list of phenotypes of some children over whom there is a legal
dispute.
Child 1
Child 2
Child 3
Child 4
Blood Types
O
MN
A
N
A
MN
AB
MN
Rh+
Rh+
RhRh-
58. Which children could belong to a couple in which the woman has blood type A, N, Rh+ and the man
has blood type O, M, Rh+?
A.
B.
C.
D.
Children 1 and 3
Children 1 and 4
Children 2 and 3
Children 2 and 4
Jun 99,7
Use the following to answer the next question
A program to detect carriers of -thalassemia (a mild blood disorder) found the incidence of the
disease to be 4% in a particular population. A recessive allele found on an autosomal chromosome
causes -thalassemia.
Numerical Response
20. What is the frequency of the recessive -thalassemia allele in the gene pool of this population?
(Record your answer as a value form 0 to 1 rounded to one decimal place in the numerical-response
section on the answer sheet.)
Answer: __________
Jun 99,37
Use the following information to answer the next four questions
Sickle cell anemia is caused by the sickle cell allele (Hb S) of a gene that contributes to
hemoglobin (Hb) production. The abnormal hemoglobin (hemoglobin-S) produced causes red
blood cells to become deformed and block capillaries. Tissue damage results. Affected
individuals homozygous for the sickle cell gene rarely survive to reproductive age.
Heterozygous individuals produce both normal hemoglobin and a small percentage of
hemoglobin-S. These individuals are more resistant to malaria than individuals who are
homozygous for the allele for normal hemoglobin (Hb A). Their red blood cells are prone to
sickling when there is a deficiency of oxygen.
59. If a man and a woman who are both heterozygous for the alleles Hb A and HbS have a child, the
probability that the child would not be heterozygous is
A.
B.
C.
D.
0.00
0.25
0.50
0.75
Jun 99,8
Use the following additional information to answer the next question
Numerical Response
21. The phenotype of III-6 is unknown. What is the probability that this individual is a carrier of the sickle
cell allele?
(Record your answer as a whole number percentage in the numerical-response section on the answer
sheet.)
Answer: ________ %
Jun 96,38
Use the following additional information to answer the next two questions
The malaria-causing microorganism Plasmodium falciparum is injected by mosquitoes into the
bloodstream of humans. Historically, the frequency of the Hb S in Africa relates directly to the
presence of malaria-causing organisms. In western Africa, the frequency of the HbS allele in
the gene pool is 0.15. In central Africa, the frequency is 0.10, and in southern Africa the
frequency is 0.05.
60. What is the frequency of the HbA allele in the human gene pool in western Africa?
A.
B.
C.
D.
0.72
0.85
0.90
0.95
Jun 99,39
61. Which of the following conclusions can be drawn from all the information provided on sickle cell
anemia?
A.
B.
C.
D.
The sickle cell gene will eventually disappear because of its interaction with malaria
Malaria causes heterozygous individuals to be less fertile than homozygous individuals
In Africa, sickle cell anemia will disappear since it is lethal in the homozygous condition
In Africa, carriers for sickle cell anemia have an advantage over homozygous individuals
Jun 99,40
Use the following information to answer the next question
A bacterium has been found that produces a form of plastic polyhydroxybutyrate (PHB).
Genes from this bacterium have been transferred into a weed called Arabidopsis thaliana.
These weeds now produce a biodegradable plastic.
- from Poirier, et al., 1997
62. The technology of transferring a gene from bacterium into a green plant is based on the principle that
A.
B.
C.
D.
all genes carry the same genetic information
all genes have the same basic chemical components
the genotypes of the bacterium and green plant are the same
the phenotype of an organism is not altered when one gene is exchanged for another
Jan 00,28
Use the following information to answer the next four questions
The flowering plant, Mirabilis jalapa (M.jalapa) may have branches with all white leaves, all
green leaves, and all variegated leaves (leaves with green and white patches) on the same plant.
Leaf colour is dependent on the colour of plastids present in cytoplasm. As in the case of other
plants, pollen (containing sperm nuclei) contributes chromosomes but almost no cytoplasm to the
zygote. The ovule contributes both chromosomes and cytoplasm to the zygote. The following data
of offspring phenotypes were collected from crosses between flowers from various branches.
