Download Chapter 3 Mendelism: The Basic Principles of Inheritance

3823) Which of the following genes have been
implicated in hereditary forms of breast cancer?
BRCA1, b) BRCA2, c) RB, d) phMSH2, e) BRCA1
and BRCA2
1123) Which of the following properties is not
characteristic of a malignant cell?
Answer: e
a) It no longer undergoes cell cycle regulation and
growth arrest.
b) It has undergone mutations in a c-onc and in other
genes.
c) It cannot leave its site of growth in the tumor.
d) It may have undergone a chromosome
Answer: c
rearrangement.
e) All of these are correct.
4353) Which of the following can be determined using a cloned
gene?
1. A gene’s nucleotide sequence
2. A gene’s function
3. A gene’s amino acid sequence
a) 1, b) 2, c) 3, d) 1 and 2, e) All of these
Answer: e
1009) What method allows bacteria to protect endogenous
restriction sites from being cleaved by restriction enzymes?
a) Methylation
b) Amylation
c) Glycosylation
d) Methylation and Amylation
e) None of these
Answer: a
Chapter 3
Mendelism: The Basic Principles
of Inheritance
© John Wiley & Sons, Inc.
Chapter 23
--The theory of allele frequency
--Random genetic Drift
Chapter 24
--Genetic variation
--Molecular
--Speciation
Chapter Outline
Mendel’s Study of Heredity
Applications of Mendel’s Principles
Testing Genetic Hypotheses
Mendelian Principles in Human
Genetics
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The Garden Pea: true-breeding
Pisum Sativum
--anther: sperm-pollen
--ovary: eggs
Petals of the flowers are
close down
--self-fertilization
----inbreeding and uniform
Cross-fertilization
Two plants with different
characteristics
(Tall and short)
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Monohybrid Crosses:
single phenotype
Heritable (genetic) factors (genes)
--Dominant (expressed)
--Recessive (latent or not
expressed)
(Alleles)
--parental strain: two copies of a
gene (diploid and homozygous)
--gametes has a single copy
(haploid)
--offspring strain: two different copies
of a gene (heterozygous)© John Wiley & Sons, Inc.
hybrid
Phenotype: physical appearance
Genotype: allelic constitution of each strain
D allele of dominant gene
d allele of recessive gene
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Ratio of progeny ~3:1
Vocabulary for Monohybrid Crosses
 True-breeding
 Cross-fertilize
 Monohybrid cross
 Dominant
 Recessive
 Gene, genetic factor ( fragment of
DNA) that determines a
characteristic.
 Allele,alternative forms of a gene
 Parental
 Filial
 Segregate
 Characteristic, an attribute or
feature of a particular gene
 Homozygous, an individual
organism possessing two of the
same alleles at a locus
 Heterozygous, an individual
organism possessing two
different alleles at a locus
 Genotype, set of alleles that
an individual organism
possesses
 Phenotype (trait), the
appearance or manifestation of
a character
 Locus, specific place on a
chromosome occupied by an
allele
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Symbolic Representation of a
Monohybrid Cross
2
2
75%
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Mendel’s Principles
 The Principle of Dominance: In a
heterozygote, one allele may conceal
the presence of another. (This means the
dominant allele is observed in the phenotype)
 The Principle of Segregation: In a
heterozygote, two different alleles
segregate from each other during the
formation of gametes. (This means that the pair
of alleles encoding the traits in each parental plant had
separated or segregated from one another during meiosis)
 The Principle of Dominance: In a
heterozygote, one allele may conceal the
presence of another.
--genetic function
--control phenotype
--regulate physiological consequences
 The Principle of Segregation: In a
heterozygote, two different alleles
segregate from each other during the
formation of gametes.
--genetic transmission
-- may control phenotype
--may regulate physiological consequences
Dihybrid Crosses
How are genes transmitted?
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Diploid
2
Haploid
25%
75%
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Comparison of Observed and
Expected Results in the F2
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Mendel’s Principles
The Principle of Independent
Assortment: The alleles of different
genes segregate,(assort) independently
of each other (The traits in the offspring of this crosses did not
always match the combinations of traits in the parental organisms).
In humans, diploid cells contain 46 chromosomes,
23 female chromosomes
23 male chromosomes
During meiosis, the pairs of similar homologous chromosome are divided in half to
form haploid cells, and this separation is random.
Recombination scrambles pieces of maternal and paternal genes, which ensures that genes
assort independently from one another, except genetic linkage
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The Principle of
Dominance
The Principle of
Segregation
Applications of Mendel’s
Principles
Mendel’s principles can be used to
predict the outcomes of crosses between
different strains of organisms.
Three methods to predict outcomes
– The Punnett Square Method
– The Forked-Line Method
– The Probability Method
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The Punnett Square Method
one or two genes
Allele
A
a
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The Forked-Line Method for an Intercross
two or more genes
Trihybrid heterozygous cross ….
--independent sorting
--monohybrid cross…Dd x Dd....Ratio 3:1
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The Forked-Line Method for a Testcross
Heterozygous vs homozygous recessive
Dd
D
x dd
d
d Dd dd
d Dd dd
Ratio 1:1
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Probability
 The probability of an event is the frequency of that event in the sample.
 Express the likelihood of the occurrence of a particular event.
 For a coin toss:
– The probability of heads is 1/2. =0.5
– The probability of tails is 1/2.
 For two heterozygotes (Gg) producing an offspring:
– The probability of the incorporation of G in each gamete in GG zygote is
1/2 and 1/2. 1/2 x 1/2=1/4
– The probability of the incorporation of G or g in Gg is 1/2 because there two
possible ways “Gg or gG”... 1/4+1/4= 1/2 but not in the context of the total
probability (4)= 1/4
– The probability of gg is 1/4. (1/2 x1/2)
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Two rules of probability
The Multiplicative (Product) Rule
 If the events A and B are independent, the
probability that they will occur together, denoted
P(A and B), is P(A)  P(B).
 States that the probability of two or more
independent events occurring together is
calculated by multiplying their independent
probabilities
 Example: probability of drawing the ace of hearts [an
ace (A) AND a heart (H)], P(A and H)
P(A) = 4/52 P(H) = 1/4
P(A)  P(H) = 4/52  1/4 = 1/52
The Additive (Addition) Rule
 If the events A and B are independent, the
probability that at least one of them occurs,
denoted P(A or B), is given by P(A) + P(B)
 States that the probability of any one of two or or
mutually exclusive events is calculated by adding
their probabilities
 Example: Probability of a drawing EITHER an ace
OR a king, P(A or H)
P(A) = 4/52 P(H) = 4/52
P(A or H) = (4/52)  (4/52) = 8/52
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What is the change of zygote
to be AA?
1/2 x 1/2= 1/4
What is the change of zygote
to be aa?
1/2 x 1/2= 1/4
Phenotype
Recessive 1/4
Dominant 1/2 +1/4=3/4
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What fraction of progeny will be homozygous
for all four recessive genes?
1/4 x 1/4 x 1/4 x 1/4= 1/256
What fraction of progeny will be homozygous
for all four genes?
D and d
1/2 x 1/2 x 1/2 x 1/2= 1/16
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Mathematical definition of probability
P: a/n
where
a = the outcome of interest
n = the number of possible outcomes.
Examples
• P(getting a head on a coin toss) =
1/2
• P(rolling a 1 on a die) = 1/6
• P(drawing the ace of spades) = 1/54
• P(dominant offspring from the cross
Aa x Aa) = 3/4
• If P = 0, the event is impossible
• P(rolling a 7 on a standard die) = 0
• If P = 1, the event is certain
• P(rolling a number less than 10) = 1
Definition of binomial formula
(hybrid-heterozygous)
P: n! ps qt
s!t!
•
•
•
•
•
•
n = total number of events
s = number of type A events
t = number of type B events
p = probability of type A event
q = probability of type B event
Always true: s + t = n, p + q = 1
! Factorial
Example 3!= 3x2x1
Testing Genetic Hypotheses
Hypothesis: a well-formulated
scientific idea
Data collected from observations or
from experimentation enable scientists
to test hypotheses.
Genetics: Are the results of a cross
consistent with a hypothesis?
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The Chi-Square Test
Predict Expected numbers based on hypothesis.
Determine the degrees of freedom. (n-1: different
expected phenotype)
Compare the 2 statistic to the critical value
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Example:
Mendel’s Dihybrid Cross
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Example: A Dihybrid Cross
with Campion (Hugo deVries)
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P: 0.05%
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Comparison to the Critical Value
Mendel’s Dihybrid Cross:




