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Welcome to Genetics:
Unit 2 Seminar!
Please feel free to chat with your
classmates!
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Agenda
• Brief Review of Unit Material
• Self Assessment Question
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Alleles

Alleles are pairs of
genes that carry
the same traits and
are found at the
same locations on
pairs of
chromosomes.
From Father
From Mother
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Autosomal Inheritance

Dominant – only one allele of a gene necessary to
express the trait
GENES

Recessive – both alleles of a gene must be
identical to express that trait
GENES
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Genetic Terms

Homozygous Trait - Both genes for that
trait are the same.


A pea plant with two genes for tallness.
Heterozygous Trait - Both genes for that
trait are not the same.

A pea plant with one gene for tallness and one for
shortness.
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Human Molecular Genetics
• Homozygous - a diploid cell or organism carry two
identical copies of a gene
• Heterozygous - a diploid cell or organism having two
different alleles of a particular gene
heterozygous
homozygous
75%
25%
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Human Molecular Genetics
• Genotype – entire genetic constitution of an individual
cell or organism, usually with emphasis on the particular
alleles at one or more specific loci.
GENES
• Phenotype – the detectable physical and physiological
characteristics of a cell or organism determined by its
genotype; also, the specific trait associated with a
particular allele.
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Practice Problem 1
• Beak color is an autosomal trait in
chickens. Red beaks are dominant
over white beaks. The allele for red
beaks is R. The allele for white beaks
is r.
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Practice Problem 1
• Claude is a chicken that has a red beak and has
a genotype of Rr. If this chicken mates with a
white-beaked chicken named Filette, what will the
baby chickens look like? What would be their
phenotypes and genotypes?
Claude
Filette
Rr
rr
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Practice Problem 1
X
Claude
Rr
?
Filette
rr
What is their Genotype?
What is their Phenotype?
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Practice Problem 1
X
Claude
Rr
Filette
rr
Genotype is Rr
Phenotype is red
beak
50%
Genotype is rr
Phenotype is
white beak
50%
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Practice Problem 2
• John is homozygous dominant for the
T allele decides to have children with a
Jane who is homozygous recessive.
• We will say the T allele is a Tall gene:
• T = tall
• t = short
*Note: In reality, there is no tall gene.
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Practice Problem 2
• John is homozygous dominant for the T
allele decides to have children with a Jane
who is homozygous recessive. First what is
John and Jane’s genotypes?
John
Jane
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Practice Problem 2
• John is homozygous dominant for the T
allele decides to have children with a Jane
who is homozygous recessive. First what is
John and Jane’s genotypes?
John
TT
Jane
tt
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Practice Problem 2
• What will their children be like? Again,
include the children's genotypes and
phenotypes.
John
TT
X
Jane
tt
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Practice Problem 2
John and Jane’s Children
Both are tall
John Junior
Tt
Jane Junior
Tt
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Self Assessment Quiz: Question 1
• 1. Alternative forms of a gene are called:
- Alleles
- Antigens
- Chromosomes
- Gametes
- Genotypes
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Self Assessment Quiz: Question 1
Answer
• 1. Alternative forms of a gene are called:
- Alleles Correct Answer
- Antigens
- Chromosomes
- Gametes
- Genotypes
*Alleles are pairs of genes that carry the same traits
and are found at the same locations on pairs of
chromosomes.
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Self Assessment Quiz: Question 2
• Parents heterozygous for a recessive allele
are called :
–
–
–
–
–
Codominant
Carriers
Hybrids
True breeding
Pleiotropic
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Self Assessment Quiz: Question 2
• Parents heterozygous for a recessive allele
are called :
–
–
–
–
–
Codominant
Carriers CORRECT ANSWER
Hybrids
True breeding
Pleiotropic
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Self Assessment Quiz: Question 3
• Probabilities are calculated using the
multiplication rule when they:
- Are equally likely
- Are independent
- Are mutually exclusive
- Occur disproportionately
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Self Assessment Quiz: Question 3 Answer
• Probabilities are calculated using the
multiplication rule when they:
- Are equally likely
- Are independent Correct Answer
- Are mutually exclusive
- Occur disproportionately
• Aa X Aa what are changes of being aa
genotype
• ½X½=¼
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Self Assessment Quiz: Question 4
• Probabilities are calculated using the
addition rule when they:
–
–
–
–
Are equally likely
Are independent
Are mutually exclusive
Occur disproportionately
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Self Assessment Quiz: Question 4
Answer
• Probabilities are calculated using the
addition rule when they:
–
–
–
–
Are equally likely
Are independent
Are mutually exclusive CORRECT ANSWER
Occur disproportionately
* Rule of addition is that the probability of an event
that can occur in two or more independent ways
is the sum of the separate probabilities of the
different ways.
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Self Assessment Quiz: Question 5
•
Mating two organisms produces a 3:1 ratio of the
phenotype in progeny. The parental genotypes
are :
–
–
–
–
Aa × Aa
Aa × aa
AA × aa
AA × AA
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Self Assessment Quiz: Question 5
Answer
•
Mating two organisms produces a 3:1 ratio of the
phenotype in progeny. The parental genotypes
are :
–
–
–
–
Aa × Aa CORRECT ANSWER
Aa × aa
AA × aa
AA × AA
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Monohybrid Genetic Cross
• Genetic cross : Aa X Aa produces A and a
gametes from each parent
• Punnett square shows four possible outcomes =
AA, Aa, aA, and aa
• Three combinations = AA, Aa, and aA produce
plants with round seeds and display a round
phenotype
