Term 2 Lecture 14: Additional Factors Affecting Inheritance

Question 1

Principles that modify Mendelian Laws

Answer 1

• incomplete and codominance
• penetrance and expressivity
• multiple alleles
• epistasis - gene interactions
• sex influenced inheritance

Question 2

Dominance

Answer 2

-Additional factors at a single locus can affect the genetic outcome and phenotype
- Genes at the same locus - two versions of the same gene, each defined as an allele

Types of dominance

Complete e.g. in plants red or white flowers

Incomplete e.g. in plants flowers in shades of pink - varying levels of expression also seen in eye and hair colour

Codominance e.g. in cystic fibrosis - phenotype of heterozygote includes phenotypes of both homozygous

• level of phenotype may affect dominance
• phenotypes can be observed at different levels: anatomical, physiological and molecular level

Question 3

Codominance example

Answer 3

Cystic fibrosis (CFTR gene)
- CFTR gene encodes the gated channel cell membrane protein CFTR that regulates flow of chloride (Cl-) ions in and out of the cell.
- homozygote patients experience a build up of chloride ions in the cell, producing thick mucous in the lungs and digestive tract causing disease symptoms
- this is an example of an allele exhibiting codominance at the molecular level and recessive at a physiological level
- the healthy allele produces sufficient functional CFTR protein so that the heterozygote carriers appear unaffected and are asymptomatic

Question 4

Other effecting factors: penetrance

Answer 4

Penetrance: the percentage of individuals that have a particular genotype that is expressed and observed in the phenotype e.g. human phenotype polydactyly (extra digits)

Incomplete penetrance: you may have the gene but it may not be observed/ may be only partly expressed

Question 5

Other affecting factors: Lethal alleles

Answer 5

• cause death at a n early stage of development therefore some genotypes/ phenotypes do not appear amongst progeny
• presence causes unexpected genotypic and phenotypic ratios in progeny
  E.g. homozygoous dominant allele for blonde fur in mice (YY) is a lethal allele and causes death in the developing embryo whereas heterozygous (Yy) blonde mice carry to term (we don’t know why this is)

Question 6

Codominance where multiple alleles are involved

Answer 6

E.g. human blood type
3 alleles exist A, B and O
You can be A,AB,B or O type

Due to the antigens on the blood cells and consequently the antibodies that the body makes in response your body will only accept certain types of blood by transfusion.

I^A >i A dominant to O
I^B >i A dominant to O
I^A = I^B so A and B are codominant

A person with AB blood will produce A and B antigens making them a universal acceptor

A person with O blood type aka ii produces no antigens and is therefore a universal donor.

To determine an individuals blood type spots of blood are applied to a plate containing test for: A(anti B) B (anti A) AB( no antibodies) and O (A&B antibodies)
Red blood cells that do not interact with the recipient antibody remain evenly dispersed showing donor and recipient are compatible.
Where red blood cells react with the recipient antibodies they clump - donor and recipient not compatible

Question 7

Gene interaction takes place when genes at multiple loci determine a single phenotype

Answer 7

• effect of genes at one locus depends on the presence of genes at other loci
• genes exhibit independent assortment but do not act independently in their phenotypic expression
• the products of genes at different loci that interact to produce new phenotypes e.g. gene interactions that cause pepper colours result in a 9:3:3:1 ratio but observed phenotype doesn’t fit known genotypes

Question 8

Epistasis

Answer 8

One gene masks or interferes with the expression of another gene
Producing a modified dihybrid phenotype ratio:

Recessive epistasis: 9:3:4
Dominant epistasis: 12:3:1
Duplicate recessive epistasis: 9:7

Gene interactions take place when genes at multiple loci determine a single phenotype

Question 9

Recessive epistasis: example: ABO blood type

Answer 9

9:3:4 F2 ratio

1) the dominant H allele encodes an enzyme that converts an intermediate compound into H
2) compound H is required for addition of a terminal sugar
3) genotypes at the ABO locus determine the type of terminal sugar to be added
4) which determines blood type
5) people with the Bombay phenotype are homozygous for a recessive mutation (hh) that fails to convert the intermediate into H
6) blood type O can result from the absence of a terminal sugar on compound H
7) or from the absence of compound H due to double recessive homozygous hh

Question 10

Dominant epistasis

Answer 10

12:3:1 F2 generation

1) plants with the genotype ww produce enzyme 1 which converts compound A (colourless) to compound B (green)
2) dominant allele W inhibits conversion of A to B
3) plants with genotype Y_ produce enzyme 2 which converts compound B to compound C (yellow)
4) plants with genotype yy do not encode a functional form of enzyme 2

