is the study of heredity

is the process in which traits are passed from parents to offspring

Historical ViewGregor Mendel: 1866 – first publicationOscar Hertwig: 1876 – discovered meiosisThomas Morgan: 1910 – sex-linkage in

Drosophila Murray Llewellyn Barr: 1948 – Barr bodyMary Lyon: 1960 – inactivation of X

chromosome in calico cats

Gregor Mendel

(1822-1884)

Austrian monk who formulated fundamental laws of heredity in early 1860s

Experimented with peas

Over seven years, he made crosses with 24,034 plants

Mendel used sweet peas for his experiments. WHY? Cheap Easy to grow Could pick traits that

were easy to follow Rapid generation time Lots of offspring Can self-fertilise or be

cross-fertilised

Traits (observable characteristic) studied by Mendel:

A controlled cross between two parents

Mendel’s Crosses:Pollen from one parent was

transferred to the stigma of the other parent. Parental generation = P

Resulting offspring = first filial generation or F1

Mendel’s workplace

If F1 plants self pollinate, produce second filial generation or F2

Basic points revised:Diploid (2n) : total number of

chromosomes2n = 4

n = 2

The human diploid number is 46 44 autosomes 2 sex chromosomes

But chromosomes are pairedThe unpaired chromosome complement is

called haploid (n)

How many are autosomes out of the 23 chromosomes? 22

• have the same length and carry the same gene sequences

• alleles: alternative forms of a gene reside at the same locus on homologous

chromosomes

Homologous chromosomes:

What is a ‘locus’?

Locus (plural-loci):

position of a gene on a chromosome

Homologous chromosomes Non homologous

chromosomes

AllelesHomozygous :

having the same two alleles at that gene locus

AA or aa

Genotype: RR aa BbHeterozygous

R a b

R a B

Gene loci

Recessiveallele

Dominant allele

Homozygousfor thedominant allele

Homozygousfor therecessive allele

Heterozygous : having dissimilar

alleles at that gene locus

Aa

Genotype & Phenotype

genotype: total set of alleles of an individualDescribe the following:

homozygous dominantheterozygoushomozygous recessive

phenotype: outward appearance of an individual

Question: [MAY, 2006]

This question is concerned with genetic inheritance.

1. How does the genotype of an organism determine its phenotype? (1)The genotype determines the proteins made by the body which in turn regulate reactions. Products of reactions determine characteristics expressed.

Recessive & Dominant Alleles

Recessive allele: expressed only when present in the homozygous condition

Dominant allele: expressed both in the homozygous and heterozygous

What is meant by:

An autosomal dominant allele: the allele is carried on an autosome & is

expressed in both in the homozygous and heterozygous condition

A recessive sex-linked allele: the allele is carried on a sex chromosome

& is expressed only in the homozygous condition

Human karyotypeAutosomes set 1 -22

Sex chromosomes set 23

OverviewA) Monohybrid InheritanceB) Dihybrid InheritanceC) The test crossD) Autosomal dominant and autosomal recessive

pedigree chartsE) LinkageF) Sex determinationG) Alleles and their interactions – multiple alleles,

incomplete dominance, codominanceH) Gene interactions – polygenic inheritance, epistasis

Monohybrid inheritance:inheritance of a single characteristic

F1 offspring of a monohybrid cross of true-breeding strains resemble only one of the parents.

WHY?

Smooth seeds (allele S) are completely dominant to wrinkled seeds (allele s).

ss

Ss

SS

F1 x F1 crosses :

Mendel discovered that traits that disappear in the F1 generation reappear in the F2 generation in a 3:1 ratio.

Which is the dominant allele?

P generation(true-breedingparents)

F1 generation

F2 generation

Purple flowers White flowers

All plants havepurple flowers

Fertilisationamong F1 plants(F1 F1)

3/4 of plantshave purple flowers

1/4 of plantshave white flowers

Allele for purple colour

What is meant by ‘true-breeding?

