BIO260 Lectures:

 

Lecture 2.5 ¡V Epistasis

Lecture 8.5 ¡V Genes, Phenotypes, and Mutations

Chapter 20, P. 688-698 ¡V Evolution of Traits

 

Lecture 2.5:

 

Epistasis:

Ø      Interactions between genes

Ø      The effect of one gene on another, which affects an organism¡¦s phenotype

Ø      Problems involving this are usually associated with poly-gene systems

 

Types of Gene Interaction (refer to P. 58):

1)      None

Ø      There is no interaction between genes

Ø      9:3:3:1 ratio

 

2)      Complementary Gene Action:

Ø      Dominant allele of each of the genes is required to produce one of the phenotypes

v     Dominant alleles might code for enzymes in a linear enzyme pathway

Ø      9:7 ratio

 

3)      Recessive Epistasis:

Ø      The homologous recessive of a gene masks the effects of another gene

Ø      9:3:4 ratio

 

4)      Dominant Epistasis Type 1:

Ø      The homologous dominant of a gene marks the effects of another gene

Ø      12:3:1 ratio

 

5)      Dominant Epistasis Type 2:

Ø      The homologous dominant of a gene marks the effects of the dominant allele of another gene

Ø      13:3 ratio

 

Method of Obtaining Probability of a Certain Phenotype:

Ø      Probabilities of specific allele combinations can be obtained through product rule

Ø      For example: To obtain AA Bb:

v     AA occurs at a probability of ¼, Bb occurs at a probability of ½

v     Using the product rule, we get: (¼)(½) = 1/8

v     This is a much more efficient way than to use a Punnet square

Ø      A more practical example from text book (P. 56)

v     Labrador retriever is an example of Recessive Epistasis

v     Gold phenotype is expressed when Gene 2 is homozygous recessive regardless of the status of Gene 1

v     Brown phenotype is expressed when Gene 2 is not homozygous recessive and Gene 1 is homozygous recessive

v     Black phenotype is expressed when Gene 1 and 2 are not homozygous recessive

v     Probability of homozygous dominant or recessive: ¼

v     Probability of heterozygous: ½

v     Probability of not homozygous recessive: 1 ¡V ¼ = ¾

¡±         Gold phenotype probability = (1)(¼) = 4/16

¡±         Brown phenotype probability = (¾)(¼) = 3/16

¡±         Black phenotype probability = (¾)(¾) = 9/16

 

Heterogeneous Traits:

Ø      Mutations of different genes might possibly give rise to an identical trait

v     i.e. Deafness

 

Complementation Test:

Ø      Only useful for recessive heterogeneous trait

Ø      Used to determine whether or not two different individuals of the same phenotype (in one aspect) shares the same mutation

Ø      It is done by mating the two individuals and observing their offsprings

Ø      If all the offsprings do not have that trait, the parents do not share the same mutation and it is called ¡§Complementation¡¨

Ø      If all the offsprings share the same mutation, the parents share the same mutation and it is called ¡§Noncomplementation¡¨

Ø      Logically, there should not be a case such that some offsprings have the mutation while others do not unless there is a spontaneous mutation

 

Summary:

Ø      Genes can interact to form different phenotypes

Ø      Dominant alleles of two interacting genes can be important for producing a particular phenotype

Ø      One gene¡¦s effect can mask another¡¦s

Ø      A phenotype can arise from different genes

Ø      Even more different phenotype can be produced if we include the factors of codominance or incomplete dominance

 

Breeding Studies:

Ø      Can be used to determine what and how genes are associated with a trait as well as the presence or absence of epistasis

Ø      Usually cannot be done on humans (we use pedigrees instead)

 

Example: Mice

Ø      We mate a brown mouse with a white mouse.

Ø      The F2 generation is 9 Black : 3 Brown : 4 White. (very close to 2:1:1 ratio)

Ø      To determine whether this is recessive epistasis (2 genes) or incomplete dominance of 1 gene, we cross the brown and the white in F2

Ø      If the offsprings are¡K

v     All black è Incomplete dominance

v     Half black, half brown è Recessive epistasis

 

Factors affecting phenotype expression:

Ø      A genotype does not always express the same phenotype due to:

1)      Environment

v     Conditionally Lethal Alleles (CLA):

¡±         Alleles that are lethal under some conditions

v     Permissive:

¡±         Conditions that are viable to organisms with CLA

¡±         At times, individuals with CLA is almost indistinguishable from a wild-type under such conditions

v     Restrictive:

¡±         Conditions that are lethal to organisms with CLA

2)      Modifier genes è Genes that have an effect on a phenotype, modify the effects of other genes

3)      Chance

v     Random events can cause change to phenotype through change in environment and degree of exposure to environmental agents.

v     For example:

¡±         Cosmic rays, random mistake in cellular machinery

Ø      A phenotype is not always completely expressed and this factor is measured by:

1)      Penetrance è % of population with expected phenotype

2)      Expressivity è How intense the trait is expressed (i.e. affects one eye or two eyes, intensity of colour)

v     Variation in penetrance and expressivity can be a result of chance and/or modifier genes

v     Penetrance and expressivity cannot be derived from Mendel¡¦s principles but can be determined by observation and statistics

