Heredity

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 occupyidentical positions on corresponding chromosomes

Recap 2

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 colourin 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 genefor brown fur

Dominance 3

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 thedominant gene is always in capitals

Symbols 4

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

Alleles 5

A black male mouse (BB) is mated (crossed) with afemale brown mouse (bb)

In gamete production by meiosis, the alleles areseparated

Sperms will carry one copy of the B allele

Ova will carry one copy of the b allele

When the sperm fertilizes the ovum, thealleles B and b come together in the zygote

F1

6

B

BB

B

b

b

b

b

B

b

meiosis

meiosis

fertilization

All offspring willbe black (Bb)

sperm mother cell

ovum mother cell

zygote

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

They are all black because the allele for black coat colour isdominant 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 allelesproduce the same effect. Both alleles produce black coats.

The bb mice are also homozygous for coat colour. Both allelesproduce a brown coat colour

The next slide shows what happens when the two heterozygotes are mated and produce young

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B

b

B

b

B

b

B

b

B

B

B

b

B

b

b

b

BB

Bb

Bb

bb

sperm mother cell

ovum mother cell

meiosis

Possible combinationsFertilizationsperms

ovazygotes

F2 9

A neater way of working out the possible combinations is to use a Punnett Square*

b

B

B b

Draw a grid

Enter the alleles in the gametes

Enter the possible combinations

female gametes

malegametes

BB Bb

Bb bbThese are the F2 generation

Punnett square 10

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

Although the BB and Bb mice look identical, the Bb mice will notbreed true. When mated together there is a chance that 1 in 4 of theiroffspring 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 mightexpect 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

3:1 ratio 11

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 isblack. 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

Some terminology12

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?

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

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

1 CC 2 Cc and 1 cc, a ratio of 3 green to 1 white seedling

You would expect the cross to produce 72 green to 24 white seedlings (3:1)

c

C

c

C

cc

CC Cc

Cc

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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 chromosomescalled the X 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 gametesreceive only one: either an X or a Y.

Sex chromosomes 15

X

Y

X

X

X

Y

X

X

X

X

X

X

X

Y

X

Y

sperm mother cell

ovum mother cell

meiosisfertilization

female

female

male

male

Sex ratio 16

Very few human characteristics are controlled by a single gene

Characteristics such as height or skin colour are controlled byseveral genes acting together

Those characteristics which are controlled by a single geneare usually responsible for inherited defects (see slide 19)

Single gene effects 17

An exception is the inheritance of the ABO blood group

The IA allele produces group A The IB allele produces group B

The IO allele produces group O IO is 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)

ABO blood groups 18

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 thehands or feet

Sickle cell anaemia (recessive)The red blood cells becomedistorted if the oxygen concentration falls. They tend to block small blood vessels in the joints

Genetic defects19

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 normalwoman, 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

D d

d

d

Dd

Dd

dd

ddWhat are the probabilities if both parents are affected?

Genetic counselling (Genetic defects)20

If two normal parents have an affected child, they must both be heterozygous (Nn) for the recessive allele n

NN Nn

Nn

N n

N

n nn

A nn parent would have cystic fibrosis

A NN parent would produce only normalchildren

Since the parents are now known to beheterozygous it can be predicted that theirnext child has a I in 4 chance of inheritingthe disease

This chance applies to all subsequent children*

Cystic fibrosis (recessive)21

Hb = haemoglobin

HbA is the allele for normal haemoglobinHbS 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 HbAHbS produces sickle cell ‘trait’ because HbA

is incompletely dominant to HbS

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

Sickle cell anaemia (recessive)

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Heterozygous recessive individuals do not usually exhibit any disease symptoms but because their offspring may inheritthe disease, the heterozygotes are called ‘carriers’

HbA

HbA HbAHbA HbAHbS

HbAHbSHbS

HbS

HbSHbS

carriers

Similarly, individuals with the genotype Nn are carriers forcystic fibrosis

Carriers 23

It is sometimes possible to work out the genotypes of parents andto track the inheritance of an allele by studying family trees

= normal female = affected female

= normal male = affected male

Parents have normal phenotypes but produce

an affected child

For this to happen, both parents must have heterozygousgenotypes (Nn) for the characteristic

Family trees 24

If one of the parents is homozygousfor a dominant allele, all the childrenwill be affected

If one parent is heterozygous for a dominant allele and the other is homozygous recessive, there isa chance that half their children willbe affected

If both parents are heterozygous fora recessive allele, there is a chance that one in four of their children will be affected

AA

Aa aa

Aa Aa

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grandparents

parents

children

cystic fibrosis

What can you deduce about the genotypes of the grandparents fromthis family tree?

marriage marriage

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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 eitherNN or Nn genotypes

If they were both NN, none of their children or grandchildren couldbe 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 oneNN and the other Nn

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If both parents have the Dd genotype there is a 75% chanceof their having affected children, but the DD individual isunlikely to survive

D

D DD Dd

Dd dd

d

d

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Question 1

Which of the following are heterozygous genotypes?

(a) Aa

(b) bb

(c) nn

(d) Bb

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Question 2

A B C

A b c

Which of these genes are alleles?

(a) A and A

(b) A and B

(c) B and C

(d) B and b

chromosomes

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Question 3

Which of the following processes separates homologous chromosomes ?

(a) mitosis

(b) cell division

(c) meiosis

(d) fertilization

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Question 4

Which of the following terms correctly describes the genotype bb ?

(a) homozygous dominant

(b) heterozygous dominant

(c) homozygous recessive

(d) heterozygous recessive

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Question 5

What is the likely ratio of affected children born to parentsboth of whom are heterozygous for cystic fibrosis ?

(a) 1 affected: 3 normal

(b) 3 affected: 1 normal

(c) 2 affected: 2 normal

(d) all affected

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Question 6

Which of the following phenotypes corresponds to the Genotype IAIO ?

(a) Blood group A

(b) Blood group B

(c) Blood group O

(d) Blood group AB

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Question 7

What is the expected ratio of offspring from a black rabbit Bb and a white rabbit bb ?

(c) 50% white; 50% black

(a) 3 black: 1 white

(b) 1 black: 3 white

(d) all black

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Question 8Which of these Punnett squares correctly representsa cross between two heterozygous individuals ?

A a

A

a

A a

A

a

A a

A

a

a

A

a

AA

AA

AAaa

aa

a

Aa Aa

aa

Aa aa

aa

Aa

AA

Aa

Aa

Aa

(a) (b)

(c) (d)

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Question 9

A married couple has a family of 6 boys.What are the chances that the next child will be a girl ?

(d) 1:1

(a) 6:1

(b) 1:6

(c) 3:1

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Question 10

Which of the following is a ‘carrier’ genotype for a disease caused by a recessive gene ?

(a) nn

(b) NN

(c) Nn

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Question 11

If normal parents have a child with cystic fibrosis

(a) one of them must be heterozygous

(b) both of them must be heterozygous

(c) one of them must be homozygous

(d) both of them must be homozygous

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Answer

Correct

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Answer

Incorrect

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