P generation(true-breedingparents)

F1 generation

F2 generation

Purple flowers White flowers

All plants havepurple flowers

Fertilisationamong F1 plants(F1 F1)

3/4 of plantshave purple flowers

1/4 of plantshave white flowers

Alleles are the same

Reginald Crundall Punnett (1875-1967)

The monohybrid ratio:the ratio of dominant phenotypes to

recessive phenotypes of 3:1

Mendel’s first law of heredity or the Law of segregation

(separation):

When any individual produces gametes, the alleles separate, so that each gamete receives

only one member of the pair of alleles.

Gametes:

OverviewA) Monohybrid Inheritance

B) Dihybrid InheritanceC) The test crossD) Autosomal dominant and autosomal recessive

pedigree chartsE) LinkageF) Sex determinationG) Alleles and their interactions – multiple alleles,

incomplete dominance, codominanceH) Gene interactions – polygenic inheritance, epistasis

Dihybrid Inheritance

the inheritance of two separate traits in a single cross

for example: RRYY x rryy

Let:R represent round seed r represent wrinkled seedY represent yellow seedy represent green seed

Seed shape

Seed colour

Parental Phenotypes: Round, yellow seeds

Wrinkled, green seeds

Parental genotypes: RRYY rryyx

x

R R Y Y r r y y

Gametes: R Y

F1 genotypes:

F1 phenotypes: 100% round, yellow seeds

x r y

R r Y y

How to write the gametes:

R rYy

If two RrYy plants are crossed, the F2 generation would be:-

RY

F1 genotypes: RrYy x RrYyGametes: Ry ryrY RY Ry ryrYx

♂♀ RY Ry rY ry

RY

Ry

rY

ry

RY

F1 genotypes: RrYy x RrYyGametes: Ry ryrY RY Ry ryrYx

♂♀ RY Ry rY ry

RY R

Ry

rY

ry

RY

F1 genotypes: RrYy x RrYyGametes: Ry ryrY RY Ry ryrYx

♂♀ RY Ry rY ry

RY RR

Ry

rY

ry

RY

F1 genotypes: RrYy x RrYyGametes: Ry ryrY RY Ry ryrYx

♂♀ RY Ry rY ry

RY RRY

Ry

rY

ry

RY

F1 genotypes: RrYy x RrYyGametes: Ry ryrY RY Ry ryrYx

♂♀ RY Ry rY ry

RY RRYY

Ry

rY

ry

RY

F1 genotypes: RrYy x RrYyGametes: Ry ryrY RY Ry ryrYx

♂♀ RY Ry rY ry

RY RRYY RRYy RrYY RrYy

Ry RRYy RRyy RrYy Rryy

rY RrYY RrYy rrYY rrYy

ry RrYy Rryy rrYy rryy

What is the phenotypic ratio?

RY

F1 genotypes: RrYy x RrYyGametes: Ry ryrY RY Ry ryrYx

♂♀ RY Ry rY ry

RY RRYY RRYy RrYY RrYy

Ry RRYy RRyy RrYy Rryy

rY RrYY RrYy rrYY rrYy

ry RrYy Rryy rrYy rryy

R - round seed r - wrinkled

seedY - yellow seedy - green seed

F2 phenotypes:

round yellow : R_Y_ round green : R_yy wrinkled yellow : rrY_ wrinkled green : rryy

RY

F1 genotypes: RrYy x RrYyGametes: Ry ryrY RY Ry ryrYx

♂♀ RY Ry rY ry

RY RRYY RRYy RrYY RrYy

Ry RRYy RRyy RrYy Rryy

rY RrYY RrYy rrYY rrYy

ry RrYy Rryy rrYy rryy

F2 phenotypes:

round yellow : R_Y_ round green : R_yy wrinkled yellow : rrY_ wrinkled green : rryy

R - round seed r - wrinkled seedY - yellow seedy - green seed

RY

F1 genotypes: RrYy x RrYyGametes: Ry ryrY RY Ry ryrYx

♂♀ RY Ry rY ry

RY RRYY RRYy RrYY RrYy

Ry RRYy RRyy RrYy Rryy

rY RrYY RrYy rrYY rrYy

ry RrYy Rryy rrYy rryy

F2 phenotypes:

round yellow : R_Y_ round green : R_yy wrinkled yellow : rrY_ wrinkled green : rryy

