BIO 340 EXAMS 2 SET ANSWERS AND QUESTIONS
SET A+
✔✔AB, Ab, aB, ab - ✔✔What are the types of gametes that can be produced by an
individual with the genotype AaBb?
✔✔4 - ✔✔How many different types of gametes could an organism with the genotype
AABbCc produce?
✔✔Both OS and oS - ✔✔Genetic inheritance is being studied in a certain species of
plant in which orange flower color (O) is dominant to white (o) and round leaf shape (S)
is dominant to oval (s). A true-breeding orange-flowered, round-leaved male plant (Plant
A) is mated with a true-breeding white-flowered, oval-leaved female plant (Plant B) to
produce dihybrid offspring plant (Plant C). Assuming normal Mendelian genetics and
independent inheritance of these two traits, which gametes might be produced by Plant
C?
✔✔1/4 - ✔✔In guinea pigs coat color is determined by a single gene with two alleles. A
guinea pig from a true-breeding black strain is mated with a guinea pig from a true-
breeding white strain. The F1 progeny are all black. Two of the F1 progeny are mated
with each other.
What proportion of the F2 progeny is expected to be white?
✔✔1/3 - ✔✔In guinea pigs coat color is determined by a single gene with two alleles. A
guinea pig from a true-breeding black strain is mated with a guinea pig from a true-
breeding white strain. The F1 progeny are all black. Two of the F1 progeny are mated
with each other.
What proportion of the black F2 progeny is expected to be homozygous?
✔✔Hh x Hh - ✔✔The presence of short hairs on leaves is controlled by dominant allele
H. Plants with genotype hh have smooth leaves. Two parental plants with unknown
genotype were crossed. What are the parental genotypes if F1=
32 Hairy and 11 smooth?
, ✔✔hh x hh - ✔✔The presence of short hairs on leaves is controlled by dominant allele
H. Plants with genotype hh have smooth leaves. Two parental plants with unknown
genotype were crossed.
What are the parental genotypes if F1=
0 Hairy and 24 smooth?
✔✔There should be 1 green seed for every 1 yellow seed. - ✔✔Mendel imagined that
seed color in pea plants is influenced by a gene with two alleles in which the allele for
yellow seeds (Y) is dominant over the allele for green seeds (y). What would Mendel
predict the seed colors would be from a cross of a male plant with the mixed Yy
genotype with a female plant with the yy genotype?
✔✔AArr and aaRr - ✔✔In apples, the allele A is dominant for a big size apple and the
allele R is dominant for red color. You cross one tree that produces big yellow apples
and another tree that produces small red apples. Half of the offspring trees produce big
red apples and half produce big yellow apples. What are the genotypes of the parents?
(Pay attention to the phenotypic description of the parents so you can easily eliminate
the genotypes that do not match these descriptions.)
✔✔50% - ✔✔You have flipped a standard quarter 10 times and had four head and six
tails come up. With that experience, what will be the chances of you getting a heads on
the next flip?
✔✔0.05 - ✔✔If a student randomly guesses at five multiple-choice questions during a
BIO 340 exam, find the probability that the student gets exactly three correct. Each
question has five possible choices. (In this case, number of trials (n) = 5, number of
successes (X) = 3, and the numerical probability of success (p) = 1/5, since there is one
chance in five of guessing a correct answer.)
✔✔1 in 16 - ✔✔What is the probability of having 4 girls in a row?
✔✔0.33 - ✔✔In a cross between two heterozygous plants (Tt x Tt), what is the
probability of a plant being homozygous given that it is tall?
✔✔1/32 - ✔✔What is the probability that the following pair of parents will produce the
indicated offspring?
TTRrYy x TtRrYy -> TTrrYY (Use a Punnett Square to solve for one gene at a time
(e.g. TT x Tt) in order to find the probability of the specific genotype for that gene in the
offspring. Repeat for all the genes and use the multiplication rule to get the answer for
the complete genotype.)
✔✔Both heterozygous for the alkaptonuria gene. - ✔✔John has alkaptonuria; it "runs in
the family". John and his grandmother are both homozygous for alkaptonuria. John's
grandmother has two brothers and a sister. One of the brothers had alkaptonuria and
his daughter is John's mother. Based on this information, John's parents are:
SET A+
✔✔AB, Ab, aB, ab - ✔✔What are the types of gametes that can be produced by an
individual with the genotype AaBb?
