genetics chapter 2 part 1 (1)

Genetics Chapter 2 Part 1

Dr. Tricia Hardt Smith

hardtta@vcu.edu

State of genetics in early State of genetics in early 1800’s1800’s

What is inherited?

How is it inherited?

What is the role of chance in heredity?

Johann Gregor Mendel(1822-1884)

Born to simple farmers in the Czeh Republic

1843: Augustinian monastery (Brno)

1851-53: University of Vienna; Physics Institute

Mathmatics, chemistry, entomology, paleotology, botany, plant physiology

1856-1863: Pea plant breeding experiments

1866: Published his findings

1900: Three botanists (De Vries, von Tschermak & Correns) independently conduct the same experiments, come across Mendel’s paper and draw attention to his work.

Mendelian genetics

• Mendel’s work unnoticed until 1900’s

• Introduced concept of “units of inheritance”

• When correlated with cytological data → Transmission genetics was born

Mendel’s workplace

Fig. 2.5

Chapter 2 Opener

Why pea plants?

• Reproduce well.• Each seed is a new individual, can

measure the characteristics of a large number of offspring after one breeding season

• Grow to maturity in single season

Easy to grow and hybridize artificially

Mendel’s Approach

• Mendel obtained 34 different varieties of peas from local suppliers and examined the characteristics of each

• He identified 14 strains representing seven specific traits each with two forms that could be easily distinguished. He spent two years making sure these varities bred true.

• He worked with these strains for 5 years, determining how each character was inherited

Jos A. Smith

1900 - Carl Correns, Hugo deVries, and Erich von

Tschermak rediscover and confirm Mendel’s laws.

Mendel published in 1866, was not

appreciated in his lifetime.

Mendel’s Approach Followed the Modern Scientific Method

1. Make initial observations about a phenomenon or process

2. Formulate a testable hypothesis

3. Design a controlled experiment to test the hypothesis

4. Collect data from the experiment

5. Interpret the experimental results, comparing them to those expected under the hypothesis

6. Draw a conclusion and reformulate the hypothesis if necessary

One of Mendel’s strengths was his

careful experimental

design

Five Critical Experimental Innovations

• There were five features of Mendel’s breeding experiments that were critical to his success

• Controlled crosses

• Use of pure breeding strains

• Selection of dichotomous traits

• Quantification of results

• Use of replicate (repeated), reciprocal, and test crosses

• Luck?

Controlled Crosses Between Plants

• Pea plants are capable of self-fertilization and artificial cross-fertilization

• Self-fertilization occurs naturally

• Cross-fertilization involves removing the anthers from a flower and introducing pollen of the desired type with a small brush

From Peirce Genetics

Pure-Breeding Strains to Begin Experimental Crosses

• Mendel took 2 years prior to beginning his experiments to establish pure-breeding (or true-breeding) strains

• Each experiment began with crosses between two pure-breeding parental generation plants (P generation) that produced offspring called F1 (first filial generation)

Monohybrid Crosses

Smooth Seeds

Wrinkled Seeds

Female Male

Monohybrid Cross: a cross-pollination involving two true-breeding lines that differ for only one trait

“P”

“F1”Progeny:

All progeny had smooth seed!“First Filial generation”

“Parental generation”

Parents:

All progeny had same PHENOTYPE: “the form that is shown”

Monohybrid Crosses

Smooth Seeds

Wrinkled Seeds

Female Male

Two possible Hypotheses

Hypothesis 2: The child’s phenotype is determined by the mother’s phenotype

Hypothesis 1: The smooth phenotype is “dominant” to the wrinkled phenotype

“P”

“F1”How could you

differentiate between these possibilities?

Mendel Made Reciprocal CrossesReciprocal Cross: Repeating a particular genetic cross

but with the sexes of the two parents switched

All F1 had smooth seed.

Smooth Seeds

Wrinkled Seeds

Female Male

- The smooth trait is “dominant” to the wrinkled trait

Conclusion

- Phenotype is not determined by the mother’s phenotype

“P”

“F1”

• The trait shown by the F1 offspring was called the dominant phenotype (round peas, e.g.)

• The other trait not apparent in the F1 was called the recessive phenotype (wrinkled)

• When F1 were crossed, 75% of the resulting F2 had the dominant trait, but the recessive trait reappeared in the other 25%

Alleles

• Mendel’s results rejected the blending theory of heredity

• Theorized that plants carry two discrete hereditary units for each trait, alleles; a plant receives one of these in the egg and the second in pollen

• Together the two alleles for each trait determine the phenotype of the individual

Alleles

Phenotype

Homozygous and Heterozygous Individuals

• Pure-breeding individuals, like Mendel’s parent plants, have identical copies of the two alleles for a trait (homozygous individual)

• The F1 plants had different alleles from each parent and were heterozygous

Homozygous (TT & tt)

Heterozygous (Tt)

• A 3:1 phenotypic ratio is predicted for the F2 produced by a monohybrid cross

• A 1:2:1 genotypic ratio is also predicted (¼ G/G, ½ G/g, ¼ g/g)

Now that we have a theory, we can do

some real predicting!

Now that we have a theory, we can do

some real predicting!