R - round seed r - wrinkled seedY - yellow seedy - green seed

RY

F1 genotypes: RrYy x RrYyGametes: Ry ryrY RY Ry ryrYx

♂♀ RY Ry rY ry

RY RRYY RRYy RrYY RrYy

Ry RRYy RRyy RrYy Rryy

rY RrYY RrYy rrYY rrYy

ry RrYy Rryy rrYy rryy

F2 phenotypes:

9 round yellow : R_Y_3 round green : R_yy3 wrinkled yellow : rrY_1 wrinkled green : rryy

Dihybrid Ratio

The phenotype ratio: 9 : 3 : 3 : 1

1) Parents are BOTH heterozygous for both genes.

This ratio indicates:

2. The two genes are on separate chromosomes.

x

Mendel’s second law or the Law of independent assortment

Alleles of different genes assort independently of one another during gamete formation

When does Mendel’s second law apply and when does it not ?

It applies to genes that lie on separate chromosomes.

It does not apply when genes lie on the same chromosome.

1. In the edible pea, tall (T) is dominant to short (t), and green pods (G) are dominant to yellow pods (g). For the following crosses:

i) List the types of gametes;ii) Write the offspring genotypes;iii) Write the phenotypic ratios.

a) TtGg x TtGgb) TtGg x TTGGc) TtGg x Ttgg

a) Tall – TShort – tGreen pods – GYellow pods – g

TG Tg tG tg

TG

Tg

tG

tg

a) Tall – TShort – tGreen pods – GYellow pods – g

TG Tg tG tg

TG TTGG TTGg TtGG TtGg

Tg TTGg TTgg TtGg Ttgg

tG TtGG TtGg ttGG ttGg

tg TtGg Ttgg ttGg ttgg

a) Tall – TShort – tGreen pods – GYellow pods – g

F1 phenotypes:

tall green : T_G_ tall yellow : T_gg short green: ttG_ short yellow : ttgg

TG Tg tG tg

TG TTGG TTGg TtGG TtGg

Tg TTGg TTgg TtGg Ttgg

tG TtGG TtGg ttGG ttGg

tg TtGg Ttgg ttGg ttgg

a) Tall – TShort – tGreen pods – GYellow pods – g

TG Tg tG tg

TG TTGG TTGg TtGG TtGgTg TTGg TTgg TtGg TtggtG TtGG TtGg ttGG ttGgtg TtGg Ttgg ttGg ttgg

F1 phenotypes:

9 tall green : T_G_ 3 tall yellow : T_gg 3 short green : ttG_ 1 short yellow : ttgg

b) Tall – TShort – tGreen pods – GYellow pods – g

TG Tg tG tg

TG

b) Tall – TShort – tGreen pods – GYellow pods – g

TG Tg tG tg

TG TTGG TTGg TtGG TtGg

F1 phenotypes:

all tall green

Tg tg

TG

Tg

tG

tg

c) Tall – TShort – tGreen pods – GYellow pods – g

Tg tg

TG TTGg TtGg Tg TTgg Ttgg tG TtGg ttGg tg Ttgg ttgg

c) Tall – TShort – tGreen pods – GYellow pods – g

F1 phenotypes:

3 tall green : T_G_ 3 tall yellow : T_gg 1 short green : ttG_ 1 short yellow : ttgg

OverviewA) Monohybrid InheritanceB) Dihybrid Inheritance

C) The test crossD) Autosomal dominant and autosomal recessive

pedigree chartsE) LinkageF) Sex determinationG) Alleles and their interactions – multiple alleles,

incomplete dominance, codominanceH) Gene interactions – polygenic inheritance, epistasis

Test CrossWhy is it done?

is a breeding experiment carried out to determine an unknown genotype within one breeding generation

Test CrossHow is it done?

it is done by crossing an organism of unknown genotype with a homozygous recessive organism

Test Cross How is a

conclusion drawn?

the phenotypic ratios among offspring are different, depending on the genotype of the unknown parent

xbb

50% black : 50% brown

x

100% black

bb

BbbbBb

Test Cross: another exampleBlack: B – BB, BbBrown: b - bb

Test cross with a double recessive

LetS – smooth (seed shape)s – wrinkled (seed shape)Y – yellow (seed colour)y – green (seed colour)

SSYYSSYySsYY SsYy

Smooth seeds, yellow colour

1) What is the genotype ?

