Biology 22: Genetics and
Molecular Biology
Spring Semester 2008
Enrollment
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Biology 22 is currently full.
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Please sign-in on the sheet to your left (next to
your name) if you are in the class.
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Please sign-in on the sheet to your right if you
want to add the class; include lab section.
Enrollment
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If you want to add, you must show up on the first lab day to see if a spot is
available.
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If a places are available, names will be chosen on Friday in lab.
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All decisions to determine if places are available and the handing out of “add
numbers” will be completed by this Friday.
Syllabus
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Please read through carefully.
Brief discussion of key points
Summary of key points:
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Bio 21 (or equivalent) & Chem 11 pre-requisites
English 1 advisory
Bio 22 covers genetics and molecular
No make-up exams or assignments
Ch 3 Basic Principles of Heredity
Outline
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Mendelian Genetics & monohybrid crosses
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Predicting Genetic Outcomes
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Mendel’s 1st Law
Punnett Squares
Formulas for more complex problems
Other genetic concepts
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Test Cross
Incomplete dominance
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IN LAB ON FRIDAY
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Di-hybrid Crosses
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Chi-Square Test (you will apply this test to the data collected).
Gregor Mendel
1822-1884
(Or the other way around!)
MONOHYBRID CROSS
Reciprocal cross
homozygotes
F1 filial
heterozygotes
F2
3:1 ratio
MONOHYBRID CROSS
But why did Mendel observe…
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The disappearance of trait in F1?
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The re-appearance of trait in 25% of F2?
Mendel reasoned that:
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2 alleles exist
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Alleles are either
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Round
Dominant
Recessive
wrinkled
Thus Mendel proposed:
1. Each trait is governed by two factors – now called genes.
2. Genes are found in alternative forms called alleles.
3. Some alleles are dominant and mask alleles that are recessive.
Mendel’s
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st
1
Law
Principle of Segregation
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A diploid organism has 2 alleles for a characteristic.
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The alleles separate and the gamete receives only one
of these alleles.
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Alleles separate in equal proportions.
Review concept: Segregation during meiosis
Be able to identify
Homologous chromosomes
Alleles
PUNNETT SQUARE
1) a method to predict genotypic and phenotypic ratios
(backcross example)
Round seeds versus wrinkled seeds
Parentals:
RR x rr
R
R
R
R
r
r
r
r
Rr
Rr
Rr
F1 x F1:
Rr x Rr
Rr
R
½R
½r
r
R
½R
½ r
¼ RR ¼ Rr
¼ Rr
¼ rr
r
2) Rules of probability to predict outcomes
1/6 x1/6 =1/36
1/6 + 1/6=2/6=1/3
either/or
and
Independent events
ordered
Mutually exclusive
Genetic applications
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What is the P that two heterozygous parents
have 3 children ALL with albinism? ¼ x ¼ x ¼ = 1/64
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What is the P that they have 3 children, 1 with
albinism and 2 with normal pigmentation?
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ANN or NAN or NNA
9/64 + 9/64 + 9/64 = 27/64
Binomial Rule
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What is the P that they have 5 children 2 with albinism
and three normal pigmentation? “unordered event”
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P = n!/s!t! asbt
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P = 5!/2!3! (1/4)2 (3/4)3 = 0.26
A “Test Cross”
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What is a test cross?
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Known recessive genotype X unknown genotype
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What do we learn from a test cross?
Intermediate expression of traits.
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Incomplete dominance
Know these terms
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Gene
Allele
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Dominant
Recessive
Locus
Genotype
Homozygote
Heterozygote
Phenotype
Incomplete Dominance
Character
End of lecture
click to practice some
problems.
Practice Problems
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Try your skills at the following problem.
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First some rules & concepts are reviewed.
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Problem
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Powerpoint will be updated with answers and
posted on ecompanion.
Note: When solving genetics
problems
1.
Convert parental phenotypes to genotypes
1.
Use Punnett Square to determine genotypes of offspring
1.
Convert offspring genotypes to phenotypes
Using Probability in Genetic
Analysis
1. Probability (P) of an event (E) occurring:
P(E) = Number of ways that event E can occur
Total number of possible outcomes
Eg. P(Rr) from cross Rr x Rr
2 ways to get Rr genotype
4 possible outcomes
P(Rr) = 2/4 = 1/2
Using Probability in Genetic
Analysis
2. Addition Rule of Probability – used in
an “either/or” situation
P(E1 or E2) = P(E1) + P(E2)
Eg. P(Rr or RR) from cross Rr x Rr
2 ways to get Rr genotype
1 way to get RR genotype
4 possible outcomes
P(Rr or RR) = 2/4 + 1/4 = 3/4
Using Probability in Genetic
Analysis
3. Multiplication Rule of Probability –
used in an “and” situation
P(E1 and E2) = P(E1) X P(E2)
Eg. P(wrinkled, yellow) from cross RrYy x RrYy
P(rr and Y_) = 1/4 x 3/4 = 3/16
Using Probability in Genetic
Analysis
4. Conditional Probability:
Calculating the probability that
each individual has a
particular genotype
INTRODUCTION TO PROBLEM
Phenylketonuria (PKU) is a
genetic disorder that is
characterized by an inability of
the body to utilize the essential
amino acid, phenylalanine.
PROBLEM
Jack and Jill do not have PKU,
however each has a sibling with the
disease. What is the probability that
Jack and Jill will have a child with
PKU?
What is the probability they will not
have a child with PKU?
PROBLEM
What is the probability they will not
have a child with PKU?
Using Probability in Genetic
Analysis
5. Ordered Events: use Multiplication Rule
For Jack and Jill, what is the probability that
the first child will have PKU, the second child
will not have PKU and the third child will have
PKU?
Using Probability in Genetic
Analysis
6. Binomial Rule of Probability
Out of 3 children born to Jack and Jill,
what is the probability that 2 will have PKU?
The End