Ø      Phenocopy:

v     Change in phenotype as a result of environmental agents

v     This change often mimics the effects of a mutant gene

v     Not inheritable since genotype is not changed

v     Ex. Chemically induced mental retardation

Ø      Some what of an opposite to Phenocopy:

v     Sometimes, individuals with deleterious mutant genes can have much lower expressivity and/or penetrance due to environment

¡±         Ex. PKU victims can live as normal individuals if their diet has a minimal amount of phenylalanine

¡±         Ex. People in danger of heart disease (genetically) can lessen the expressivity/penetrance if exercise regularly and have healthy diet

 

Continuous/Quantitative Traits:

Ø      A result of many genes (polygenic)

Ø      Possibly involve complex forms of epistasis

 

Lecture 8.5:

 

Genes vs. Proteins:

Ø      Genes encode proteins/polypeptides

Ø      Mutant genes can be translated into drastically different proteins

v     i.e. Hemoglobin A and Hemoglobin S:

¡±         Difference in just one amino acid residue (Glu-6-Val)

¡±         HbS = sickle-cell-anemia, HbA = normal

¡±         Great structural difference between the two

¡±         Example of missense mutation

Ø      Multiple genes can encode a quaternary (multimer) protein:

v     Each subunit can be encoded by a different gene

v     Alteration in one subunit by a mutant gene can affect function of multimer

¡±         i.e. HbA and HbS

Ø      Phenotype and dominance/recessiveness of alleles can be traced back to genes

Ø      Mutations in a gene can vary in effect

v     Modification of amino acid residues in active site usually give rise to significant consequences whereas changes in non-structured regions of the protein might not

Ø      Some mutations do not affect amino-acid composition (neutral mutations), but still generate abnormal phenotype by affecting polypeptide production:

v     Possibly something to do with the regulatory proteins not recognizing the change recognition sequence

v     Recall: There are degenerates in codons. Change in bases may not always show in proteins

Ø      It might take more than one gene to give rise to a specific phenotype

 

Missense Mutation:

Ø      Mutation that causes a substitution of an amino acid (or codon)

 

Null Mutations:

Ø      Tend to be recessive

Ø      Can¡¦t synthesize the right protein because of modification or deletion of the gene

Ø      In heterozygous genotype, the wild-type allele can make up for the null allele by synthesizing some of the proteins.

v     If that single wild-type allele can produce above the threshold amount (amount needed to fulfill normal biochemical requirements of the cell, usually ½ the normal amount), then the heterozygote will be of wild-type phenotype

 

Hypomorphic Mutations:

Ø      Hypomorphic alleles have reduced production of a particular protein

Ø      Heterozygotes usually produce more than half the amount of wild-type thus the effect is often noticeable only in homozygous recessive

 

 

Incomplete Dominance:

Ø      Can be a result of null or hypomorphic mutations where the phenotype varies continuously with amount of protein produced

v     i.e. Pigments in flowers

 

Haploinsufficiency:

Ø      Situation when one wt allele is not enough to produce the wt phenotype.

Ø      Phenotype is extremely sensitive to amount of a particular protein

Ø      Not common

Ø      Heterozygotes tend not to show wt phenotype

 

Hypermorphic Mutations:

Ø      Generate an excess amount of protein such that the surplus affects the genotype

 

Dominant Negative Alleles:

Ø      Encode proteins that inhibit activity of other proteins

 

Neomorphic Mutations:

Ø      Produce proteins with a new function or produce normal proteins at inappropriate time/place (ectopic expression)

 

Chapter 20, P. 688-698 ¡V Evolution of Traits:

 

Main Idea:

Ø      Evolution occurs when there is selection

Ø      Pathogens and parasites develop resistance through evolution

v     Pathogen drug resistance is often evolved from incomplete medical treatments where pathogens are not completely eliminated by antibiotics (selective pressure)

¡±         The remaining ones might have developed partial resistance to drug and will multiply once more in absence of the antibiotic creating a lot of partial-resistant bacteria

¡±         If that antibiotic is taken again, complete resistance can be evolved.

¡±         This is kind of like a multi-staged evolution

Ø      Fitness Cost:

v     There are usually trade-offs for resistant traits

v     These trade-offs can be due to a variety of physical factors such as reduced metabolic efficiency

v     In absence of selective agent (drug), the resistant strains will be less favoured than the wild-type

Ø      Ecological Factors of Applying Insecticides:

v     Let¡¦s assume, if a concentrate dose of insecticide is applied, both the target and its predators/parasites are affected

v     The target (prey) will much more likely to develop resistance than the predator because:

¡±         The predator is also killed

¡±         Its food source is limited due to massive death of its prey

v     The predator population can even die out because of the above reasons

¡±         This will further favour the regeneration of prey population since the predators are now decimated

Ø      Phenotypic Variation for a Continuous Trait:

v     Mean = Average quantity of the trait

v     Variance = (Sum of Individual Deviation from Mean)2/(Number of Individuals)

v     Total Phenotype Variance (VP) = Genetic Variance (VG) + Environmental Variance (VE)

v     Graphs n P.693:

¡±         (VE)1/2 = Dx of the curve at ½ ymax, in a curve for genetically identical population

¡±         (VG)1/2 = Dx of the between curves for genetically identical population and genetically diverse population, at ½ ymax,

v     Heritability (h2) = VG/VP