Dominance - PowerPoint Presentation

Slide1

Dominance

Inheritance

Review

Incomplete

Inheritance

Co-dominance

Inheritance

SBI3USlide2

•

When

one

(1)

allele

is

stronger

(so

dominant)

than

the

other

allele

in

heterozygous

genotype

,

resulting

in

phenotype

of

the strong (dominant)

Examples:

a) For hair texture trait  Curly hair is dominant (C)

 Straight hair is recessive (c)

 Therefore, Cc genotype translates to Curly hair phenotype

b) For dimple trait  Having dimple is dominant (D)

 Not having dimple is recessive (d)

 Therefore, Dd genotype translates to having dimple phenotype

c) For hitchhiker thumb trait  Having the hitchhiker thumb is dominant (_____)



Not

having

the

hitchhiker

thumb

is

_______

(___)



Therefore,

______

genotype

translates

to

_______Slide3

Sample

problem:

Dwarfism,

is

a

dominant

trait

.

If

both

parents

with

dwarfism

phenotype

have

normal

looking

offspring,

a)

determine

the

genotype of the parents b)

the percentage of their child looking normal c)

the percentage of their grandchildren from the normal offsprings

looking dwarfish given that their life partners are also normal Answers:

a) P: Dd x Dd (because if either

one of the parents has DD, none of their children

will look normal)DD = Dd = dd =142

4

1

4

=

25%

(dwarfism)

=

50%

(dwarfism)

=

25%

(normal)

c)

Zero

percent

(0%)



F

1

=

dd

x

dd

b)

D

d

D

DD

Dd

d

Dd

ddSlide4

Sample

problem:

Breast

cancer

has

been

tracked

to

associate

with

genes

BRCA1

&

BRCA2

,

which

are

tumour

suppressors

and

function

to

inhibit

the growth of

tumour (cell cycle genes) It is a recessive trait.

That means, if the gene is negatively mutated and

the offsprings carry both mutated alleles, she is more likely to get breast cancer.

a)

Determine

the

phenotype

of

the

genotype

variants

for

this

trait

b)

If

the

mother

is

heterozygous

for

the

trait

and

the

father

has

normal

alleles,

how

many

of

their

6

children

will

i)

be

carrier

of

the

cancer

trait

(heterozygous)

ii)

be

homozygous

for normal BRCA1iii) develop breast cancer

Answers:

a) BRCA1 + BRCA1  normal; least risk for cancer BRCA1 + brca1  carrier of the cancer trait brca1 + brca1  develop cancer

b) i) 6 ii) 6

2424

= 3 children = 3 children

iii) zero

BRCA1

BRCA1

BRCA1BRCA1

BRCA1 brca1 BRCA1 BRCA1 BRCA1 brca1

BRCA1

brca1Slide5

•

Also

known

as

Blending

Inheritance

•

When

two

(2)

alleles

are

equally

dominant

(same

strength),

they

interact

to

produce

a new, third, phenotype This means, the heterozygous genotype has its own

phenotype Three (3) phenotypes are produced from such cross

• Although

Mendel did not observe such pattern of inheritance withpeas, there are

many examples found in nature.When the

heterozygousgenotype producesa different/newphenotypeSlide6

A

red

hibiscus

is

crossed

with

a

white

hibiscus

What

are

their

genotypes?

P:

(H

R

H

R

)

x

(H

W

HW)F1: HRHR = ¼ = 25%  Red

HR HW = ½ = 50%  PinkHW

HW

= ¼ = 25%  WhiteF2: Predict the phenotype ratio

for self-pollinated hybridhibiscus.Slide7

In

guinea

pigs,

colour

of

coat

is

determined

by

at

least

three

alleles.



Yellow

coat

is

determined

by

the

homozygous genotype YY, White by the homozygous genotype WW, and

Cream by the heterozygous genotype YW.Determine the expected

genotype

and phenotype ratio of the F1generation which would result from a cross

between:a) two cream coloured guinea pigs;

•b) a yellow coated and a cream coated animal

Your Turn:Slide8

…

Solutions

Let

G

YGY represent yellow

coatLet GWGW represent white coat

Let GYGW represent cream coat a) P generation phenotypes:

genotypes:cream x cream GY

GW x GYGWG

YGYGWG

YGY GY GWGW GYGW GW G

WF1 : genotypesGYGY = ¼ = 25%

(yellow)GYGW = ½ = 50% (cream)GW GW = ¼ = 25% (white)Therefore, the expected genotypic ratio

is: 1 : 2 : 1 the expected phenotypic ratio is: 1 : 2 : 1b) P generation phenotypes: genotypes:cream x yellow GYG

W x GYGYGYGYGYGYGY GY GYGW GYGW G

YGWF1 : genotypesGYGY = ½ = 50% (yellow)GYGW = ½ = 50% (cream)

Therefore, the expected genotypic ratio is: 1 : 1 the expected phenotypic ratio is: 1 : 1Slide9

1.