Source of pollen (male)
White branch
Green branch
Variegated branch
Source of ovule (female)
White branch
Green branch
White offspring
Green offspring
White offspring
Green offspring
White offspring
Green offspring
-from Griffiths, 1993
63. These data indicate that, regardless of its branch source, pollen has no effect on the leaf colour of
resulting offspring. A reasonable explanation for this observation is that
A.
B.
C.
D.
leaf colour is a codominant trait
leaf colour is a dominant – recessive trait
cell organelles or cytoplasm are active only in pollen
cell organelles or cytoplasm contain genetic information
Jan 00,29
Use the following additional information to answer the next question
Several geneticists studied M. jalapa plants with deep crimson flowers and M. jalapa plants
with yellow flowers. Cross-pollinating these plants produced plants with scarlet-red flowers
(F1 generation).
These F1 plants were allowed to self-pollinate, and the resulting seeds produced M. jalapa
plants with three different flower colours. Data similar to the following were collected for
flower colour:
140 deep crimson
310 scarlet-red
160 yellow
-from Engels, 1975
64. With respect to the alleles for flower colour, these results indicate
A.
B.
C.
D.
X – linked inheritance
gene – linked inheritance
dominant – recessive inheritance
incomplete dominance inheritance
Jan 00,30
Use the following additional information to answer the next two questions
A different variety of homozygous M. jalapa produces flowers that are light crimson. Purebreeding genotypes and phenotypes are:
RPRP – deep crimson
RR – light crimson
rr – yellow
When two pure-breeding P1 plants are cross-pollinated, only scarlet-red-flowered offspring
(RPr) are produced.
When another pair of pure-breeding P1 plants are cross-pollinated, only orange-flowered
offspring (Rr) are produced.
-from Engels, 1975
65. The likely genotypes of the P1 plants for these two crosses is represented in row
Row
A
B
C
D
P1 genotypes
scarlet-red-flowered offspring
RPR x rr
RPRP x rr
RPr x RPr
RPRP x RR
P1 genotypes
orange-flowered offspring
RR x rr
RR x rr
Rr x Rr
RPR x Rr
Jan 00,31
66. Which of the following phenotypes is the predicted flower colour of M. jalapa with the genotype RPR?
A.
B.
C.
D.
Yellow
Orange
Crimson
Scarlet-red
Jan 00,32
Use the following information to answer the next two questions
Feather colour in parakeets is controlled by two genes. For one pigment gene, the B allele
produces blue colour and the b allele does not produce any colour. For the other pigment
gene, the Y allele produces yellow colour and the y allele does not produce any colour. Any
genotype containing at least one B allele and one Y allele will produce a green parakeet.
67. Which of the following parental genotypes could produce offspring with the four different colour
patterns?
A.
B.
C.
D.
BBYy x BbYy
BbYY x Bbyy
bbYY x bbyy
Bbyy x bbYy
Jan 00,33
68. Which is the probability of obtaining a blue parakeet when two green heterozygous parakeets are
crossed?
A.
B.
C.
D.
0
3/16
1/4
9/16

Jun 00,33
Use the following information to answer the next two questions
Cystic fibrosis is a recessive Mendelian trait in the human population. A symptom of cystic
fibrosis is the production of large amounts of mucin protein. New studies indicate that
although the cystic fibrosis condition is present at birth, increased mucin production is
preceded by an infection with the bacterium Pseudomonas aeruginosa. Individuals who are
not affected by cystic fibrosis produce a natural antibiotic, defensin, that kills the
Pseudomonas aeruginosa and eliminates the stimulus for increased mucin production.
Defensin is destroyed by a high chloride content in the tissues of individuals with cystic
fibrosis as a result of faulty chloride-channel proteins.
-from Sternberg, 1997
69. The allele that causes cystic fibrosis most likely results in a faulty amino acid sequence for the
A.
B.
C.
D.
channel proteins
mucin molecules
defensin molecules
Pseudomonas bacteria
Jun 00,34
70. In a normal individual, the population of Pseudomonas aeruginosa exhibits which of the following
population growth curves following initial infection of the individual?
Jun 00,35
Use the following information to answer the next three questions
Amniocentesis is a common prenatal procedure used to obtain cell to test for genetic
abnormalities that lead to disorders such as Down syndrome, cystic fibrosis, and hemophilia.
