Beyond Mendelism

Fig. 12.1 Allelic forms of a gene

1. Single-gene inheritance :a. Deviations from complete dominance and recessiveness

b. Multiple alleles

c. One gene determine more than one trait

2. Multifactorial inheritance

Extension to Mendel

Deviations from complete dominance and recessiveness

Fig. 3.3

Incomplete dominance

CodominanceF1 hybrid display the traits of both parents

spotted dotted

Multiple allelesComplete set of known alleles of one

gene is called an allelic series

Fig. 3.5

ABO blood types are determined by three alleles of one gene

Fig. 3.6

Establishing thedominance relationsbetween multiple alleles

Mutations are the source of new allelesWild-type allele: frequency more than 1%

Mutant allele: frequency less than 1%

Monomorphic(One wild-type allele)

ABO blood type: polymorphic

agouti

black/yellowblack

Two alleles with recessive lethal Recessive lethal alleles

Table 3.1

Sickle-cell anemia

Mutant -globin aggregates to form long-fiber

Pleiotropy

Pleiotropy of sickle-cell anemia: dominance relation vary

Cells break down

Oxygen drops

Phenotype at level of expression of anemiaHbA is dominant

Phenotype at cell shape level

HbA/HbA Normal shape

HbS/HbS Sickled

HBS/HbA Partially sickled***

High altitudes: HbA & HBs are incompletely dominant

Se level: A dominant

Phenotype at protein level

At this molecular level, alleles HbA and HBs are codominant

Phenylketonuria (PKU)

Phenylketonuria• Autosomal recessive disease- single gene disease• Defective allele of gene coding for liver enzyme

phenylalanine hydroxylase (PAH)• Phe in diet converted into phenylpyruvic acid,

transported to brain via bloodstream • Impedes normal development, mental retardation• Many mutations at different sites (allelic series):

Near active site (null) or away from it with residual function (leaky)– Normally functioning wild type (P)

– All defective recessive mutations, null and leaky (p)

Phenylketonuria• Underlying complexity of genetic system

involved:• Some cases of elevated phenylalanine level and

its symptoms: Not associated with PAH locus (other genes involved)

• Some cases: People with PKU (elevated phenylalanine) no abnormal cognitive development

• Hence involvement of other genes + environment (gene for tetrahydrobiopterin synthesis)

Phenylketonuria• Many steps in pathway from Phe ingestion to

impaired cognitive development– Amount of Phe in diet– Transportation of Phe to appropriate sites in

liver– Liver: PAH + cofactor, tetrahydrobiopterin – To affect cognitive development: Excess

phenylpyruvic acid must be transported to brain via bloodstream, pass through BBB

– Inside brain: Developmental processes must be susceptible to action of phenylpyruvic acid

Multifactorial InheritanceInteraction of genes

Identifying the interacting genes that contribute to a particular biological property

1. Treat cells with mutagens (UV): Set of mutants with abnormal expression of property under study

2. Test the mutants to determine how many gene loci are involved, which mutations are alleles of same gene

3. Combine mutations pairwise by means of crosses to form double mutants to see if they interact (gene product interaction). Specific ratio of progeny

Complementation

• Harebell: Blue WT, three white-petaled mutants, homozygous pure-breeding strains $, £, and ¥

• Phenotypically alike, genetically identical ?

• Mutant condition determined by recessive allele of a single gene

Harebell plant (Campanula species)

Are they three alleles of one gene, or of two genes, or of three genes?

Check for complementation

Interaction of genes1. Complementation:• Production of a wild-type phenotype when two

recessive mutant alleles are brought together in the same cell

• Complementation test: By intercrossing two individuals that are homozygous for different recessive mutants

• Observe whether progeny have wild-type phenotype– If recessive mutations are alleles of same gene: no

wild-type progeny

Complementation

• Two recessive mutations in different genes wild-type function provided by respective wild-type alleles

• Cross F1 dihybrid plants• F2 9:7 (blue:white)

– modification of the dihybrid 9 :3:3:1

• Complementation is a result of cooperative interaction of wild-type alleles of two genes

• Plant will have white petals if it is homozygous for the recessive mutant allele of either gene or both genes

• Blue phenotype: At least one dominant allele of both genes (both are needed to complement each other and complete sequential steps in the pathway)

• Three of genotypic classes will produce same phenotype, so overall only two phenotypes result. Different steps in a biochemical pathway

Homozygous mutation

Theoretical notation

Cross between

F1 Enzyme 2 functional but no substrate

Other crosses: wild-type alleles for both enzymes

Three phenotypically identical whitemutants—$, £, and ¥—are intercrossed

Mutations in the same gene (such as $ and £) cannot complement, because the F1 has one gene with two mutant alleles. Pathway is blocked and flowers are white

When mutations are in different genes (suchas £ and ¥), complementation of wild-type alleles of each gene occurs in F1 heterozygote. Pigment is synthesized and flowers are blue

Complementation: Gene interaction from different pathways

• Inheritance of skin coloration in corn snakes• Natural color: Repeating black and orange

camouflage pattern• Orange pigment: o+ (presence of orange

pigment) and o (absence of orange pigment).• Black pigment: b+ (presence of black pigment)

and b (absence of black pigment).

Genotypes

• Natural: o+/– ; b+/–.• Black: o/o ; b+/–• Orange: o+/– ; b/b• Albino: o/o ; b/b

Independence of interacting genes 9:3:3:1

Epistasis• Ability of a mutation at one locus to override a mutation

at another in a double mutant• When a mutant allele of one gene masks expression of

alleles of another gene and expresses its own phenotype– Overriding mutation: epistatic– Overridden mutation: hypostatic

• Genes in same cellular pathway

• Epistatic mutation of gene earlier in pathway than that of hypostatic

White (w/w) Magenta (m/m) unlinked

WT

9:3:4= Epistasis, white epistatic to magenta

Petal pigment synthesis in blue-eyed Mary (Collinsia parviflora)

(a) Rose (b) pea (c) walnut (d) single

Domestic chickens comb shapes

Comb shape: Two independently assorting genes (R and P) each with two alleles

Wyandotte chickens: Rose combs (RRpp)

Brahma chickens: Pea combs (rrPP)

F1 hybrids: Walnut (RrPp)

F2: of 9:3:3:1 walnut:rose:pea:single

Figure 4.17b

The crosses of Bateson and Punnett

Fig. 3.13

Epistatic: the effect of one gene hides the effect of the other gene

Recessive epistasis

Addition of A or B sugars

H allele is epistatic to the I gene

Suppressors

• Suppressor: Mutant allele of one gene that reverses effect of a mutation of another gene, resulting in a wild-type or near wild-type phenotype• Use a mutant, expose to mutagen and screen descendents for WT

Difference from epistasis:

• Suppressor cancels expression of a mutant allele and restores corresponding wild-type phenotype• Two phenotypes segregateF2 ratio: 13:3

Red: WT

Pd: Purple

su suppressespd (red WT)

Pd+ (Red)Pd/su (Red)Pd (Purple)

Penetrance & Expressivity

Penetrance: percentage of individuals with a given allele who exhibit the phenotype associated with that allele

Expressivity measures degree to which a given allele is expressed at phenotypic level (intensity of phenotype)

Penetrance & Expressivity

Reasons for not expressing trait:1.Influence of environment: Phenotype of

mutant individual raised in one set of circumstances may match phenotype of a wild-type individual raised in a separate set of circumstances

2. Influence of other genes: Modifiers, epistatic genes, or suppressors in rest of genome may act to prevent expression of typical phenotype