Howdy!

My

name

is

Bob

Howard,

and

I

own

20

purebred

red

cows.

Something

strange

happened

several

months

ago.

During

a

violent

storm,

all

of

the

fences

that

separate

my

cattle

from

my

neighbors

cattle

blew

down.

During

the

time

that

the

fences

were

down,

three

bulls,

one

from

each

neighbor,

had

access

to

my

cows.

For

awhile,

I

thought

that

none

of

the

bulls

found

my

cows,

but

over

the

months,

I

have

come

to

the

conclusion

that

all

of

my

cows

are

expecting

calves.

One

of

the

bulls

is

the

father.

Which

bull

is

it?

A

local

college

professor

told

me

to

use

a

little

genetics

detective

work

to

figure

out

who

the

father

is.

He

told

me

to

collect

information

about

each

of

the

bulls,

and

to

read

articles

about

genetics

and

Gregor

Mendel's

experiments

in

genetics.

So,

I

did

exactly

what

he

said.

I

compiled

the

information.

Now,

I

need

your

help

to

make

sense

of

the

data

and

to

figure

out

who

the

father

is.

After

reading

through

the

information,

maybe

you

can

tell

me

why

my

red

cows

had

9

roan

calves

and

11

red

calves.

I

don’t

really

understand

how

this

happened.

When

you

have

determined

which

bull

is

the

father,

please

tell

me

the

answer.Slide10

This

is

Rocky

.

He

is

a

2,200

pound

Red

bull.

The

colour

of

Rocky&’s

calves,

if

mated

with

a

red

cow,

can

be

determined

by

using

a

Punnett

square.

His

offspring

will

also

be

unique

in

colour

compared

to

the

other

two

bulls.

This

is

Rufus

.

He

is

a

1,920

pound

White

bull.

The

colour

of

Rufus’

calves

can

also

be

figured

out,

if

he

is

mated

with

a

red

cow, by using aPunnett square.

The colour of

his

calves will

also differ

from the other twobulls offspring.

This

is

Ferdinand

.

He

is

2,000

pound

Roan

bull.

The

laws

of

genetics

tell

us

that

the

offspring

he

produces

will

probably

be

different,

in

colour,

than

the

other

two

bulls’

offspring.

Using

a

Punnett

square,

you

can

see

the

gene

combinations,

for

colour,

that

Ferdinand’s

offspring

could

have

if

he

mates

with

a

red

cow.

The

colour

of

Ferdinand’s

calves

has

to

do

with

probability.

Who

did

it?

…

a

bonus

mark

for

first

correct

answer

posted

on

Wiki

discussionSlide11

C

o

-

d

o

m

i

nanceSlide12

•

When

the

dominancy

(strength)

of

the

allele

changes

unpredictably



That

is,

sometimes

one

allele

is

dominant

;

other

times

it

is

not

 This results in

the

re-appearance

of

the

both

parental

phenotypes

which

causes

the

offspring

phenotype

to

look

patchy

!

•

The

genotypes

are

not

blended

and

they

still

obey

Mendel’s

law

of

segregation,

resulting

in

a

mixture

of

the

phenotype.

Haemoglobin

The

gene

for

haemoglobin

Hb

has

two

co-dominant

alleles: i) HbA (the

normal gene)

ii) HbS (the mutated gene)Heterozygous:HbAHbSis good for protecting from Malarianormal red blood cellsickled red

blood cellSlide13

Cat

coat:

orange

(

OO

)

x

black

(oo)

=

To

r

t

oi

s

e

s

h

e

ll

maternal

‘O’

allele

paternal

‘o’

allele

mosaic

adultSlide14

http://gslc.genetics.utah.edu/units/basics/blood/types.cfm

Phenotype

Genotype

Picture

Antibodies

Notes

Type

A

Type

B

I

A

I

A

or

I

A

i

I

B

I

B

or

I

B

i

B

–

Antibodies

A

-

Antibodies

Type

AB

I

A

I

B

No

antibodies

Universal

Recipient

Type

O

ii

A

–

Antibodies

B

–

Antibodies

Universal

Donor