SY sy

Parents: SSYY x ssyyGametes:

F1:SsYy

(100% smooth & yellow )

x

S – smooth (seed shape)s – wrinkled (seed shape)

Y – yellow (seed colour)y – green (seed colour)

1) Genotype is homozygous for both alleles (100%)

SY sy

Parents: SSYY x ssyyGametes:

F1:SsYy

(100% smooth & yellow )

x

2) What is the genotype?

SY

Parents: SsYy x ssyyGametes:

F1:

Sy sysY sy

SsYy Ssyy ssYy ssyy

1/4 smooth :& yellow

1/4 smooth :& green

1/4 w rinkled :& yellow

1/4 w rinkled& green

x

S – smooth (seed shape)s – wrinkled (seed shape)

Y – yellow (seed colour)y – green (seed colour)

2) Genotype is heterozygous for both alleles (1:1:1:1)

SY

Parents: SsYy x ssyyGametes:

F1:

Sy sysY sy

SsYy Ssyy ssYy ssyy

1/4 smooth :& yellow

1/4 smooth :& green

1/4 w rinkled :& yellow

1/4 w rinkled& green

x

3) What is the genotype ?

SY

Parents: SSYy x ssyyGametes:

F1:

Sy sy

SsYy Ssyy

1 smooth :& yellow

1 smooth& green

x

S – smooth (seed shape)s – wrinkled (seed shape)Y – yellow (seed colour)y – green (seed colour)

3) Genotype is homozygous for smooth and heterozygous for yellow (1:1)

SY

Parents: SSYy x ssyyGametes:

F1:

Sy sy

SsYy Ssyy

1 smooth :& yellow

1 smooth& green

x

SY

Parents: SsYY x ssyyGametes:

F1:

sY sy

SsYy ssYy

1 smooth :& yellow

1 w rinkled& yellow

x

4) What is the genotype?

S – smooth (seed shape)s – wrinkled (seed shape)

Y – yellow (seed colour)y – green (seed colour)

4) Genotype is heterozygous for smooth and homozygous for yellow (1:1)

SY

Parents: SsYY x ssyyGametes:

F1:

sY sy

SsYy ssYy

1 smooth :& yellow

1 w rinkled& yellow

x

OverviewA) Monohybrid InheritanceB) Dihybrid InheritanceC) The test cross

D) Autosomal dominant and autosomal recessive pedigree charts

E) LinkageF) Sex determinationG) Alleles and their interactions – multiple alleles,

incomplete dominance, codominanceH) Gene interactions – polygenic inheritance, epistasis

some human traits are controlled by a single gene

some of these exhibit: dominant inheritance recessive inheritance

PEDIGREE ANALYSIS

Pedigree analysis is used to track inheritance patterns in families

What does a PEDIGREE CHART show?

the phenotypes of individuals in several generations of a family, and provides a basis for attempting to determine their genotype

What does ‘wild type’ mean?

The most common phenotype in the

population

K ey:A A = affectedA a = affectedaa = norm al

= norm al fem ale

= norm al m ale

= affected fem ale

= affected m ale

A a A a

aa A a aa aaaaA ?

A a aaA a aa aaaa

P aren tal generation

F1 generation

F2 generation

Autosomal dominant disorders

Autosomal dominant disorders

1. Affected children usually have an affected parent.

A A

Autosomal dominant disorders

2. Heterozygotes (Aa) are affected.

A A

Autosomal dominant disorders

3. Two affected parents can produce an unaffected child.

A A

Autosomal dominant disorders

4. Two unaffected parents will not have affected children.

A A

aa A ?