The test is usually offered between the 15th and 18th weeks of pregnancy to women who have
an increased risk of having children with genetic abnormalities.
Use the following additional information to answer the next question
Pedigree of a Family with Cystic Fibrosis
Note: Cystic fibrosis in this family is caused by a recessive allele that is found on
chromosome 7.
71. Prior to performing amniocentesis, a genetic counsellor collected pedigree information regarding the
incidence of cystic fibrosis within this family. The row that indicates the genotypes of individuals I-1,
I-2 and II-2 is
Row
A.
B.
C.
D.
I-1
Aa
AA
XAY
XAY
I-2
Aa
aa
XAXA
XAXa
II-2
aa
Aa
X aY
XAY
Jun 00,36
71. Down syndrome is a trisomy disorder that can be caused by the presence of three copies of
chromosome 21. Which of the following chromosome combinations identifies Down syndrome?
A.
B.
C.
D.
46 chromosomes consisting of 45 autosomes and 1 sex chromosome
46 chromosomes consisting of 44 autosomes and 2 sex chromosomes
47 chromosomes consisting of 45 autosomes and 2 sex chromosomes
47 chromosomes consisting of 44 autosomes and 3 sex chromosomes
Jun 00,37
72. A genetic abnormality such as Down syndrome can be diagnosed by using the cells obtained during
amniocentesis to create a
A.
B.
C.
D.
karyotype
therapeutic gene
DNA fingerprint
recombinant vector
Jun 00,38
Use the following information to answer the next two questions
In humans, the allele for normal blood clotting, H, is dominant to the allele for hemophilia, h.
The trait is X-linked.
73. A female hemophiliac marries a man who is not a hemophiliac. The row that indicates the probability
of this couple having a child that is a hemophiliac and the sex that the child would be is
Row
A.
B.
C.
D.
Probability
0.25
0.25
0.50
0.50
Sex of Affected Child
male
either female or male
male
either male or female
Jun 00,6
Numerical Response
22. A woman who is not a hemophiliac has a father who is a hemophiliac. If this woman marries a man
who is a hemophiliac, what is the probability of them having a hemophiliac son?
Answer: __________
(Record your answer as a value from 0 to 1, round to two decimal places in the numerical-response
section on the answer sheet.)
Jun 00,39
Use the following information to answer the next four questions
Two different genes control the expression of kernel colour in Mexican black corn: black
pigment gene B and dotted pigment gene D. Gene B influences the expression of gene D.
The dotted phenotype appears only when gene B is in the homozygous recessive state. A
colourless variation occurs when both genes are homozygous recessive.
After pure-breeding black-pigmented plants were crossed with colourless plants, all of the
offspring were black-pigmented.
-from Griffiths et al., 1993
74. The genotypes of the parents of these F1 offspring could be
A.
B.
C.
D.
BBDD x bbdd
BbDD x bbdd
Bbdd x bbDD
bbDD x BBdd
Jun 00, 40
75. Plants of the F1 generation are suspected of being heterozygous for both genes. A test cross of
colourless plants with the heterozygote plants should produce a phenotypic ratio in the offspring of
A.
B.
C.
D.
1:0
3:1
2:1:1
1:1:1:1
Jun 00,7
Numerical Response
23. What is the probability of dotted offspring being produced from the test cross described above?
Answer: __________
(Record your answer as a value from 0 to 1, rounded to two decimal places, in the numericalresponse section on the answer sheet.)
Jun 00,8
Numerical Response
24. If the total number of offspring produced in the test crosses was 1,024 plants, how many plants would
be expected to be black pigmented?
Answer: _________
(Record your answer as a whole number in the numerical-response section on the answer sheet.)
Jan 01,29
Use the following information to answer the next two questions
In tomato plants, purple stems (P) are dominant to green stems (p), and red tomatoes (T) are
dominant to yellow tomatoes (t). The two genes are located on separate chromosomes.
A purple-stemmed, red-tomato plant is crossed with a purple-stemmed, yellow-tomato plant.
They produce:
28 purple-stemmed, red-tomato plants
31 purple-stemmed, yellow-tomato plants
11 green-stemmed, red-tomato plants
9 green-stemmed, yellow-tomato plants
76. The genetic composition of the parents is
A.
B.
C.