✔✔4 - ✔✔How many different types of gametes could an organism with the genotype
AABbCc produce?
✔✔Both OS and oS - ✔✔Genetic inheritance is being studied in a certain species of
plant in which orange flower color (O) is dominant to white (o) and round leaf shape (S)
is dominant to oval (s). A true-breeding orange-flowered, round-leaved male plant (Plant
A) is mated with a true-breeding white-flowered, oval-leaved female plant (Plant B) to
produce dihybrid offspring plant (Plant C). Assuming normal Mendelian genetics and
independent inheritance of these two traits, which gametes might be produced by Plant
C?
✔✔1/4 - ✔✔In guinea pigs coat color is determined by a single gene with two alleles. A
guinea pig from a true-breeding black strain is mated with a guinea pig from a true-
breeding white strain. The F1 progeny are all black. Two of the F1 progeny are mated
with each other.
What proportion of the F2 progeny is expected to be white?
✔✔1/3 - ✔✔In guinea pigs coat color is determined by a single gene with two alleles. A
guinea pig from a true-breeding black strain is mated with a guinea pig from a true-
breeding white strain. The F1 progeny are all black. Two of the F1 progeny are mated
with each other.
What proportion of the black F2 progeny is expected to be homozygous?
✔✔Hh x Hh - ✔✔The presence of short hairs on leaves is controlled by dominant allele
H. Plants with genotype hh have smooth leaves. Two parental plants with unknown
genotype were crossed. What are the parental genotypes if F1=
32 Hairy and 11 smooth?
, ✔✔hh x hh - ✔✔The presence of short hairs on leaves is controlled by dominant allele
H. Plants with genotype hh have smooth leaves. Two parental plants with unknown
genotype were crossed.
What are the parental genotypes if F1=
0 Hairy and 24 smooth?
✔✔There should be 1 green seed for every 1 yellow seed. - ✔✔Mendel imagined that
seed color in pea plants is influenced by a gene with two alleles in which the allele for
yellow seeds (Y) is dominant over the allele for green seeds (y). What would Mendel
predict the seed colors would be from a cross of a male plant with the mixed Yy
genotype with a female plant with the yy genotype?
✔✔AArr and aaRr - ✔✔In apples, the allele A is dominant for a big size apple and the
allele R is dominant for red color. You cross one tree that produces big yellow apples
and another tree that produces small red apples. Half of the offspring trees produce big
red apples and half produce big yellow apples. What are the genotypes of the parents?
(Pay attention to the phenotypic description of the parents so you can easily eliminate
the genotypes that do not match these descriptions.)
✔✔50% - ✔✔You have flipped a standard quarter 10 times and had four head and six
tails come up. With that experience, what will be the chances of you getting a heads on
the next flip?
✔✔0.05 - ✔✔If a student randomly guesses at five multiple-choice questions during a
BIO 340 exam, find the probability that the student gets exactly three correct. Each
question has five possible choices. (In this case, number of trials (n) = 5, number of
successes (X) = 3, and the numerical probability of success (p) = 1/5, since there is one
chance in five of guessing a correct answer.)
✔✔1 in 16 - ✔✔What is the probability of having 4 girls in a row?
✔✔0.33 - ✔✔In a cross between two heterozygous plants (Tt x Tt), what is the
probability of a plant being homozygous given that it is tall?
✔✔1/32 - ✔✔What is the probability that the following pair of parents will produce the
indicated offspring?
TTRrYy x TtRrYy -> TTrrYY (Use a Punnett Square to solve for one gene at a time
(e.g. TT x Tt) in order to find the probability of the specific genotype for that gene in the
offspring. Repeat for all the genes and use the multiplication rule to get the answer for
the complete genotype.)
✔✔Both heterozygous for the alkaptonuria gene. - ✔✔John has alkaptonuria; it "runs in
the family". John and his grandmother are both homozygous for alkaptonuria. John's
grandmother has two brothers and a sister. One of the brothers had alkaptonuria and
his daughter is John's mother. Based on this information, John's parents are:
