_____ equal segregation
_____ penetrance
_____ co-dominance
_____ recessive epistasis
_____ haploinsufficient
_____ independent assortment
_____ lethal allele
_____ compound heterozygote
_____ 9:3:3:1 ratio
_____ 3:1 ratio
_____ imprinting
_____ incomplete dominance
_____ testcross
_____ chi square goodness of fit
_____ complete dominance
_____ expressivity
_____ heterozygote
A. genes on different chromosomes assort independently during meiosis
B. the heterozygote has the same phenotype as the homozygous dominant
C. statistical test to determine if observed results are expected results
D. the heterozygote shows an intermediate phenotype between the two homozygotes
E. genotypic ratio observed in the progeny of a di-hybrid cross
F. an allele that causes death when present in certain genotypes
,G. genotypic frequency observed in the progeny of a mono- hybrid cross
H. degree or intensity with which a genotype is expressed
I. mating to an individual homozygous for recessive alleles
J. homozygous recessive genotype at one gene masks the phenotype of another gene
K. percentage of individuals with a genotype who express the expected phenotype
L. both alleles are fully expressed in the heterozygote
M. phenotypic frequency observed in the progeny of a mono- hybrid cross
N. an individual with two different mutant alleles of the same gene
O. one parental allele is epigenetically silenced via DNA methylation
P. The two alleles separate during gamete formation
Q. heterozygous for the A and a alleles
R. one functional allele is not enough for normal phenotype
Explain the three laws of inheritance and their connections to meiosis.
1. What is Mendel’s first law?
Answer: Mendel’s first law states that alleles segregate into gametes during
meiosis. This discovery came from his monohybrid experimental crosses.
2. In a hypothetical organism, two unlinked genes are being studied: gene A (alleles
A and a) and gene B (alleles B and b). An individual with the genotype AaBb is
test-crossed with an aabb individual.
A. State the expected phenotypic ratio of the offspring, according to
Mendel's Law of Independent Assortment.
1:1:1:1
, B. Explain the precise meiotic event that is the physical basis for the Law of
Independent Assortment, detailing how it ensures all possible allele
combinations are equally likely in the gametes. Specify the relevant phase
of meiosis and the behavior of the chromosomes.
Solve problems associated with monohybrid crosses.
3. For gene A, in a diploid organism, four units of the protein product are needed for
normal function. Each wild-type allele produces five units. If a mutation causes a
null allele (no protein product is produced), is this allele dominant or recessive?
Explain your reasoning.
4 units of gene A protein= normal function.
One wild-type allele produces 5 units of protein.
A/A genotype produces a total of 10 units of protein, more than is needed for normal
function.
a/a genotype produces 0 units of protein, there isn’t any A protein produced so this
genotype will have a mutant phenotype
A/a genotype produces 5 units of protein.
The A/a genotype produces enough protein (5 units) to allow normal function (you only
need 4 unit of protein). A heterozygote (A/a) which contains one wild-type allele (A) and
one null allele (a) and would be phenotypically normal so this mutant allele is recessive
because you need two mutant alleles (a/a) to display the mutant phenotype.
4. For gene B, in a diploid organism, eight units of the product are needed for
normal function. Each wild-type allele produces five units. If a mutation causes a
null allele (no protein product is produced), is this allele dominant or recessive?
Explain your reasoning.
8 units of gene B protein= normal function
One wild-type allele produces 5 units of protein
, B/B genotype produces 10 units of B protein, more than enough for normal function.
b/b genotype produces 0 units of protein, there isn’t any B protein produced so this
genotype will have a mutant phenotype
B/b genotype produces 5 units of protein. This genotype doesn’t produce the needed 8
units of proteins for full function so this genotype will be phenotypically mutant. In this
case the null allele for the B gene is dominant.
5. The human piebald trait is a defect in melanocyte development characterized by
patches of skin and hair lacking pigmentation. The piebald gene is located on
human chromosome four. The condition results from the inability of
pigment-producing cells to migrate properly during development. Two parents
with piebald spotting have two children. One child has piebald spotting, and the
other child has normal pigmentation.
a. Is the piebald spotting trait dominant or recessive? Explain why you came
to this conclusion.
You need to analyze both possibilities. If the trait is recessive, then the parents would
both be homozygous recessive. Therefore the parents would be the following: p/p X p/p
, these parents can only produce children of genotype p/p and all children would be
phenotypically piebald. This is not the result observed in this situation.
Therefore, the trait must be dominant. In this situation, a child would only need to inherit
one piebald allele.
b. What are the genotypes of the parents?
both parents are heterozygous
Describe Mendel’s pea experiments.
6. Mendel crossed a pure-breeding tall pea plant with a pure-breeding short pea
plant.
a. What phenotype did he observe in the F1 generation?
All tall
_____ penetrance
_____ co-dominance
_____ recessive epistasis
_____ haploinsufficient
_____ independent assortment
_____ lethal allele
_____ compound heterozygote
_____ 9:3:3:1 ratio
_____ 3:1 ratio
_____ imprinting
_____ incomplete dominance
_____ testcross
_____ chi square goodness of fit
_____ complete dominance
_____ expressivity
_____ heterozygote
A. genes on different chromosomes assort independently during meiosis
B. the heterozygote has the same phenotype as the homozygous dominant
C. statistical test to determine if observed results are expected results
D. the heterozygote shows an intermediate phenotype between the two homozygotes
E. genotypic ratio observed in the progeny of a di-hybrid cross
F. an allele that causes death when present in certain genotypes
,G. genotypic frequency observed in the progeny of a mono- hybrid cross
H. degree or intensity with which a genotype is expressed
I. mating to an individual homozygous for recessive alleles
J. homozygous recessive genotype at one gene masks the phenotype of another gene
K. percentage of individuals with a genotype who express the expected phenotype
L. both alleles are fully expressed in the heterozygote
M. phenotypic frequency observed in the progeny of a mono- hybrid cross
N. an individual with two different mutant alleles of the same gene
O. one parental allele is epigenetically silenced via DNA methylation
P. The two alleles separate during gamete formation
Q. heterozygous for the A and a alleles
R. one functional allele is not enough for normal phenotype
Explain the three laws of inheritance and their connections to meiosis.
1. What is Mendel’s first law?
Answer: Mendel’s first law states that alleles segregate into gametes during
meiosis. This discovery came from his monohybrid experimental crosses.
2. In a hypothetical organism, two unlinked genes are being studied: gene A (alleles
A and a) and gene B (alleles B and b). An individual with the genotype AaBb is
test-crossed with an aabb individual.
A. State the expected phenotypic ratio of the offspring, according to
Mendel's Law of Independent Assortment.
1:1:1:1
, B. Explain the precise meiotic event that is the physical basis for the Law of
Independent Assortment, detailing how it ensures all possible allele
combinations are equally likely in the gametes. Specify the relevant phase
of meiosis and the behavior of the chromosomes.
Solve problems associated with monohybrid crosses.
3. For gene A, in a diploid organism, four units of the protein product are needed for
normal function. Each wild-type allele produces five units. If a mutation causes a
null allele (no protein product is produced), is this allele dominant or recessive?
Explain your reasoning.
4 units of gene A protein= normal function.
One wild-type allele produces 5 units of protein.
A/A genotype produces a total of 10 units of protein, more than is needed for normal
function.
a/a genotype produces 0 units of protein, there isn’t any A protein produced so this
genotype will have a mutant phenotype
A/a genotype produces 5 units of protein.
The A/a genotype produces enough protein (5 units) to allow normal function (you only
need 4 unit of protein). A heterozygote (A/a) which contains one wild-type allele (A) and
one null allele (a) and would be phenotypically normal so this mutant allele is recessive
because you need two mutant alleles (a/a) to display the mutant phenotype.
4. For gene B, in a diploid organism, eight units of the product are needed for
normal function. Each wild-type allele produces five units. If a mutation causes a
null allele (no protein product is produced), is this allele dominant or recessive?
Explain your reasoning.
8 units of gene B protein= normal function
One wild-type allele produces 5 units of protein
, B/B genotype produces 10 units of B protein, more than enough for normal function.
b/b genotype produces 0 units of protein, there isn’t any B protein produced so this
genotype will have a mutant phenotype
B/b genotype produces 5 units of protein. This genotype doesn’t produce the needed 8
units of proteins for full function so this genotype will be phenotypically mutant. In this
case the null allele for the B gene is dominant.
5. The human piebald trait is a defect in melanocyte development characterized by
patches of skin and hair lacking pigmentation. The piebald gene is located on
human chromosome four. The condition results from the inability of
pigment-producing cells to migrate properly during development. Two parents
with piebald spotting have two children. One child has piebald spotting, and the
other child has normal pigmentation.
a. Is the piebald spotting trait dominant or recessive? Explain why you came
to this conclusion.
You need to analyze both possibilities. If the trait is recessive, then the parents would
both be homozygous recessive. Therefore the parents would be the following: p/p X p/p
, these parents can only produce children of genotype p/p and all children would be
phenotypically piebald. This is not the result observed in this situation.
Therefore, the trait must be dominant. In this situation, a child would only need to inherit
one piebald allele.
b. What are the genotypes of the parents?
both parents are heterozygous
Describe Mendel’s pea experiments.
6. Mendel crossed a pure-breeding tall pea plant with a pure-breeding short pea
plant.
a. What phenotype did he observe in the F1 generation?
All tall
