dihybrid punnett squares what exactly is a dihybrid cross? predicting the outcomes from crossing 2...

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Dihybrid Punnett Squares Dihybrid Punnett Squares What exactly is a dihybrid cross? Predicting the outcomes from crossing 2 traits. Why do we use them? Instead of doing to monohybrid crosses, you can combine them to see all the possible combinations from that particular mother and father.

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Dihybrid Punnett SquaresDihybrid Punnett SquaresWhat exactly is a dihybrid cross?

Predicting the outcomes from crossing 2 traits.

Why do we use them?Instead of doing to monohybrid

crosses, you can combine them to see all the possible combinations from that particular mother and father.

Dihybrid CrossDihybrid CrossTraits: Seed shape & Seed colorTraits: Seed shape & Seed colorAlleles:Alleles: R round r wrinkled Y yellow y green

RrYy x RrYy

RY Ry rY ryRY Ry rY ry RY Ry rY ryRY Ry rY ry

All possible gamete combinationsAll possible gamete combinations

Dihybrid CrossDihybrid Cross

RYRY RyRy rYrY ryry

RYRY

RyRy

rYrY

ryry

Dihybrid CrossDihybrid Cross

RRYY

RRYy

RrYY

RrYy

RRYy

RRyy

RrYy

Rryy

RrYY

RrYy

rrYY

rrYy

RrYy

Rryy

rrYy

rryy

Round/Yellow: 9

Round/green: 3

wrinkled/Yellow: 3

wrinkled/green: 1

9:3:3:1 phenotypic ratio

RY Ry rY ry

RY

Ry

rY

ry

Incomplete DominanceWhen the heterozygous condition

is intermediate between dominant and recessive

Happens because one allele is defective, example: R=red flowers, R’= white flowers so RR’ = pink flowers (R codes for enzyme to produce red pigment, R’=no enzyme)

Incomplete dominance

White

Red

F1 generation: 1 Red: 2 Pink: 1 white

RR x R’R’

R’R’ x R’R’

CodominanceCodominance

Heterozygote expresses both alleles’ conditionsEx. A black rooster bred with a

white hen produces a black and white checkered chicken.

Usually uses both capital letters (black= B, white=W , checkered = BW)

Codominance practice: cross a black rooster with a white hen

B

B

WW

Results:

100% checkered (BW)

BW BW

BW BW

Key:

B = black

W = white

BW = checkered

Now let’s cross a checkered hen with a checkered rooster.

B

W

WBResults:

25% Black

50% Checkered

25% White

(1:2:1)

BB BW

BW WW

CodominanceCodominance

Two allelesTwo alleles are expressed are expressed ((multiple allelesmultiple alleles) in ) in heterozygous heterozygous individualsindividuals..

Example:Example: blood type blood type

1.1. type Atype A = I= IAAIIAA or I or IAAii2.2. type Btype B = I= IBBIIBB or I or IBBii3.3. type ABtype AB= I= IAAIIBB

4.4. type Otype O = ii= ii

Blood type quick factsBlood type quick factsRed blood cells are called erythrocytes

Proteins on their surfaces are called antigens, controlled by genes

Antigens make antibodies to foreign substances, which includes RBCs with different antigens on their surface

4 phenotypes: A, B, AB, O3 alleles: IA, IB, i

• The lA allele is dominant to i, so

inheriting either the lAi alleles or the lA lA alleles from both parents will give

you type A blood.

Phenotype APhenotype A

• Surface molecule A is produced.

Surface molecule A

• The lB allele is also dominant to i.

Phenotype BPhenotype B

• To have type B blood, you must

inherit the lB allele from one parent and

either another lB allele or the i allele

from the other.

• Surface molecule B is produced.

Surface molecule B

• The lA and lB alleles are codominant.

Phenotype ABPhenotype AB

• If you inherit the lA allele from one parent and the lB allele from the other, your red

blood cells will produce both surface molecules and you will have type

AB blood.

Surface molecule B

Surface molecule A

Phenotype OPhenotype O

•No antigens produced

Blood Group

Antigens Antibodies Can receive

from

Can give to

A A B A or O A or AB

B B A B or O B or AB

AB A and B None A, B, AB, O

AB

O None both O A, B, AB, O

Codominance ProblemCodominance Problem

Example:homozygous male Type B (IBIB) x

heterozygous female Type A (IAi)

IAIB IBi

IAIB IBi

1/2 = IAIB

1/2 = IBi

IB

IA i

IB

Another Codominance Another Codominance ProblemProblem

Example:Example: male Type O (ii) x female type AB (IAIB)

IAi IBi

IAi IBi

1/2 = IAi1/2 = IBi

i

IA IB

i