D.
PpTt and PPTT
PPTt and PpTT
PpTt and PpTt
PpTt and Pptt
Jan 01,30
77. One of the green-stemmed, red-tomato plants was crossed with another tomato plant. One of the
offspring was a purple-stemmed, yellow-tomato plant. If this offspring were crossed with a greenstemmed, yellow-tomato plant, then the possible phenotype of phenotypes of the offspring would be
A.
B.
C.
D.
green-stemmed, yellow-tomato plants
green-stemmed, yellow-tomato plants and purple-stemmed, yellow-tomato plants
green-stemmed, yellow-tomato plants; purple-stemmed, yellow-tomato plants; and
purple-stemmed, red-tomato plants
green-stemmed, yellow-tomato plants; purple-stemmed, yellow-tomato plants; purplestemmed, red-tomato plants; and green-stemmed, red-tomato plants
Jan 01,31
Use the following information to answer the next two questions
Gene Loci for a Tomato Plant
From Griffiths et al., 1993
78. During meiosis, which of the following pairs of genes has the greatest chance of being separated by
crossing over?
A.
B.
C.
D.
(m) and (d)
(ne) and (p)
(m) and (lc)
(p) and (o)
Jan 01,32
Use the following additional information to answer the next question
Cross-over frequencies for some genes on a tomato plant:
Genes
Cross-Over Frequency
normal leaf (M) and tall plant (D)
12%
normal leaf (M) and normal tomato (O)
33%
normal leaf (M) and simple inflorescence (S)
64%
tall plant (D) and normal tomato (O)
21%
tall plant (D) and simple inflorescence (S)
52%
79. The cross-over frequency between genes O and S is
A.
B.
C.
D.
6%
29%
31%
97%
Jan 01,33
Use the following information to answer the next three questions
80. In the dihybrid cross between the two black mice, the C allele codes for
A.
B.
C.
D.
black colour
brown colour
colour absent
colour present
Jan 01,6
Numerical Response
25. What is the expected phenotypic ratio that results from a cross between two black mice heterozygous
for both genes?
Phenotypic Ratio: __________ : __________ : __________
Coat Colour: Black
Brown
White
(Record your three-digit answer in the numerical-response section on the answer sheet.)
Jan 01,7
Numerical Response
26. What is the expected phenotypic ratio resulting from a cross between a bbCc female mouse and BbCc
male mouse?
Phenotypic Ratio: __________ : __________ : __________
Coat Colour: Black
Brown
White
(Record your three-digit answer in the numerical-response section on the answer sheet.)
Jan 01,34
Use the following information to answer the next question
Farmers who raise sheep for wool try not to produce offspring with black wool. Black wool is
very brittle and difficult to dye; therefore, white wool is more desirable. If a farmer purchases
a white ram, he will generally carry out a test cross to determine whether the ram is
heterozygous and homozygous for white wool. White wool (W) is dominant to black wool (w).
81. If the ram is heterozygous for white wool, the expected phenotypes of the offspring of the farmer’s test
cross would be
A.
B.
C.
D.
all black
all white
1/2 black and 1/2 white
3/4 black and 1/4 white
Jan 02,5
Use the following information to answer the next three questions
Description and Symbols Used to Represent One Type of Coat Colour in Horses
1
DNA sequence for coat colour
2
TT, Tt
Tt
3
T
t
4
Tobiano (white spotting pattern)
Not tobiano (no white spotting
pattern)
Numerical Response
27. Using the number above, match these descriptions and symbols with the term below to which they
apply
Description or
Symbol Number: __________ ___________ ___________ ___________
Term:
gene
allele
phenotype
genotype
(Record all four digits of your answer in the numerical-response section on the answer sheet.)
Jan 02,28
82. What are the genotypes for coat colour of two horses that are predicted to produce offspring in a 1:1
genotypic ratio?
A.
B.
C.
D.
Tt and tt
Tt and Tt
Tobiano and tobiano
Tobiano and not tobiano
Jan 02,6
Numerical Response
28. Given that the diploid number for horses is 64, what is the number of chromosomes found in a horse’s
somatic cell and what is the number of chromosomes found in a horse’s gamete cell?
Number of
Chromosomes: _____ _____ , _____ _____
Cell type: somatic cell
gamete cell
(Record all four digits of your answer in the numerical-response section on the answer sheet.)
