Beyond Mendelism
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Transcript of Beyond Mendelism
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