Biology 22: Genetics and Molecular Biology

Spring Semester 2008

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Enrollment

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Syllabus Please read through carefully. Brief discussion of key points Summary of key points:

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

Mendelian Genetics & monohybrid crosses Mendel’s 1st Law

Predicting Genetic Outcomes

Punnett Squares Formulas for more complex problems

Other genetic concepts

Test Cross Incomplete dominance

IN LAB ON FRIDAY

Di-hybrid Crosses

Chi-Square Test (you will apply this test to the data collected).

Gregor Mendel

1822-1884

homozygotes

F1 filial

MONOHYBRID CROSS

F2 3:1 ratio

(Or the other way around!)

Reciprocal cross

heterozygotes

MONOHYBRID CROSS

The disappearance of trait in F1?

The re-appearance of trait in 25% of F2?

But why did Mendel observe…

Mendel reasoned that: 2 alleles exist

Alleles are either Dominant Recessive

wrinkled

Round

1. Each trait is governed by two factors – now called 1. Each trait is governed by two factors – now called genesgenes..

2. Genes are found in alternative forms called 2. Genes are found in alternative forms called allelesalleles. .

3. Some alleles are 3. Some alleles are dominantdominant and mask alleles that are and mask alleles that are recessiverecessive..

Thus Mendel proposed:Thus Mendel proposed:

Mendel’s 1st Law Principle of Segregation

A diploid organism has 2 alleles for a characteristic.

The alleles separate and the gamete receives only one

of these alleles.

Alleles separate in equal proportions.

Review concept: Segregation during meiosis

Be able to identify

Homologous chromosomes

Alleles

PUNNETT SQUARE

(backcross example)

1) a method to predict genotypic and phenotypic ratios

Round seeds versus wrinkled seeds

Parentals: Parentals: RR x rrRR x rr

R R r r R R r r

RR

RR

r r r r

F1 x F1: F1 x F1: Rr x RrRr x Rr

R r R r R r R r

½ R½ R

½ r½ r

½ R ½ r ½ R ½ r ¼ RR¼ RR ¼ Rr¼ Rr

¼ Rr¼ Rr ¼ rr¼ rr

RrRr RrRr

RrRr RrRr

2) Rules of probability to predict outcomes

Independent eventsordered

Mutually exclusive

and

either/or

1/6 x1/6 =1/361/6 + 1/6=2/6=1/3

Genetic applications What is the P that two heterozygous parents

have 3 children ALL with albinism?

What is the P that they have 3 children, 1 with albinism and 2 with normal pigmentation? ANN or NAN or NNA

9/64 + 9/64 + 9/64 = 27/64

¼ x ¼ x ¼ = 1/64

What is the P that they have 5 children 2 with albinism and three normal pigmentation? “unordered event”

P = n!/s!t! asbt

P = 5!/2!3! (1/4)2 (3/4)3 = 0.26

Binomial Rule

A “Test Cross” What is a test cross?

Known recessive genotype X unknown genotype

What do we learn from a test cross?

Intermediate expression of traits.

Incomplete dominance

Know these terms Gene Allele

Dominant Recessive

Locus Genotype Homozygote Heterozygote Phenotype Incomplete Dominance Character

End of lectureclick to practice some

problems.

End of lectureclick to practice some

problems.

Practice Problems Try your skills at the following problem.

First some rules & concepts are reviewed.

Problem

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:1. Probability (P) of an event (E) occurring:

P(E) = P(E) = Number of ways that event E can occurNumber of ways that event E can occur

Total number of possible outcomesTotal number of possible outcomes

Eg. P(Rr) from cross Rr x RrEg. P(Rr) from cross Rr x Rr 2 ways to get Rr genotype 2 ways to get Rr genotype 4 possible outcomes 4 possible outcomes P(Rr) = 2/4 = 1/2 P(Rr) = 2/4 = 1/2

Using Probability in Genetic Analysis

2. Addition Rule of Probability – used in an “either/or” situation   

Eg. P(Rr or RR) from cross Rr x Rr Eg. P(Rr or RR) from cross Rr x Rr 2 ways to get Rr genotype 2 ways to get Rr genotype 1 way to get RR genotype 1 way to get RR genotype 4 possible outcomes 4 possible outcomes P(Rr or RR) = 2/4 + 1/4 = 3/4 P(Rr or RR) = 2/4 + 1/4 = 3/4

P(EP(E11 or E or E22) = P(E) = P(E11) + P(E) + P(E22))

Using Probability in Genetic Analysis

3. Multiplication Rule of Probability – used in an “and” situation  

 

Eg. P(wrinkled, yellow) from cross RrYy x RrYy Eg. P(wrinkled, yellow) from cross RrYy x RrYy

P(rr and Y_) = 1/4 x 3/4 = 3/16 P(rr and Y_) = 1/4 x 3/4 = 3/16

P(EP(E11 and E and E22) = P(E) = P(E11) X P(E) X P(E22))

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 Jack and Jill do notdo not have PKU, have PKU, however however eacheach has a sibling with the has a sibling with the disease. What is the probability that disease. What is the probability that Jack and Jill will have a child with Jack and Jill will have a child with PKU?PKU?

What is the probability they will not What is the probability they will not have a child with PKU?have a child with PKU?

PROBLEM

What is the probability they will What is the probability they will notnot

have a child with PKU?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 For Jack and Jill, what is the probability that the first child will have PKU, the second child the first child will have PKU, the second child will not have PKU and the third child will have will not have PKU and the third child will have PKU?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