Genetics Chapter 2

8/13/2019 Genetics Chapter 2

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Comparison of Mitosis and Meiosis

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Chapter 2

Mendelian Genetics


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Mendels Experimental Design

Why are pea plants

so useful?

Self-fertilization(selfing)

Cross-fertilization

(cross-breeding)

discrete,

nonoverlapping traits

- flower and pod color,

height

Pisum sativum

(common garden pea)


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Plant Anatomy

anatomy:

Pistil

Style

Ovary

anatomy:

Stamen

Anther (pollen-producing)

Filament (forsupport)


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7 Phenotypes (traits) of Pea Plants

Pod location

Seed and pod

color and shape


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Flower color

How did Mendel use these characteristics in designing

experiments?

unambiguously described each plant

Stem length(used only 2 extremes:

tall or short)

7 Phenotypes (traits) of Pea Plants


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Crossing plants with only 1 variable

trait

Chose homogeneous plants(pure-breeding: first grew

them for 2 years)

Cross tall and dwarf plants

(P1or parental generation)

Offspring - F1or first filial

generationhybrids

(monohybrids : different in

only 1 trait)


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Examine the traitsof the F1generation

all were tall : dominant

recessive trait was masked

Examine the phenotypes

of the F2generation

3:1 ratio

tall (787)

dwarf (277)

Crossing plants with only 1 variable

trait

(monohybrid)


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Mendels proposed mechanism

a certain inherited unit (gene)

produced 1 trait

Ratios of offspring produced :

each trait is composed of 2factors , one dom inant and 1

recessive alleles

-Homozygotes (pure-bred) ;

heterozygotes (hybrids)

Difference in genotypesand

phenotypesin the P1, F1, and

F2generations


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Mendels First Principle: The Law

of Segregation

Each allele is a different form of a gene that will

separate and randomly distribute into a gamete

- each gamete has an equal probability of receiving

either allele

Fertilization = fusion of two gametes

The Law of Segregation explains

F1genotype: HETEROZYGOTE (hybrid)

F1phenotype: DOMINANT (recessive

trait masked)


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The Law of Segregation: The

Schematic Approach


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The Law of Segregation: Use of

the Punnett Square

critical assumption :

all gametes listed occur (segregate) at equalprobability

every offspring (box) is equally likely


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The Law of Segregation: using

Probability to Predict Genotypes and

Phenotypes


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With only phenotypes, Mendel confirmed

the Law of Segregation by selfing F2plants

F2generationPhenotypic ratio 3:1 dominant to

recessive (recessive trait

reappears)

Genotypic ratio 1:2:1

Self-fertilize F2to produce

F3generation

Predict what you would

expect in

F2: dwarf X dwarf

F2: tall X tall


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Testing the Law of Segregation

How can you determine the genotype of an F2

offspring that expresses the dominant phenotype?

Testcross: Mate the organism with the dominant

phenotype with an organism that expresses therecessive phenotype.


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What Kind of Data Is Obtained from

the Test Cross?

1. Tall (homozygote)

X dwarf

(homozygous

recessive)

2. Tall (heterozygote)

X dwarf

(homozygous

recessive)

Two possible outcomes that are phenotypically distinct :


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P1phenotypes: yellow, round x

green, wrinkled

F1phenotype: yellow, round

(dihybrid: heterozygous for 2 genes)

F2phenotypes:315round, yellow

108round, green

101wrinkled, yellow

32wrinkled, green

divide each category by number in

smallest group:

9.84 : 3.38 : 3.16 : 1.00 (~9:3:3:1)

use of a Punnett square

Crossing plants with 2 variable traits

(dihybrid)


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Punnett square prediction of

genotypes and phenotypes

critical assumption :

all gametes listed occur (segregate) at equalprobability

alleles of 2 different genes are inherited (assorted) independently

(round,

yellow)

(round,

green)

(wrinkled,

yellow)

(wrinkled,

green)

Th L f I d d t


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The Law of Independent

Assortment

Alleles for 1 gene segregate independently of thealleles for other genes

Isolate individual phenotypes :

1. seed shape : 315+108 round

101+ 32 wrinkled423 : 133

round : wrinkled

(3.18:1.00) expect 3:1 for 1 gene

2. seed color : 315+101 yellow108 + 32 green

416 : 140

yellow : green

(2.97:1.00) expect 3:1 for 1 gene

F2phenotypes:

315round, yellow108round, green

101wrinkled, yellow

32wrinkled, green

equalsegregationof alleles for each of the 2 genes

T t f I d d t A t t


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Test of Independent Assortment :

Testcross the Dihybrid Plant

Cross: F1WwGg X wwggPredict the phenotypes in

progeny

1:1:1:1 Round Yellow

Round Green

Wrinkled Yellow

Wrinkled Green

Mendels results verified this and confirmed the Law of

Independent Assortment : alleles for 1 gene segregate

independently of the alleles for other genes


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Multihybrid Crosses

with n genes (each containing 2 segregating alleles),there will be :

differentgametes

differentphenotypes

differentgenotypes

2n

2n

3n

frequency of homozygous recessives = n


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Genetic nomenclature

botany and mammalian genetics :

uppercase letters : dominant alleles,

lowercase letters : recessive alleles

W : round w : wrinkledG : yellow g : green

phenotypes :

shape WW, Ww : roundww : wrinkled

color GG, Gg : yellow

gg : green

Wild t d M t t Ch t i ti i


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Wild-type and Mutant Characteristics in

