Lecture 5

Extensions into Mendel’s Laws II

Example of Additive Interactions

seed coat in lentils

Tan x Grey

F1 are brown; F2 9 brown, 3 tan, 3 gray, 1 green

9/3/3/1 ratio in F2 lentils

suggests two independently assorting genes for seed coat color (operating in independent pathways)

Biochemical Explanation of Lentil Seeds

2 genes controlling the same trait function additively in independent pathways; gray (one pathway) + tan (other pathway) → brown

Epistasis

a gene interaction in which an allele of one gene masks the effects of another gene’s alleles

The allele that does the masking

epistatic to the other gene

The gene that is masked

hypostatic to the other allele

Epistasis can be

recessive or dominant

Recessive Epistasis

must be homozygous recessive in order to mask other allele

Dominant Epistasis

one copy of an allele masks the other gene → dominant negative

Labrador Color

recessive epistasis; coat color is determined by two genes

Labrador retriever color controlled

alternative independently assorting alleles of two different genes, E and B

Gene B determines

black and brown

Gene E

recessive allele of E (e) is epistatic to B and determines yellow

Recessive Epistasis in Labs

9/3/4 ratio in F2 progeny of dihybrid crosses indicates recessive epistasis (9 B-E-, 3 bb-E-, 4 B-ee, bb ee)

Two Independent Genes vs Recessive Epistasis

9/3/3/1 v. 9/3/4; independent pathways v. additive pathway

Genotype ee

masks the effect of all B genotypes

Biochemical Explanation for Labradors

Protein E generates eumelanin from a colorless precursor and Protein B deposits eumelanin

Labrador B-

deposits eumelanin densely (black)

Labrador bb

deposits eumelanin less densely (brown)

Labrador ee animals

cannot make eumelanin (yellow)

Homozygosity for the h Bombay allele

epistatic to the l gene determining ABO blood types; recessive epistasis

All people of A, B, or O phenotype

carry at least on dominant wild type H allele and produce some H

Bombay phenotype with hh genotype

do not make any H at all and may appear to be type O → without H substance there is nothing for A or B sugar to attach to

Sweet peas

recessive epistasis; purple F1 progeny are produced by crossing two pure breeding white lines → dihybrid cross generates 9/7 ratio in F2 progeny

Complementation

each recessive allele is complemented by wild type allele

Reciprocal Recessive Epistasis

homozygosity for the recessive allele of either gene results in a white phenotype → two genes work in tandem to make purple sweet pea flowers; a dominant allele of each gene must be present to produce that color; 9/7 phenotype ratio

Biochemical Explanation for Sweet Pea Color

one pathway has two reactions catalyzed by different enzymes; at least one dominant allele for both genes is required for purple pigment, homozygous recessive for either or both genes results in no pigment

Heterogenous Traits

have the same phenotype but are caused by mutation in different genes; deafness in humans caused by mutations, about 50 genes

Complementation Testing

used to determine if a particular phenotype arises from mutation in the same or separate genes; can be applied ONLY with recessive phenotypes

Deafness in Humans

genetic heterogeneity; mutations in many genes can cause deafness

Ocular-cutaneous Albanisms is

heterogeneity; recessive

Summer Squash

12/3/1 ratio in F2 progeny of dihybrid crosses indicates dominant epistasis I; 12 A-B-, aa B-, 3 A- bb, 1 aa bb

Summer Squash Dominant B allele

causes white color and mask any combination of A and a alleles

Biochemical Explanation for Summer Squash

protein B is dominant allele that prevents pigment deposition, epistatic to any A allele; b is the normal allele, with B being the dominant negative

Chicken Feather Color

13/3 ratio in F2 progeny of dihybrid crosses indicates dominant epistasis II; B codes for dominant negative melanosome protein → stops melanosome

Chicken Feather Color B and A

B is epistatic to A; color requires at least one copy of A and the absence of B

Redundant genes control

leaf development in maize; redundant genes result in 15/1 phenotypic ratio

Proteins encoded by redundant

perform nearly the same function

Maize A and B

dominant alleles A and B specify proteins that function in independent pathways to instruct cells to become part of the leaf; a and b specify no proteins

Maize pathways

are both sufficient; only plants that lack both dominant alleles have thin leaves