Mendel and the Gene IdeaChp 14

Blending Model vs. Particulate Model

•genetic material mixes like paint

•Parent’s traits inseparable

•Discreet inheritable units

•Traits retain separate identities

Figure 14-01

Gregor Mendel

•Austrian Monk

•Did not know about DNA, genes, or chromosomes!

•Tried to prove particulate model of inheritance using pea plants

Web Lab: Mendel and his Peas

Advantages of working with pea plants:•Many different varieties

•Character = inheritable feature (GENE)•Trait = variation of that character (ALLELE)

•Can control plant matings (paintbrush)

LE 14-2

Removed stamensfrom purple flower

Transferred sperm-bearing pollen fromstamens of whiteflower to egg-bearing carpel ofpurple flower

Carpel Stamens

Parentalgeneration(P)

Pollinated carpelmatured into pod

Planted seedsfrom pod

Examinedoffspring:all purpleflowers

Firstgenerationoffspring(F1)

•Started with true-breeding varieties = HOMOZYGOUS•If allowed to self-pollinate, all offspring had the same traits as their parent plant

•Then, cross-pollinated two different true-breeding varieties = HYBRIDIZATION

Hybridization

LE 14-5_1

Appearance:Genetic makeup:

Gametes:

P Generation

F1 Generation

PurpleflowersPP

Whiteflowerspp

pP

Purple flowersPp

Appearance:Genetic makeup:

Monohybrid Cross – tracks a single character/gene

hybrid

Mendel’s Law of Segregation: Parental alleles separate (segregate) during gamete formation• occurs in Metaphase I of Meiosis I• only one allele per gamete

LE 14-5_2

Appearance:

P Generation

Genetic makeup:

Gametes

F1 Generation

Appearance:Genetic makeup:

Gametes:

F2 Generation

Purple flowersPp

P p1 21 2

P p

F1 sperm

F1 eggsPP Pp

Pp pp

P

p

3 : 1

PurpleflowersPP

Whiteflowerspp

P p

gametes

gametesPunnett Square shows possible offspring after fertilization

LE 14-6

Phenotype

Purple

Purple3

Purple

Genotype

PP(homozygous

Pp(heterozygous

Pp(heterozygous

pp(homozygous

1

2

1

Ratio 1:2:1

White

Ratio 3:1

1

R r

Rr

F1

F2 R rRr

rrRr

3 round: 1 wrinkled1 RR : 2 Rr : 1 rr

Law of Dominance: The dominant allele (trait) is fully expressed and the recessive allele has no noticeable effect

Recessive ≠ Bad

Test Cross

•Used to determine the genotype of an individual with a dominant phenotype

•Cross it with a homozygous recessive and observe offspring ratios

T T T

t

t

t

t

t

100% tall 100% Tt

Tt

Tt Tt

Tt

1 tall : 1 dwarf 1 Tt : 1 tt

Tt

Tt

tt

tt

Law of Independent Assortment: Each allele pair segregates independently of other allele pairs during Meiosis (Metaphase I).

Dihybrid Cross: tracks two genes (characters)

Tall purple

TtPp

TP Tp tP tp

TP

TP

Tp

Tp

tP

tP

tp

tp

TTPP

TTPp

TTPp

TtPP

TtPP

TtPp

TtPp

ttpp

TTpp

TtPp

TtPp

ttPP

Ttpp

Ttpp

ttPp

ttPp

9 tall purple3 tall white

3 dwarf purple1 dwarf white

12:4 or 3:112:4 or 3:1

Rules of Probability & Genetics

• Probability ranges from 0 – 1 • Rule of Multiplication: Used to determine

the chance of two or more independent events occurring together

1. Determine probability of each independent event2. Multiply probabilities together

Ex: What is the probability (chance) that when 2 coins are flipped they will both end up heads?

½ x ½ = ¼ (or .25)

½ ½

½

½

¼

¼

¼

¼

Multiplication RuleProbability Problems

• What is the probability (chance) that 2 dice will both show a 4 when rolled?

1/6 x 1/6 = 1/36

Ex (F1): T t P p x T t P p

F2: t p t p

½ x ½ x ½ x ½ = 1/16

t t p p

½ ½ ½ ½

T t

T TT Tt

t Tt tt

P p

P PP Pp

p Pp pp

Rule of Addition

• What is the probability (chance) that the sum of the numbers shown on 2 dice will equal 5?

• Rule of Addition: used to determine the probability of an event that can occur in two or more different ways/combinations– Calculate probability for each possible

combination/way using Rule of Multiplication– Add the probabilities for each separate

combination to get the total probability

Rule of Addition

• What is the probability (chance) that the sum of the numbers shown on 2 dice will equal 5?

