Outline for today’s lecture (Ch. 14, Part I)

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Outline for today’s lecture (Ch. 14, Part I) Ploidy vs. DNA content The basis of heredity ca. 1850s Mendel’s Experiments and Theory Law of Segregation Law of Independent Assortment Introduction to Probability

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Outline for today’s lecture (Ch. 14, Part I). Ploidy vs. DNA content The basis of heredity ca. 1850s Mendel’s Experiments and Theory Law of Segregation Law of Independent Assortment Introduction to Probability. Reminder: Homologous chromosomes. Pair at meiosis (all pairs) - PowerPoint PPT Presentation

Transcript of Outline for today’s lecture (Ch. 14, Part I)

Page 1: Outline for today’s lecture (Ch. 14, Part I)

Outline for today’s lecture (Ch. 14, Part I)

• Ploidy vs. DNA content

• The basis of heredity ca. 1850s

• Mendel’s Experiments and Theory

– Law of Segregation

– Law of Independent Assortment

• Introduction to Probability

Page 2: Outline for today’s lecture (Ch. 14, Part I)

Reminder: Homologous chromosomes

-Pair at meiosis (all pairs)

-Same sequence (except sex chromosomes)

Page 3: Outline for today’s lecture (Ch. 14, Part I)

Ploidy vs. DNA content in Meiosis

4

2

1

“C”

(D

NA

per

chr

omos

om

e)

G1 S G2 Meiosis I Meiosis II

Diploid Haploid

Diploid: Contains two chromosomes from a homologous pairone from each parent

Haploid: Contains only one chromosome from a homologous pair

Page 4: Outline for today’s lecture (Ch. 14, Part I)

The Nature of Heredity, ca. 1859

• Observation: Offspring generally intermediate in phenotype (“trait value”) between those of parents

– Obvious example: Human children with one African and one Nothern European parent

• Proposed explanation: “Blending Inheritance”– Genetic material miscible, like paint:

• Black + White = Gray

• Tall + Short = Medium

• Etc.

Page 5: Outline for today’s lecture (Ch. 14, Part I)

Blending Inheritance: A logical difficulty

• Variation reduced every generation

• Ultimate consequence is a homogeneous population

– At odds with reality

• How to explain variation?

– “Sports” (Mutation in modern parlance)

Page 6: Outline for today’s lecture (Ch. 14, Part I)

Gregor Mendel: The Origin of Genetics

• Austrian farm boy, entered Augustinian monastary in 1843

• Attended University of Vienna in early 1850s

– Learned two things about science• Do experiments• Analyze your data (i.e., mathematically)

• ~1857, began an experimental program to investigate the basis of inheritance (i.e., heredity) with peas

Page 7: Outline for today’s lecture (Ch. 14, Part I)

Mendel’s Experiments

• Peas were a fortuitous study organism for several reasons:

– Many variable characters (e.g., flower color, seed shape, seed color, etc.)

– Many varieties that bred “true” for particular traits (e.g., purple flowers, round seeds, etc.)

– Easy to do controlled crosses, both “self” and outcross

Page 8: Outline for today’s lecture (Ch. 14, Part I)

Mendel’s Experiments – Choice of characters

• Used only discrete characters, i.e., “either-or”, not continuous

freq

uen

cy

0.5

1.0

0.0

ColorBlack

White

Height

freq

uen

cy

0.5

1.0

0.0

2.0 3.0 4.0

Page 9: Outline for today’s lecture (Ch. 14, Part I)

Mendel’s Experiments – Breeding design

1. Start with lines that breed true for different traits, e.g., purple and white flowers

2. First generation of a cross is called P (“parental”)

3. Offspring are F1 (“filial”)

4. Grand-offspring are F2

Self

Self

Self

Self

P

F1

F2

Page 10: Outline for today’s lecture (Ch. 14, Part I)

Mendel’s Experiments – Breeding design

XP

F1

F2

1. Cross two true-breeding lines (purple, white)

2. Self F1s

3. Observe phenotypes of MANY F2 offspring and COUNT THEM

Self

Page 11: Outline for today’s lecture (Ch. 14, Part I)

Mendel’s Experiments – Results

XP

F1

1. Cross two true-breeding lines (purple, white)

2. F1s ALL PURPLE

• What do we expect if “blending inheritance”?

Page 12: Outline for today’s lecture (Ch. 14, Part I)

Mendel’s Experiments – Breeding design

XP

F1

F2

1. Cross two true-breeding lines (purple, white)

2. F1s ALL PURPLE

3. Self F1s

4. F2s are NOT all purple – – 705 purple– 224 white– i.e., ~ 3:1 purple : white

Self

Page 13: Outline for today’s lecture (Ch. 14, Part I)

Mendel’s Experiments – Conclusions

XP

F1

F2

1. “Heritable Factor” (i.e., gene) for white flowers did not disappear in the F1, but only the purple “factor” affected flower color.