Punnett Square

• The alleles (in gametes) carried by one parent are arranged along the top of the square and those of the other parent, down the side

• The results expected from random fusion of the gametes are placed within the square

R r

R

r

RR Rr

Rr rr

Punnett Square

Mendel’s Results Revisited: F1

Smooth Seeds

Wrinkled Seeds

“AA”

“aa”

Gametes possible: “A” or “A”

Gametes possible: “a” or “a”

“Aa”

Smooth

Aa Aa

AaAa

a a

A

A

♂ Gametes

♀ G

ame

tes

Use a “Punnett Square” to determine all possible progeny genotypes

♂♀

Explains why all progeny were smooth

Genotype?Genotype?

Br rr

A

B

C

D

E

What is the predicted cross of homozygous recessive red and heterozygous dominant brown?

A.All brownB.3 brown, 1 redC.2 brown, 2 redD.1 brown, 3 red

E.All red

All bro

wn

3 brown, 1

red

2 brown, 2

red

1 brown, 3

red

All red

20% 20%20%20%20%

Br rr

A

B

C

D

E

Br

Br

rr

rr

B r

r

r

Red + Brown = Blond! (sometimes?)Not all traits are dominant/reccessive!

More in upcoming lectures!

Punnett Square Practice Problems!

Chapter 2: Problems 2 & 3

Mendel’s First Law

• Mendel used his theory of particulate inheritance to formulate the law of segregation (Mendel’s first law)

• Alleles are separated into gametes. Gametes randomly combine to create progeny in predictable proportions.

• Hypothesis!: Mendel expected that half of the gametes of heterozygous F1 individuals would carry the dominant allele and half the recessive

• How can we test this?

Genotype?Genotype?

Conclusion: all F1 plants are heterozygous!

Conclusion: all F1 plants are heterozygous!

The Test Cross

• Allows to distinguish genotype of individual expressing dominant phenotype by crossing it with homozygous recessive individual

What other predictions can we test?

• Mendel’s hypothesis predicts that F2 plants with the dominant phenotype can be homozygous or heterozygous

• The heterozygous state (2/3) is twice as likely as the homozygous state (1/3)

• HOW? Mendel used a self-fertilization experiment to test the predictions of the hypothesis

F3 generation

• Hypothesis confirmed:

• 1/3 of plants were homozygous and breed true

• 2/3 of heterozygous F2 plants generated a 3:1 ratio of dominant:recessive phenotype among their progeny

WHAT HAPPENS IF WE STUDY TWO TRAITS?

Will the presence of one charastics affect the prescent of another?

Dihybrid-Cross Analysis of Two Genes

• To study the simultaneous transmission of two traits, Mendel made dihybrid crosses between organisms that differed for two traits

• He began each cross with pure-breeding lines (e.g., RRGG and rrgg) and produced F1 that were heterozygous for both traits (e.g., RrGg).

• If assortment is random, four gametes should be equally likely in the F1 (e.g., RG, Rg, rG, rg)

2.3 Dihybrid and Trihybrid Crosses

• How can we calculate the crosses of two or more traits at the same time?

• Dihybrid Punnet Square

• Forked Diagram

An Aid to Prediction of Gamete Frequency

• The forked-line diagram is used to determine gamete genotypes and frequencies

Let’s give it a try!

• Self Fertilization of a heterozygous yellow, round pea?

• Round (R) is dominant to wrinkled (r)

• Yellow (G) is dominant to green (g) • What does the dihybrid

Punnet square look like?

• What does the forked diagram look like?

F2 ?

Independent Assortment of Alleles from the RrGg × RrGg Cross

• Mendel predicted that alleles of each locus unite at random to produce the F2, generating

• round, yellow R-G- (¾)(¾) = 9/16

• round, green R-gg (¾)(¼) = 3/16

• wrinkled, yellow rrG- (¾)(¼) = 3/16

• wrinkled, green rrgg (¼)(¼) = 1/16

9:3:3:1 ratio!9:3:3:1 ratio!The dihybrid ratio: 9/16 both dominant traits, 3/16 each for two combinations

of one dominant and one recessive, and 1/16 both recessives

Mendel’s Second Law

• The 9:3:3:1 ratios generated in Mendel’s dihybrid crosses illustrate Mendel’s second law, also known as Mendel’s law of independent assortment

• The law states that during gamete formation the segregation of alleles at one locus is independent of the segregation of alleles at another locus.

• Within the 9:3:3:1 ratio, Mendel recognized two 3:1 ratios for each trait

Testing Independent Assortment by Test-Cross Analysis

• Mendel wants to test his hypothesis about independent assortment. HOW?

• He predicted that the F1 seeds were dihybrid, of genotype RrGr, and that crossing them to a plant of genotype rrgg would yield four offspring phenotypes with equal frequency

Test Cross!

Testing Independent Assortment by Trihybrid-Cross Analysis

• To test his hypothesis about independent assortment further, Mendel performed trihybrid-cross analysis

• The trihybrid cross involved three traits: round vs. wrinkled peas, yellow vs. green peas, and purple vs. white flowers

• The cross was: RRGGPP × rrggpp; the F1 were RrGgPp

How many possible combinations?

• Double check yourself! Do you see all the possible combinations of phenotypes in your answer?

• The number of possibilities can be expressed as 2n, where n = number of genes

• In a trihybrid cross, there are 8 possibilties 23 = 8!

Go try some problems!

• Chapter 2, problem 6

Questions?