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Heredity. Heredity. 2. Recap. Genes control the characteristics of living organisms. Genes are carried on the chromosomes. Chromosomes are in pairs, one from each parent. Genes are in pairs. Genes controlling the same characteristics occupy

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Heredity

Heredity

Heredity


Recap

2

Recap

Genes control the characteristics of living

organisms

Genes are carried on the chromosomes

Chromosomes are in pairs, one from each parent

Genes are in pairs

Genes controlling the same characteristics occupy

identical positions on corresponding chromosomes


Dominance

Dominance

3

The gene pairs control one characteristic

But they do not always control it in the same way

Of the gene pair which help determine coat colour

in mice, one might try to produce black fur and its

partner might try to produce brown fur

The gene for black fur is dominant to the gene

for brown fur


Symbols

Symbols

4

The genes are represented by letters

The gene for black fur is given the letter B

The gene for brown fur is given the letter b

BB

bb

The genes must have the same letter but the

dominant gene is always in capitals


Alleles

5

Alleles

The genes of a corresponding pair are called

alleles

This means alternative forms of the same gene

B and b are alleles of the gene for coat colour

B is the dominant allele

b is the recessive allele


Heredity

6

F1

A black male mouse (BB) is mated (crossed) with a

female brown mouse (bb)

In gamete production by meiosis, the alleles are

separated

Sperms will carry one copy of the Ballele

Ova will carry one copy of the b allele

When the sperm fertilizes the ovum, the

alleles B and b come together in the zygote


Heredity

B

B

B

B

B

b

b

b

b

b

All offspring will

be black (Bb)

meiosis

fertilization

sperm mother cell

ovum mother cell

zygote

meiosis


Heredity

8

The offspring from this cross are called the F1 (First Filial) generation

They are all black because the allele for black coat colour is

dominant to the allele for brown coat colour

These Bb mice are called heterozygotes. Because the B and b

alleles have different effects; producing either black or brown coat

colour The mice are heterozygous for coat colour

The BB mice are called homozygotes because the two alleles

produce the same effect. Both alleles produce black coats.

The bb mice are also homozygous for coat colour. Both alleles

produce a brown coat colour

The next slide shows what happens when the two heterozygotes

are mated and produce young


Heredity

F2

9

B

B

b

b

B

B

B

B

B

b

b

b

b

B

b

b

Fertilization

Possible combinations

sperms

BB

sperm mother cell

Bb

meiosis

ovum mother cell

Bb

bb

ova

zygotes


Punnett square

10

Punnett square

B

b

B

b

A neater way of working out the possible combinations

is to use a Punnett Square*

Draw a grid

Enter the alleles in the gametes

Enter the possible combinations

female gametes

BB

Bb

male

gametes

These are the

F2 generation

Bb

bb


3 1 ratio

3:1 ratio

11

The offspring are in the ratio of 3 black to 1 brown

Although the BBand Bb mice look identical, the Bb mice will not

breed true. When mated together there is a chancethat 1 in 4 of their

offspring will be brown

This is only a chance because sperms and ova meet at random

A litter of 5, may contain no brown mice; in a litter of 12, you might

expect 3 brown mice but you would not be surprised at anything

between 2 and 5.

The total offspring from successive matings of the heterozygotes

would be expected to produce in something close to the 3:1 ratio

For example, 6 successive litters might produce 35 black and

13 brown mice. This is a ratio of 2.7:1, near enough to 3:1


Some terminology

Some terminology

12

The offspring of the heterozgotes are the F2 generation

The genetic constitution of an organism is called its genotype

The visible or physiological characteristics of an organism are

called its phenotype

The phenotype of this mouse is

black. Its genotype is BB

BB

The phenotype of this mouse is

also black, but its genotype is Bb

Bb

The phenotype of this mouse is

brown. Its genotype is bb

bb


Heredity

13

These tobacco seedlings are the F2 generation from a cross

Between heterozygous (Cc) parents. C is the gene for chlorophyll.

cc plants can make no chlorophyll. There are 75 green seedlings present.