"Particulate" inheritance

1. Purple is “dominant” and white is “recessive”

Self

Page 14: Outline for today’s lecture (Ch. 14, Part I)

Mendel’s Experiments – Important Points

XP

F1

F2

1. Followed the pattern of inheritance for multiple generations (i.e., > 1)

– What if the experiment terminated after F1?

2. Quantitative Analysis– Many 19th century

botanists would have said “some white flowers reappeared in F2”

3. Mendel was lucky!

Self

Page 15: Outline for today’s lecture (Ch. 14, Part I)

Mendel’s Experiments in modern genetic terms

1. Alternative versions of genes account for variation in inherited characters

– Alternative versions of genes are “alleles”

– Alleles reside at the SAME genetic locus

2. Relationship between alleles, chromosomes, and DNA

– DNA at a locus varies in sequence

– Sequence variants cause different phenotypes (e.g., purple and white flowers)

Page 16: Outline for today’s lecture (Ch. 14, Part I)

Mendel’s Experiments in modern genetic terms

• Diploid individuals have homologous pairs of chromosomes, one from each parent

• An individual inherits one allele from each parent

• Alleles may be same or different

• If different, the dominant allele determines the organism’s phenotype

“Flower-color locus”

Purple allele

White allele

Page 17: Outline for today’s lecture (Ch. 14, Part I)

Mendel’s Experiments in modern genetic terms

• The two alleles at a locus segregate during gamete production

– Each gamete gets only one of the two alleles present in somatic cells

– Segregation corresponds to the different gametes in meiosis (I or II?)

Page 18: Outline for today’s lecture (Ch. 14, Part I)

Recall Meiosis I – Metaphase I

• What about crossing-over?

Page 19: Outline for today’s lecture (Ch. 14, Part I)

Mendel’s “Law of Segregation”

• If an individual has identical alleles at a locus (i.e., is true-breeding), that allele is present in all its gametes

• If an individual has two different alleles at a locus, half its gametes receive one allele, half receive the other allele

All

half

half

Page 20: Outline for today’s lecture (Ch. 14, Part I)

Genetic Terminology –

• If a diploid individual has two copies of the SAME ALLELE at a locus (i.e., it got the same allele from mom and dad), it is a HOMOZYGOTE

• If it has two different ALLELES at a locus (got a different allele from mom than from dad) it is a HETEROZYGOTE

• The genetic makeup at a locus (or loci) is the individual’s GENOTYPE

• An organism's Traits comprise its PHENOTYPE

Page 21: Outline for today’s lecture (Ch. 14, Part I)

Mendel’s Law of segregation: a test

Genotype: PP x pp

Gametes: P p

Genotype: Pp

Gametes: 1/2 P, 1/2 p

Genotype: 1/4 PP, 1/2 Pp, 1/4 pp

Phenotype: 3 purple, 1 white

XP

F1

F2

Self

Page 22: Outline for today’s lecture (Ch. 14, Part I)

Mendel’s Law of segregation: a test

• Phenotype:

• Genotype: Pp Pp

• Ova (female gametes) 1/2 P, 1/2 p• Sperm (male gametes) 1/2 P, 1/2 p

• Half of male gametes will be P. Of those, half will unite with an ovum that is P.

• Thus, the frequency of PP in the F2 is: (1/2)(1/2 ) = 1/4• Frequency of pp = (1/2)(1/2) = 1/4, • Frequency of Pp = 2(1/2)(1/2) = 1/2

X

F1

Page 23: Outline for today’s lecture (Ch. 14, Part I)

The “Punnett Square”

• Gamete genotypes of one parent given as columns

• Gamete genotypes of other parent given as rows

• Offspring genotypes given as cells in the table

• Each cell has equal frequency (here = 1/4)

Sperm genotype

Eg

g g

eno

typ

e

Male Parent

P

P

p

pPp

Pp

pppP

PpPP

Female Parent

Page 24: Outline for today’s lecture (Ch. 14, Part I)

The “Punnett Square”

• Note that in this cross there are TWO ways to get a heterozygote

• P from mom, p from Dad• p from mom, P from Dad

• Frequency of heterozygotes = 1/4 Pp + 1/4 pP = 1/2

Sperm genotype

Eg

g g

eno

typ

e

Male Parent

P

P

p

pPp

Pp

pppP

PpPP

Female Parent

Page 25: Outline for today’s lecture (Ch. 14, Part I)

The “Testcross”

• Individuals homozygous for a dominant allele have the same phenotype as heterozygotes

• To determine the genotype of an individual, cross it to a known homozygous recessive

• What is the phenotypic ratio among these offspring?