• Fourth combination = aa displays wrinkled
phenotype
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3:1
Round:wrinkle
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Self Assessment Quiz: Question 6
• A mating between organisms that are both
heterozygous for two traits is expected to produce
a 9:3:3:1 ratio in the offspring when the genes
underlying the two traits exhibit:
- Codominance
- Incomplete dominance
- Independent assortment
- Multiple alleles
- Pleiotropy
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Self Assessment Quiz: Question 6
Answer
• A mating between organisms that are both
heterozygous for two traits is expected to produce
a 9:3:3:1 ratio in the offspring when the genes
underlying the two traits exhibit:
- Codominance
- Incomplete dominance
- Independent assortment Correct Answer
- Multiple alleles
- Pleiotropy
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Dihybrid Cross
• Mendel studied inheritance of two different traits,
such as seed color (yellow vs. green) and seed
shape (round vs. wrinkled) in the same cross =
dihybrid cross
• The F1 progeny were hybrid for both characteristics,
and the phenotype of the seeds was round
(dominant to wrinkled) and yellow (dominant to
green)
• In the F2 progeny, he observed the 9 round yellow : 3
wrinkled yellow : 3 round green : 1 wrinkled green
ratio
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Dihybrid Cross
• Mendel carried out similar experiments with
other combinations of traits. For each pair of
traits, he consistently observed the 9 : 3 : 3 : 1
ratio
• He also deduced the biological reason for the
observation:
• In the F2 progeny, if the 3 : 1 ratio of round:
wrinkled, is combined at random with the 3: 1
ratio of yellow: green, it yields the 9: 3: 3: 1
ratio of a dihybrid cross
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Fig. 2.10
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Self Assessment Quiz: Question 7
• Suppose you perform a dihybrid cross that focuses on two traits
that assort independently: (1) Body color, where the black allele
(B) is dominant to the white allele (b); (2) Body size, where tall (T)
is dominant to short (t). In the P1 generation, a true -breeding
Black, Tall individual is crossed with a true-breeding white, short
individual. F1 individuals are then crossed with one another to
produce the F2 generation. What is the expected probability of
Tall, white individuals among the F2 generation?
-0
- 1/16
- 3/16
- 4/16
- 8/16
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Self Assessment Quiz: Question 7
Answer
• Suppose you perform a dihybrid cross that focuses on two traits
that assort independently: (1) Body color, where the black allele
(B) is dominant to the white allele (b); (2) Body size, where tall (T)
is dominant to short (t). In the P1 generation, a true -breeding
Black, Tall individual is crossed with a true-breeding white, short
individual. F1 individuals are then crossed with one another to
produce the F2 generation. What is the expected probability of
Tall, white individuals among the F2 generation?
-0
- 1/16
- 3/16 Correct Answer
- 4/16
- 8/16
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Fig. 2.10
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Self Assessment Quiz: Question 8
• Assuming equal sex ratios, if a mating has
already produced three girls, what is the
probability that the next child will be a girl?
-0
-1
- 1/2
- (1/2)3
- 1 – (1/2)3
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Self Assessment Quiz: Question 8
Answer
• Assuming equal sex ratios, if a mating has
already produced three girls, what is the
probability that the next child will be a girl?
-0
-1
- ½ Correct Answer
- (1/2)3
- 1 – (1/2)3
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Self Assessment Quiz: Question 9
• In pea plants, is a single-gene trait where purpleflowers is dominant to white flowers. Suppose
you find a purple-flowered pea plant. Which of
the following would you perform to determine the
genotype of this plant?
•
- Complementation test
- Independent assortment
- Mutant screen
- Testcross
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Self Assessment Quiz: Question 9
Answer
• In pea plants, is a single-gene trait where purpleflowers is dominant to white flowers. Suppose
you find a purple-flowered pea plant. Which of
the following would you perform to determine the
genotype of this plant?
•
- Complementation test
- Independent assortment
- Mutant screen
- Testcross Correct Answer
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Testcross Analysis
• Testcross = a cross, between an organism of
dominant phenotype (genotype unknown) and an
organism of recessive phenotype (genotype known
to be homozygous recessive)
• In a testcross, the relative proportion of the different
gametes produced by the heterozygous parent can
be observed directly in the proportion of
phenotypes of the progeny, because the recessive
parent contributes only recessive alleles
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Testcross Results
• AA + aa = Aa – testcross produces dominant
progeny only: parent homozygous
• Aa + aa = 1/2 Aa + 1/2 aa - testcross produces 1/2
dominant and 1/2 recessive individuals: parent
heterozygous
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Self Assessment Quiz: Question 10
• Mating two organisms produces a 3:1 ratio of the
phenotype in progeny. The parental genotypes
are :
- Aa × Aa
- Aa × aa
- AA × aa
- AA × AA
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Self Assessment Quiz: Question 10
Answer
• Mating two organisms produces a 3:1 ratio of the
phenotype in progeny. The parental genotypes
are :
- Aa × Aa Correct Answer
- Aa × aa
- AA × aa
- AA × AA
*Aa X Aa produces AA; Aa; aA; aa resulting in
phenotypes of 3:1 that is 3 dominant phenotypes
to one recessive
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Self Assessment Quiz: Question 11
• A woman with keratosis, a skin condition caused
by a rare dominant allele, marries a normal man,
and they have two children. What is the
probability that both children are normal?
-0
-1/2
- 1/4
- 1–1/4
-1
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Self Assessment Quiz: Question 11
Answer
• A woman with keratosis, a skin condition caused
by a rare dominant allele, marries a normal man,
and they have two children. What is the
probability that both children are normal?
-0
-1/2
- 1/4 Correct Answer
- 1–1/4
-1
*1/2 chance with each child. Therefore, ½ X ½ = ¼.
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Questions?
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