• Possible combinations:

1 + 4 2 + 3 3 + 2 4 + 1 1/6 x 1/6 1/6 x 1/6 1/6 x 1/6 1/6 x 1/6

1/36 + 1/36 +

1/36 + 1/36 =

4/36 ( 1/9 )

A a

A AA Aa

a Aa aa

b b

B Bb Bb

b bb bb

C C

c Cc Cc

c Cc Cc

¾ ½ 1

= 3/8

x x

A a

A AA Aa

a Aa aa

b b

B Bb Bb

b bb bb

C C

c Cc Cc

c Cc Cc

A__ B__ C__ =

A__ B__ c c =

A__ b b C__ =

a a B__ C__ =

A a

A AA Aa

a Aa aa

b b

B Bb Bb

b bb bb

C C

c Cc Cc

c Cc Cc

A__ B__ C__ = 3/8

A__ B__ c c = 0

A__ b b C__ = 3/8

a a B__ C__ = 1/8

Sum = 7/8

Text book: pages 272 - 273

• All problems (#1-17) due on

• Tonight, work on problems:

# 2, 3, 4, 7, 8, & 10

Types of Dominance

• Complete Dominance – one allele complete masks the other; the heterozygous (Rr) and homozygous dominant (RR) have the same phenotype–The dominant allele usually codes for some protein/enzyme and the recessive allele codes for a defective protein/enzyme. Both are “expressed”, but only the dominant allele is functional and observable

Types of Dominance• Codominance = two alleles expressed

separately & both affect phenotype– Ex: Red & White rhododendron

“Roan” color in cattle

CR CR

CW CR CW CR CW

CW CR CW CR CW

Types of Dominance

• Incomplete Dominance = F1 hybrids (heterozygotes) have an intermediate phenotype– Results in a THIRD

phenotype– NOT the same as blending

(F2 show all phenotypes)– 1:2:1 phenotypic and

genotypic ratios in F2

Multiple Alleles = more than two alleles/varieties

IA & IB

i

are codominant

over

Fill in Interactive Question 14.6

Blood Type practice…

• Suppose a father of blood type B and a mother of blood type A have a child of type O. What are the chances that their next child will be:– Type O?– Type B? – Type A?– Type AB?

Blood Type practice…

• Suppose a father of blood type B and a mother of blood type A have a child of type O. What are the chances that their next child will be:– Type O? ¼ – Type B? ¼ – Type A? ¼ – Type AB? ¼

IB i

IA IAIB IAi

i IBi ii

Pleiotropy

• Pleiotropy = one gene has multiple phenotypic effects– Ex: cystic fibrosis & sickle-cell disease

Sperm

BC bC Bc bc

BbCcBBCcBbCCBBCC

BbCC bbCC BbCc bbCc

BbccBBccBbCcBBCc

BbCc bbCc Bbcc bbcc

BC

bC

Bc

bc

BbCc BbCc

14

14

14

14

14

14

14

14

916

316

416

Epistasis = (“force upon”) one gene affects the expression of another gene

Ex: coat color in mice

brownwhite

M__B__

mm __ __

¾ * ¾ = 9/16

¾ * ¼ = 3/16

¼ * 1 = 4/16 (¼)

aabbcc Aabbcc AaBbcc AaBbCc AABbCc AABBCc AABBCC

AaBbCcAaBbCc

20/64

15/64

6/64

1/64

Fra

ctio

n o

f p

rog

eny

Polygenic = additive effect of 2 or more genes on one phenotype

Ex: skin color height

# alleles + 1 = # phenotypes

AaBbCc = 25 cm

Seven (6 alleles + 1 = 7 phenotypes)

Nature vs. Nurture….

Acidic soil

Basic soil

Pedigree

• Family tree showing relationships between family members and the pattern of inheritance across the generations

Figure 14-16

Aa Aa

AaAA Aa

No – either he or his wife is AA, but can’t be certain

AA aa aa Aa A_ A_ A_

Aa Aa Aa Aa aa Aa Aa A_ A_ A_ A_ A_

Aa Aa

Recessive Disorders• Carriers = heterozygotes (Aa) have the

defective allele, but are not affected by disorder– “carry” allele, but do not express it– Can pass it on to offspring (50% chance)

¼

2/3

A a

A AA Aa

a Aa aa

Recessive Disorders

• Cystic Fibrosis

• Tay-Sachs Disease

• Sickle-cell Disease

?

Aa Aa Aa Aa

aa aaA_ A_A a

A AA Aa

a Aa aa

2/3 Aa x 2/3 Aa x ¼ aa = 4/36 = 1/9

Dominant Disorders

• RARE – nature will weed out (no carriers!)

• Achondroplasia Dwarfism

• Huntington’s Disease

LE 14-17a

Amniocentesis

Amnioticfluidwithdrawn

Fetus

A sample ofamniotic fluid canbe taken starting atthe 14th to 16thweek of pregnancy.

Centrifugation

Placenta Uterus Cervix

Fluid

Fetalcells

Biochemical tests can beperformed immediately onthe amniotic fluid or lateron the cultured cells. Biochemical

tests

Severalweeks

Karyotyping

Fetal cells must be culturedfor several weeks to obtainsufficient numbers forkaryotyping.

LE 14-17b

Chorionic villus sampling (CVS)

Placenta Chorionic villi

Fetus

Suction tubeinserted throughcervix

Fetalcells

Biochemicaltests Karyotyping and biochemical

tests can be performed onthe fetal cells immediately,providing results within a dayor so.

Severalhours

A sample of chorionic villustissue can be taken as earlyas the 8th to 10th week ofpregnancy.

Karyotyping