What is the ratio of green to white seedlings? What ratio would you expect?


Heredity

Cc

CC

Cc

14

There are 21 white seedlings. This is a ratio of 75:21 or 3.57:1

C

c

You would expect

the cross to produce

72 green to 24 white

seedlings (3:1)

C

c

cc

1 CC 2 Ccand 1 cc,

a ratio of 3 green to 1 white seedling

Is 3.57:1 near enough to 3:1 ?*


Sex chromosomes

Sex chromosomes

15

In most populations of animals there are approximately equal

numbers of males and females

This is the result of a pair of chromosomes; the sex chromosomes

called theX and Y chromosomes

The X and Y chromosomes are a homologous pair but in many

animals the Y chromosome is smaller than the X

Females have two X chromosomes in their cells.

Males have one X and one Y in their cells

At meiosis, the sex chromosomes are separated so the the gametes

receive only one: either an X or a Y.


Sex ratio

Sex ratio

16

X

X

X

X

Y

Y

Y

X

X

X

X

X

X

Y

X

X

fertilization

meiosis

female

male

sperm mother cell

female

male

ovum mother cell


Single gene effects

Single gene effects

17

Very few human characteristics are controlled by a single gene

Characteristics such as height or skin colour are controlled by

several genes acting together

Those characteristics which are controlled by a single gene

are usually responsible for inherited defects (see slide 19)


Abo blood groups

ABO blood groups

18

An exception is the inheritance of the ABO blood group

The IA allele produces group A

The IB allele produces group B

The IOallele produces group O

IOis recessive to IA and IB

The group A phenotype can result from genotypes IAIA or IAIO

The group B phenotype can result from genotypes IBIB or IBIO

The group O phenotype can result only from genotype IOIO

The AB phenotype results from the genotype IAIB

The alleles IA and IB are equally dominant (co-dominant)


Genetic defects

19

Genetic defects

Cystic fibrosis (recessive) Glands of the alimentary canal produce a

thick mucus which affects breathing, digestion and susceptibility to

chest infection

Achondroplastic dwarfism (dominant)The head and trunk grow

normally but the limbs remain short

Albinism (recessive) Albinos cannot to produce pigment in their

skin, hair or iris

Polydactyly (dominant*) an extra digit may be produced on the

hands or feet

Sickle cell anaemia (recessive)The red blood cells become

distorted if the oxygen concentration falls. They tend to block

small blood vessels in the joints


Genetic counselling

Genetic counselling

D

d

d

Dd

dd

d

Dd

dd

20

(Genetic defects)

If the genotypes of the parents are known, it is possible to

calculate the probability of their having an affected child

(i.e. one with the defect)

For example if a male achondroplastic dwarf marries a normal

woman, what are their chances of having an affected child?

The father’s genotype must be Dd. (DD is not viable)

The mother must be dd since she is not a dwarf

There is a 50% probability of their having

an affected child

What are the probabilities if both parents

are affected?


Cystic fibrosis

21

Cystic fibrosis

N

n

N

n

(recessive)

If two normal parents have an affected child, they must both be

heterozygous (Nn) for the recessive allele n

A nn parent would have cystic fibrosis

A NN parent would produce only normal

children

NN

Nn

Nn

nn

Since the parents are now known to be

heterozygous it can be predicted that their

next child has a I in 4 chance of inheriting

the disease

This chance applies to all subsequent children*


Sickle cell anaemia

Sickle cell anaemia

22

(recessive)

Hb = haemoglobin

HbA is the allele for normal haemoglobin

HbS is the allele for sickle cell haemoglobin

A person with the genotype HbSHbS will suffer from

sickle cell anaemia

A person with the genotype HbAHbA is normal

The genotype HbAHbSproduces sickle cell ‘trait’ because HbA

is incompletely dominant to HbS

The heterozygote HbAHbS has few symptoms but is a ‘carrier’

for the disease


Carriers

Carriers

23

Heterozygous recessive individuals do not usually exhibit

any disease symptoms but because their offspring may inherit

the disease, the heterozygotes are called ‘carriers’