Jan 02,29
Use the following information to answer the next six questions
Cat coat colour results from the interaction of three different genes. A gene for black-based
colours is located on an autosomal chromosome. A gene for red-based colours is located on
the X chromosome. A different gene located on a separate autosomal chromosome
determines pigment density in cat hair.
The black-based gene has three possible alleles: B – black, b – chocolate, and b1 – cinnamon.
If pigmentation in cat hair is dense, the phenotypes listed below are possible
Genotype
BB, Bb, Bb1
bb, bb1
b1b1
Phenotype
black
chocolate
cinnamon
83. According to the data above, the relationship among these alleles is such that the
A.
B.
C.
D.
black allele is codominant with the chocolate and cinnamon alleles
black allele is codominant with the chocolate allele, and the chocolate allele is
codominant with the cinnamon allele
black allele is dominant over the chocolate and cinnamon alleles, and the chocolate allele
is dominant over the cinnamon allele
black allele is dominant over the chocolate and cinnamon alleles, and the chocolate and
cinnamon alleles are codominant
Jan 02,30
Use the following additional information to answer the next two questions
There are two alleles for the pigment-density gene: dense pigment (D) and dilute pigment (d).
The chart below shows the interaction of two autosomal genes affecting coat colour – the
black-based gene and the density gene.
Density gene
Black-based pigment gene
D_
dd
B_
B_D_
black colour
B_dd
blue colour
bbD_; bb1D_
chocolate colour
bbdd; bb1dd
lilac colour
b1b1D_
cinnamon colour
b1b1dd
fawn colour
bb; bb
1
b b1
1
84. A blue-coloured female cat is bred with a cinnamon-coloured male cat. The offspring produced are
black-coloured, blue-coloured, chocolate-coloured, and lilac-coloured. The genotypes of the parental
cats are indicated in row
Row
Female Cat
Male Cat
Bb1dd
b1b1Dd
A.
1
Bb dd
b1b1DD
B.
Bbdd
b1b1Dd
C.
Bbdd
b1b1DD
D
Jan 02,31
85. A black-coloured female cat with the genotype BbDd is bred with a fawn-coloured male cat. The
percentage of their offspring predicted to be chocolate-coloured is
A.
B.
C.
D.
13 %
19 %
25 %
50 %
Jan 02,32
Use the following additional information to answer the next three questions
In cats, red pigmentation is dominant to black pigmentation. The red pigment gene, which
is located on the X chromosome, has two alleles: XR and Xr. Cats with at least on XR allele
have some orange-coloured hair as a result of having the red-based pigment. Cats with only
Xr alleles have no red-based pigment. Male cats with the XR allele will be orange. However,
female cats express the genes on only one X chromosome in each cell. This expression is
random. Therefore, an orange-and-black (tortoiseshell) female cat is possible if it is XRXr.
Some genotypes and their resulting phenotypes are shown below. In all cases, pigment
density is high.
Genotype
XRYBb
XrYBb1
XRXrBb
Phenotype
Orange male cat
Black male cat
Orange-and-black female cat (tortoiseshell)
86. The phenotype of a female cat with genotype XrXrBb1 would be
A.
B.
C.
D.
a black cat
an orange cat
an orange-and-black cat
an orange, black, and cinnamon cat
Jan 02,33
87. A cinnamon-coloured male cat (XrYb1b1) is bred with an orange-coloured female cat (XRXRBB). What
possible phenotypes could be produced in the offspring?
A.
B.
C.
D.
Tortoiseshell-coloured female cats and orange-coloured male cats
Tortoiseshell-coloured female cats, black-coloured female cats, and black-coloured male
cats
Cinnamon-coloured male cats, orange-coloured female cats, and tortoiseshell-coloured
female cats
Cinnamon-coloured male cats, black-coloured male cats, black-coloured female cats,
orange-coloured female cats, and tortoiseshell-coloured female cats
Jan 02,34
Use the following additional information to answer the next question
When the three genes that code for black-based colour, red-based colour, and density combine, they
produce other coat colours in cats.
88. What is the predicted phenotype of a female cat with genotype XRXRBb1dd?
A.
B.
C.
D.
Black
Orange
Cinnamon
Cream (light orange)
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