A a A ?A a

A a A a A?A?

K eyaa = affectedA a = carrier (appears norm al)A A = norm al

A ?

aa A?aa

Autosomal recessive disorders

Albinos

Autosomal recessive disorders

1. Most affected children have normal parents.

aa A ?

A a A ?A a

A a A a A?A?

K eyaa = affectedA a = carrier (appears norm al)A A = norm al

A ?

aa A?aa

aa A ?

A a A ?A a

A a A a A?A?

K eyaa = affectedA a = carrier (appears norm al)A A = norm al

A ?

aa A?aa

Autosomal recessive disorders

2. Heterozygotes (Aa) have a normal phenotype.

Autosomal recessive disorders

3. Two affected parents will always have affected children.

Autosomal recessive disorders

4. Affected individuals with homozygous normal mates will have normal children.

aa A ?

A a A ?A a

A a A a A?A?

K eyaa = affectedA a = carrier (appears norm al)A A = norm al

A ?

aa A?aa

Autosomal recessive disorders

5. Close relatives who reproduce are more likely to have affected children.

Categories of inheritance

1. Autosomal recessive– e.g. albinism

2. Autosomal dominant– e.g. Huntington’s Disease [degeneration of

brain cells]

3. X-linked recessive– e.g. color-blindness, haemophilia

4. X-linked dominant [Very Rare]– e.g. hypophosphatemia [low level of

phosphate in blood]

When working genetics problems:

The probability that two independent events will

both occur is the product of their individual probabilities.

e.g. What is the probability that a child is a MALE with BROWN EYES if both parents are heterozygous? Brown is dominant to blue eyes.

Multiply probability of being a male by probability of having brown eyes.

Probability of having BROWN EYES.

Probability of being a MALE:

0.5 or 50%

0.75 or 75%

0.5 x 0.75 = 0.375 or 37.5%

Question: [SEP, 2011]

The pedigree diagram shown in Figure 1 below follows the inheritance of Marfan syndrome, a disorder of connective tissue in humans that affects males and females equally. The allele for Marfan syndrome, M, is dominant to the allele, m, for unaffected (normal) connective tissue. In the diagram below, individuals affected with Marfan syndrome are represented by shaded symbols whilst unaffected individuals are represented by unshaded symbols. Males are represented as squares and females as circles.

Figure 1: Pedigree diagram showing inheritance of Marfan Syndrome across three generations.

1.1 Use the diagram in Figure 1 to deduce the genotype of all the individuals represented. Insert your answers in the table below:

= affected female

= affected male

= normal female

= normal male

M: MM, Mm

m: mm

= affected male

= normal female

= normal male

M: MM, Mm

m: mm

Individual GenotypeI-1 mmI-2 MmII-1 MmII-2 mmII-3 MmII-4 mmII-5 mmIII-1 MmIII-2 MmIII-3 Mm

mm

mmmmmm

= affected female

Mm

MmMm

Mm

Mm

Mm

1.2 Assume that individual III-3 mates with an affected male. What is the probability that this cross would produce an affected female? Your working is to be shown. (2)

Individual III-3 (Mm) mates with an affected male (MM): Mm x MM = MM and Mmprobability to produce a female is: 0.5 probability to be affected is 1So, 0.5 x 1 = 0.5

Individual III-3 (Mm) mates with an affected male (Mm): Mm x Mm = MM Mm Mm mm

probability to produce a female is 0.5 probability to be affected is 0.75 So, 0.5 x 0.75 = 0.375

1.3 What is the probability that the cross referred to in the previous question would produce an unaffected male? Your working is to be shown. (2)

Individual III-3 (Mm) mates with an affected male (MM): Mm x MM = MM and Mmprobability to be unaffected (mm) is 0

Individual III-3 (Mm) mates with an affected male (Mm): Mm x Mm = MM Mm Mm mmprobability to produce a male is 0.5 probability to be unaffected (mm) is 0.25So, 0.5 x 0.25 = 0.125