05 inheritance

INHERITANCE OF TRAITS

Allele 1 Allele 2

GENE

The Genetic Lottery

• As every individual carries two alleles for a particular gene, the chance of inheriting a particular one of these alleles is 1 in 2 or ½

• So the chance of if inheriting your genotype for any trait is ½ x ½ = ¼

Mendel’s Pea Plants

TT TTtt tt

Tt Tt Tt Tt

Tt TtTT tt

Punnet Squares for:Monohybrid Crosses

T T

t

Tt

¼

Tt

¼

t

Tt

¼

Tt

¼

T t

T

TT

¼

Tt

¼

t

Tt

¼

tt

¼

1st Generation 2nd Generation

All Tt or TtAll Tall

1 TT : 2 Tt : 1 tt or ½ TT: Tt : ½ tt 3 Tall : 1 Dwarf

Genotypic ratio Phenotypic ratio

Dihybrid Crosses• Consider two traits• W = straight brow, w = widows peak• D = detatched earlobes, d = attached• If we cross 2 people who are heterozygous

for both traits, how would the punnet square look and what would be the phenotypic ratios?– NOTE: Genotypic / phenotypic ratios can be

represented as fractions, decimals, ratios or percentages

Dihybrid Crosses

• 1WWDD:1WWdd:2WWDd:2wwDd:4WwDd:2WwDD:2Wwdd:1wwDD:1wwdd

• 9straight/detached:3straight/attached:3widow/detached:1widow attached

WD Wd wD wd

WD WWDDstraight/detached

WWDdstraight/detached

WwDDstraight/detached

WwDdstraight/detached

Wd WWDdstraight/detached

WWddstraight/attached

WwDdstraight/detached

Wwddstraight/attached

wD WwDDstraight/detached

WwDdstraight/detached

wwDDwidows/detached

wwDdwidows/detached

wd WwDdstraight/detached

Wwddstraight/attached

wwDdwidows/detached

wwddwidows/attached

Dihybrid Crosses

• 1WWDD:1WWdd:2WWDd:2wwDd:4WwDd:2WwDD:2Wwdd:1wwDD:1wwdd

• 9straight/detached:3straight/attached:3widow/detached:1widow attached

WD Wd wD wd

WD WWDDstraight/detached

WWDdstraight/detached

WwDDstraight/detached

WwDdstraight/detached

Wd WWDdstraight/detached

WWddstraight/attached

WwDdstraight/detached

Wwddstraight/attached

wD WwDDstraight/detached

WwDdstraight/detached

wwDDwidows/detached

wwDdwidows/detached

wd WwDdstraight/detached

Wwddstraight/attached

wwDdwidows/detached

wwddstraight/attached

Test Crosses

• Prior to it being possible to establish an individual’s genotype via molecular means, a test cross was used.

• Mono or dihybrid test crosses involve breeding the unknown individual with a homozygous recessive.

• Eg. An unknown black cat (DD or Dd) is mated with a grey cat (dd).

• If just one grey kitten is produced, a Dd genotype of the unknown is confirmed

• The general rule is that 16 consecutive black kittens will provide reasonable confidence that the unknown is DD

Linked Genes• If two genes are said to be linked, this

means that they are located in a simillar position on the same chromosome and have a higher chance of being passed on together.

• Eg. RBC shape and RH blood group are linked– RH gene: D = Rh+, d = Rh-

– EL1 gene: E = eliptical RBCs, e = normal RBCs

Linked Genes

• The only way the alleles of linked genes can be separated is through crossing over.

• Based on the above, Sarah is far more likely to produce parental eggs (DE or de) than recombinant eggs (De or dE)

Linked Genes• Based on the spacing between linked genes, a

probability of recombination can be expressed.• In the case of RH and EL1, the probability of

recombination is 0.01 for each type

Detecting Linkage

• Do a punnet square for a test cross with a known heterozygote (aabb x AaBb).– Genotypic ratio for unlinked genes is approx. 1:1:1:1

• If the genes are linked, instead you should see a greater proportion of parental rather than recombinant genotypes.

Genotype AaBb aabb Aabb aaBb

Unlinked 25% 25% 25% 25%

Linked 44% 44% 6% 6%

Detecting Linkage• A simple calculation can be used to either

predict distance between loci or outcomes for linked genes

– So, in a litter of 16 mice, if 7 have brown, curly hair, 7 have black, straight hair, 1 has brown straight hair and 1 has black curly hair ...

– We can estimate these gene loci to be

= 12.5 map units apart

Distance between loci =

What can you predict from the following test cross results?

• Independent assortment (unlinked)

• Linked, with no crossing over

• Linked with crossing over

• Now draw a representation of the heterozygote’s chromosomes

Genotype AaBb Aabb aaBb aabb

Ratio 1.1 0.9 1.2 1.0

Genotype AaBb aabb

Ratio 1.0 1.0

Genotype AaBb Aabb aaBb aabb

Ratio 1.0 0.45 0.55 1.0

A aB b

Pedigrees – guide to symbols

Mr Jellinek’s family tree – EYE COLOUR

Can you work out the

genotypes for each

individual?

Mr Jellinek’s family tree – EYE COLOUR

bb

bb bbbb

bb

Bb

BbBb Bb

Bb

Can you work out the

genotypes for each

individual?

Bb

Is this genetic pedigree possible, if so, explain how

1 2

43

Yes it is, known genotypes are shown below

5 6 7 8

bbbb bb

bb

bb

Bb Bb

Bb Bb BbBbBbBb

Autosomal RecessiveKey Features

• Presence of affected individuals can frequently skip a generation.• In large samples there will be an even distribution of affected males and

females• It is possible for two normal individuals to have an affected child

Autosomal DominantKey Features

• It is not possible for the disease to skip a generation • Homozygous affecteds are rare as they will always have affected offspring,

meaning that lineage is often short-lived.• In large samples there will be an even distribution of affected males and females

X-Linked Recessive• That disease will frequently skip a generation as

females can be carriers• Majority of affecteds are usually male• All sons of an affected female will be affected• All daughters of an affected male will be carriers

Key features

X-Linked DominantKey Features

• Affected male will always pass on the trait to his daughters, but not his sons• In large samples there will be an even distribution of affected males and

females• Can not skip a generation

Construct this pedigree

Susan

Did yours look like this?

Jim Jean

Scott James Natasha AlanKylie

AlisonPaul

Anne Emma Colin