• What is the genotype of the unknown individual?

Sperm genotype

Eg

g g

eno

typ

e

Male Parent

P

p

?

ppp

P-

Female Parent

Page 26: Outline for today’s lecture (Ch. 14, Part I)

The Law of Independent Assortment or Why Mendel was so Lucky

• Mendel's next step was to cross plants that bred true for each of TWO traits, e.g....

• seed shape (Round or wrinkled, Round dominant; R/r)

• seed color (Yellow or green, yellow dominant; Y/y)

• Parental cross: RRYY x rryy

• F1 are Round, Yellow (RrYy)

• Self F1s...

XP

F1

Page 27: Outline for today’s lecture (Ch. 14, Part I)

The "Dihybrid Cross" - Dependent Assortment

• Hypothesis: Loci ("Traits" to Mendel) assort together ("dependently")

• If a gamete has an R allele, it also has a Y allele (recall P generation was RRYY, rryy)

• If a gamete has an r allele it also has a y allele

• What are the expected frequencies of F2 phenotypes?

Male F1 parent = RrYy

Fem

ale

F1

par

ent

= R

rYy RY ry

ry

RY

Page 28: Outline for today’s lecture (Ch. 14, Part I)

The "Dihybrid Cross" - Dependent Assortment

• Hypothesis: Loci assort together ("dependently")

• If a gamete has an R allele, it also has a Y allele (recall parents were RRYY, rryy)

• If a gamete has an r allele it also has a y allele

• What are the expected frequencies of F2 phenotypes?

Male F1 parent = RrYy

Fem

ale

F1

par

ent

= R

rYy RY ry

ry

RYRRYY

rRyY rryy

RrYy

Page 29: Outline for today’s lecture (Ch. 14, Part I)

The "Dihybrid Cross" - Dependent Assortment

• Predict 3/4 round, yellow, 1/4 wrinkled, green

• NOT WHAT MENDEL OBSERVED!

Male F1 parent = RrYy

Fem

ale

F1

par

ent

= R

rYy RY ry

ry

RYRRYY

rRYy rryy

RrYy

Page 30: Outline for today’s lecture (Ch. 14, Part I)

The "Dihybrid Cross" - Independent Assortment

• Four combinations of alleles in gametes

• All are equally likely

• Expect traits in 9:3:3:1 ratio

• THIS IS WHAT MENDEL OBSERVED

Male F1 parent = RrYy

Fem

ale

F1

par

ent

= R

rYy

RY Ry rY ry

RY

Ry

rY

ry

RRYY RRYy RrYY RrYy

RRYy RRyy RrYy Rryy

RrYY Rryy rrYy rryy

RrYY RrYy rrYY rrYy

Page 31: Outline for today’s lecture (Ch. 14, Part I)

Mendel's Laws

1. The Law of Segregation - ONE LOCUS

• If the locus is heterozygous, half the gametes get one allele, half the gametes get the other allele

2. The Law of Independent Assortment - MULTIPLE LOCI

• Alleles at each locus segregate independently of alleles at other loci

• (When is this not true? or Why was Mendel so lucky?)

Page 32: Outline for today’s lecture (Ch. 14, Part I)

Introduction to Probability Theory

• Independent Events - if the outcome of one event does not depend on the outcome of some other event

– e.g., rolls of a die, flips of a coin, segregation of loci on different chromosomes

• The probability of BOTH of two events happening is the product of the probability of each event happening independently.

– Formally, Pr(A and B) = Pr(A) x Pr(B)

– e.g., Pr(two heads on two flips) = Pr(1st flip heads) x Pr(2nd flip heads)

Page 33: Outline for today’s lecture (Ch. 14, Part I)

Introduction to Probability Theory

• Probability of EITHER of two events happening is the sum of the probability of each event happening independently

– Formally, Pr(A or B) = Pr(A) + Pr(B)

– e.g., Pr(one head on two flips) = Pr(head,tail or tail,head)

Pr(1st flip tails)*Pr(2nd flip heads) = (1/2)(1/2) =1/4Pr(1st flip heads)*Pr(2nd flip tails) = 1/4

– 1/4 + 1/4 = ½

– Pr(1-locus heterozygote) = Pr(Aa) + Pr(aA) = ¼ + ¼ = ½

Page 34: Outline for today’s lecture (Ch. 14, Part I)

For tomorrow...

• Mendelian Genetics, continued

• Pedigree analysis

• Read the rest of Ch. 14