Drosophila

Wild-type: characteristic most common in nature

Eye color: red

Wings: oblong and flat

Mutant phenotypes: alternatives to the wild-type

Eye color: for example, white

Wings: for example, dumpy (reduced wings)


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Mutations of Drosophila Melanogaster

Genetic Symbol Convention in Drosophila

Wild type allelesymbol of the mutant followed by a superscript +

genes are named after the mutant phenotype

Capital lettermutation is dominant

Lower casemutation is recessive

e.g. : wild type (red) eyes : w+ mutant (white) eyes : w

wing mutations :


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Genetic Symbolism in Drosophila

a +sign can be usedalone for the wild type

(only when referring to

1 gene at a time)

multiple mutant

alleles can exist for a

gene : wa(white

apricot) or wc(whitecrimson)

There are cases where the gene can be designated by

more than one letter: dp = dumpy wings, dp+ = wild type

P b bilit A li ti i


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Probability: Applications in

Mendelian Genetics

Use of ratios : Probability(P) = a/na= number of times an event is observed

n= total number of possible cases

Can be determined by: observation(empirical) - 1 child/10,000 born with

phenylketonuria (PKU)

- probability that next child born will have PKU = a/n= 1/10,000

nature of the event (theoretical): drawinga heart

from a deck of cards 13/52 = 1/4

P b bilit A li ti i


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Probability: Applications in

Mendelian Genetics

P=1 : event is an absolute certainty

P=0 : event is impossible

Probability of all events in an experiment : add

up to 1

B i P i i l f P b bilit


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Basic Principle of ProbabilityIf one event has cpossible outcomes, and a

second event has dpossible outcomes, thenthere are cdpossible outcomes of the two

events.

Mutually exclusive outcomes: events inwhich the occurrences of one possibility

excludes all other possibilities

Independent outcomes:events that are

not influenced by each other


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The Sum Rule

Sum rule (either-or rule)used when eventsare mutually exclusive

Example : what is the probability of pickingeither a heart or a spade from a deck of

cards?

P = 13/52 + 13/52 = 26/52 =


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The Product Rule

Productrule(and rule)when occurrenceof one event is independent of another event

-Example 1: what is the probability of

picking a heart and then a spade from a

deck of cards?

P = 13/52 x 13/52 = 169/2704 = 0.0625

-Example 2 : what is the probability of

throwing 2 quarters with both landing on

tails?

Probability Analysis for Monohybrid and


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Probability Analysis for Monohybrid and

Dihybrid Crosses

Monohybrid cross (tall phenotype): DdX Dd Genotypic Ratio : 1 DD:2 Dd:1 dd

Calculate the probability of DdX Ddproducingthe tall phenotype (DD orDd)

P = 1/4 + 2/4 =3/4

Dihybrid cross between yellow, round plants :

WwGgX WwGg

probability of producing an offspring with wrinkledandyellow seeds :

P of wrinkled phenotype = 1/4

P of yellow seeds = 3/4

P (wrinkled andyellow) = 1/4 x 3/4 = 3/16

(independent

assortment)


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The Branched-Line Method to Calculate

Probabilities: An Alternative to Punnett Squares

Branched-line analysis of dihybrid cross:AaBbXAaBb

Independently calculate the probability of each trait Use the Product Rule

U f th B h d Li M th d f


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Use of the Branched-Line Method for a

Trihybrid CrossAaBbCcXAabbCc

Calculate the probability of each phenotypic class.use the product rule with 3 independently assorting genes

Hypothesis Testing: Mendels Crosses


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Hypothesis Testing: Mendel s Crosses

Examination of tall F1plants that are self-crossed

F2plants yielded 787 tall and 277 dwarf

2.84:1 (Mendel interpreted this as 3:1)

Based on probability, can you expect some deviation?

How much deviation is acceptable?

what if F2plants yielded 709 tall and 355 dwarf (2:1 ratio) ?

Where do you draw the line?

Statisticshelps to develop Confidence Limits

- not absolute certainty, but within random chance

- allows us to determine whether results confirm or refute

a hypothesis

Chi Di t ib ti


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Chi-square Distribution

= Greek letter chi

O = observed number for a category

E= expected number for that category

= sum of calculations for all categories

F Progeny from a Monohybrid Cross


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F2Progeny from a Monohybrid Cross

P1: tall (DD) X dwarf (dd)

F1progeny: all tall (Dd)

F1self cross (Dd X Dd)

F2progeny

1064 total

787 tall

277 dwarf

What ratio did you expect?

Test 3:1 and 1:1


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Chi-square Analysis of the F2Progeny

These 2values are meaningless. They need to be

interpreted using the Probability value (p)


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Determination of Probability Values (p)

I t t ti f 2: E i th


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Interpretation of 2: Examine the

Probability Table

The rows represent the degrees of freedomDegrees of freedom equals the number of phenotypic

categories1

This column tells us : the probability is 0.05 of obtaining a 2 value of3.841 or greater bychance alone cannot reject hypothesis

- if p


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Genetics Chapter 2

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