Conclusion: genotypes W_Y_ and W_yy produce enzyme 1 but not enzyme 2 wwY_ produces both enzyme 1 and 2

The pathway begins with the involvement of the recessive alleles at locus W if the dominant allele is present the pathway won’t initiate resulting in a 12:3:1 F2 of white:yellow:green

Question 11

Duplicate recessive epistasis

Answer 11

9:7 F2 ratio

A dominant allele at the A locus is required to produce enzyme 1 which converts A to B

A dominant allele at the B locus is required to produce enzyme 2 to convert B to C ( pigment)

Albinism arises from the absence of enzyme 1 (aaB_) so compound B is never produced
or from the absence of enzyme 2 (A_bb) so compound C is never produced, or from the absence of both enzymes (aabb)

pigmented snails must produce enzymes 1 and 2 which requires genotype A_B_

Homozygous recessives block enzyme formation resulting phenotype 9:7 ratio in F2 pigmented:albino

Question 12

Modified dihybrid phenotypic rates

Answer 12

9:3:3:1 no interaction
e.g. in pea shape and colour

9:3:4 recessive epistasis
E.g. coat colour in Labradors

12:3:1 dominant epistasis
E.g. colour in squash

9:7 duplicate epistasis
E.g. albinism in snails

9:6:1 duplicate interaction

15:1 duplicate dominant epistasis

13:3 dominant and recessive epistasis

Each ratio is F2 based on a dihybrid cross (AaBbxAaBb)

Question 13

Gene interaction takes place when genes at multiple loci determine a single phenotype

Answer 13

Complementation: determine whether mutations are at the same locus or at different loci

Test: parents homozygous for different mutations are crossed so offspring are heterozygous

If mutations are allelic (occur at the same locus) then
- heterozygous offspring have only mutant alleles
- exhibit a mutant phenotype

If mutations occur at different loci then heterozygous offspring inherit a mutant allele and a wildtype allele so some progeny exhibit wildtype phenotype

Question 14

Expression of a genotype may be influenced by environmental effects

Answer 14

E.g. temperature sensitive alleles are functional only at a certain temperature
Such as coat colour in Siamese cats - brown gene only functions below 37°C so fur growing at the bodies extremities paws face and ears is brown whilst the rest of the coat (where the cat is warmer) grows white.

Question 15

Sex influences the inheritance and expression of genes in a variety of ways

Answer 15

Sex influenced characteristics
Autosomal genes that are more readily expressed in one sex than the other

Sex limited characteristics
Autosomal genes whose expression is limited to one sex

Cytoplasmic inheritance
Cytoplasmic genes, which are usually inherited from only one parent

Question 16

Cytoplasmically inherited traits

Answer 16

1) present in females and males
2) typically only inherited from one parent (usually maternal mitochondrial inheritance)
3) reciprocal crosses give different results
4) exhibit extensive phenotypic variation even within families

Question 17

Sex influence on gene inheritance/ expression - cytoplasmic inheritance

Answer 17

Example cell contains an equal amount of mitochondria with wild type or mutated genes.
Mitochondria segregate randomly during cell division
Resulting in progeny cells that differ in numbers of mitochondria with wild type or mutated genes

Mitochondrial DNA abnormalities are linked to many hereditary human disorders

Question 18

Genetic maternal effect

Answer 18

The genotype of the maternal parent determines the phenotype of the offspring

E.g. in snails:
Dextral shelled (right hand coiled shell) male with s+s+ genome
Mates with sinistral (left hand coiled shell) female with ss genome

Gametes will be s+ and S

F1 offspring will be s+s and all will have sinistral shells as the genotype of the mother determines the phenotype of offspring

F2 generation resulting from self fertilisation of an s+s female results in offspring all with dextral shells.
This is because the mother of the F2 progeny has genotype s+ s

Offspring 1/4 s+s+ dextral
1/2 s+s dextral and 1/4 ss dextral

Question 19

Sex influenced genetics/ expression of genes summary

Answer 19

Epigenetics: phenomena due to alterations to DNA that do not include changes in the base sequence often affect the way in which DNA sequences are expressed

Involves chemical modifications e.g. DNA methylation

Genomic imprinting is a type of epigenetics differential expression of genetic material depending on whether it is inherited from the male or female parent.

E.g. insulin-like growth factor 2-IGF2
The gene is expressed only when transmitted by the male parent. In mice, if the paternal copy is deleted a small placenta and low birthrate of offspring occurs as a result (maternal allele is silent in this case, probably due to methylation)