2 = 0.51
Degrees of Freedom = 4  1 = 3
Critical Value = 7.815
Hypothesis?
DeVries’ Dihybrid Cross:




2 = 22.91
Degrees of Freedom = 4  1 = 3
Critical Value = 7.815
Hypothesis ?
© John Wiley & Sons, Inc.
Comparison to the Critical Value
Mendel’s Dihybrid Cross:




2 = 0.51
Degrees of Freedom = 4  1 = 3
Critical Value = 7.815
Hypothesis is correct
DeVries’ Dihybrid Cross:




2 = 22.91
Degrees of Freedom = 4  1 = 3
Critical Value = 7.815
Reject the Hypothesis
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Mendelian Principles in
Human Genetics
Obstacles to Human Genetic Analysis
– Incomplete family records
– Small number of progeny
– Uncontrolled environment
Despite these obstacles, many human
genetic traits have been described.
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Pedigree Conventions
Pedigrees are
diagrams that show
the relationships
among the
members of a
family.
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Inheritance of a Dominant Trait
 Every individual who carries
the dominant allele
manifests the trait.
 Every affected individual is
expected to have at least
one affected parent.
 If a dominant trait is
associated with reduced
viability or fertility, most
people who show the trait
are heterozygous, and half
their children should inherit
the condition.
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Inheritance of a Recessive Trait
 Recessive traits may
occur in individuals
whose parents are not
affected.
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Mendelian Segregation in Human Families
Binomial probabilities: affected and unaffected
Definition of binomial formula
P: n! ps qt
s!t!
•
•
•
•
•
•
n = total number of events
s = number of type A events
t = number of type B events
p = probability of type A event
q = probability of type B event
Always true: s + t = n, p + q = 1
! Factorial
Example 3!= 3x2x1
Genetic Counseling:
Non-polypoid Colorectal Cancer
--dominant mode of the disease
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Albinism
--recessive mode of the disease
1/4 albinism
3/4 no-disease
Example 1:
3 children with disease
P=(1/4)3=1/64
Example 2:
3 children
1 with disease
2 normal
P= 3/2 (3/4)2 (1/4)=0.21
Example 3:
5 children
2 with disease
3 normal
P= 10 (3/4)3 (1/4) 2 =0.26
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4093) A recessive trait is one that is:
a) Masked by a dominant trait,
if a dominant trait is present in the genotype
b) Not masked by
any other trait present in the genotype
c) Masked by another
recessive trait, if another recessive trait is present in the
Genotype
d) All of these
e) None of these
Answer: a
5097) A monohybrid cross is one in which:
a) Two traits are being studied at the same time
b) One trait is being studied
c) Two organisms are being studied at the same time
Answer: b
d) One organism is being studied
e) None of these
R
S
T
Determine P for :
1- Is R heterozygous carrier? 2/3
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