carriers

HbA

HbS

HbA

HbAHbA

HbAHbS

HbAHbS

HbSHbS

HbS

Similarly, individuals with the genotype Nn are carriers for

cystic fibrosis


Family trees

Family trees

= normal female

= affected female

= normal male

= affected male

24

It is sometimes possible to work out the genotypes of parents and

to track the inheritance of an allele by studying family trees

Parents have normal phenotypes

but produce

an affected child

For this to happen, both parents must have heterozygous

genotypes (Nn) for the characteristic


Heredity

25

AA

Aa

aa

Aa

Aa

If one of the parents is homozygous

for a dominant allele, all the children

will be affected

If one parent is heterozygous for a

dominant allele and the other is

homozygous recessive, there is

a chance that half their children will

be affected

If both parents are heterozygous for

a recessive allele, there is a chance

that one in four of their children

will be affected


Heredity

26

grandparents

marriage

marriage

parents

children

cystic fibrosis

What can you deduce about the genotypes of the grandparents from

this family tree?


Heredity

27

Cystic fibrosis is caused by a recessive gene

An affected person must therefore have the genotype nn

Since neither of the grandparents is affected, they must be either

NN or Nn genotypes

If they were both NN, none of their children or grandchildren could

be affected

If one was Nn and the other NN, then there is a chance that

50% of their children could be carriers Nn

If one of the carriers marries another carrier, there is a

1 in 4 chance of their having an affected child

The genotypes of the grand parents must be either both Nn or one

NN and the other Nn


Heredity

28

d

D

D

DD

Dd

d

Dd

dd

If both parents have the Dd genotype there is a 75% chance

of their having affected children, but the DD individual is

unlikely to survive


Question 1

29

Question 1

Which of the following are heterozygous genotypes?

(a) Aa

(b) bb

(c) nn

(d) Bb


Question 2

Question 2

A

B

C

A

b

c

30

Which of these genes are alleles?

chromosomes

(a) A and A

(b) A and B

(c) B and C

(d) B and b


Question 3

Question 3

31

Which of the following processes separates

homologous chromosomes ?

(a) mitosis

(b) cell division

(c) meiosis

(d) fertilization


Question 4

Question 4

32

Which of the following terms correctly describes

the genotype bb ?

(a) homozygous dominant

(b) heterozygous dominant

(c) homozygous recessive

(d) heterozygous recessive


Question 5

33

Question 5

What is the likely ratio of affected children born to parents

both of whom are heterozygous for cystic fibrosis ?

  • 1 affected: 3 normal

(b) 3 affected: 1 normal

(c) 2 affected: 2 normal

(d) all affected


Question 6

Question 6

34

Which of the following phenotypes corresponds to the

Genotype IAIO ?

  • Blood group A

(b) Blood group B

(c) Blood group O

(d) Blood group AB


Question 7

35

Question 7

What is the expected ratio of offspring from

a black rabbit Bb and a white rabbit bb ?

(a) 3 black: 1 white

(b) 1 black: 3 white

(c) 50% white; 50% black

(d) all black


Question 8

36

Question 8

a

a

a

a

A

A

A

A

A

A

A

a

a

a

a

Which of these Punnett squares correctly represents

a cross between two heterozygous individuals ?

(a)

(b)

aa

AA

AA

Aa

aa

AA

Aa

aa

(c)

(d)

a

Aa

AA

Aa

Aa

Aa

Aa

aa

aa


Question 9

37

Question 9

A married couple has a family of 6 boys.

What are the chances that the next child will be a girl ?

(a) 6:1

(b) 1:6

(c) 3:1

(d) 1:1


Question 10

38

Question 10

Which of the following is a ‘carrier’ genotype for a disease

caused by a recessive gene ?

(a) nn

(b) NN

(c) Nn


Question 11

Question 11

39

If normal parents have a child with cystic fibrosis

(a) one of them must be heterozygous

(b) both of them must be heterozygous

  • one of them must be homozygous

(d) both of them must be homozygous


Answer

Answer

40

Correct


Answer1

41

Answer

Incorrect


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