LECTURE 11: FROM GENE TO PHENOTYPE II Fchapter 6 Fquestions & concepts Fgenes & gene products Fallele interactions Fgene & protein interactions Fchi-square applications Fexam 1: problems

GENE & PROTEIN INTERACTIONS Fhow do you identify interacting genes? 1. generate new mutant alleles for a trait of interest by mutagenesis 2. test mutant alleles to determine if they represent alleles of multiple genes by (a) mapping & (b) complementation 3. combine mutant alleles to identify epistatic interactions

GENE & PROTEIN INTERACTIONS Fa complementation test will reveal allelic relationships 1 gene (below) dominance / recessiveness 2. semi-dominance = incomplete dominance 3. co-dominance ( e.g. : I A & I B of ABO system) 4. multiple alleles ( e.g. : I A, I B & i of ABO system) 5. conditional ( e.g. : temperature sensitive) 6. lethality 7. sterility F... or give wild type phenotypes 2 genes

Fare mutations allelic (1 gene) or not (2 genes)? Ftest by making pair-wise crosses between mutants COMPLEMENTATION TEST

 gene  genes COMPLEMENTATION TEST

F... because the 2 mutations are in different genes

Fwhat is epistasis? text (1): a situation in which the differential phenotypic expression of a genotype at one locus depends on the genotype at another locus text (2): a mutation that exerts its expression while canceling the expression of the alleles of another gene Herr Doktor Professor: interactive expression of alleles at two (or more) genes (both of the text definitions are limiting) GENE & PROTEIN INTERACTIONS

Mendelian phenotypic ratios

FMendelian dihybrid ratio GENE & PROTEIN INTERACTIONS

  Mendelian dihybrid ratio parallel pathways dominant epistasis recessive epistasis duplicate additive genes duplicate dominant genes duplicate recessive genes recessive suppression (1) recessive suppression (2)

PARALLEL PATHWAYS bw + /_; st + /_ 9 bw + /_; st/st 3 bw/bw; st + /_ 3 bw/bw; st/st 1 9 : 3 : 3 : 1

PARALLEL PATHWAYS o + /_; b + /_ 9 o + /_; b/b 3 o/o; b + /_ 3 o/o; b/b 1 9 : 3 : 3 : 1

DOMINANT EPISTASIS D/_; W/_ 9 d/d; W/_ 3 D/_; w/w 3 d/d; w/w 1 12 : 3 : 1 W  D gene expression

RECESSIVE EPISTASIS w + /_; m + /_ 9 w + /_; m/m 3 w/w; m + /_ 3 w/w; m/m 1 9 : 3 : 4

DUPLICATE DOMINANT GENES A 1 /_; A 2 /_ 9 A 1 /_; a 2 /a 2 3 a 1 /a 1 ; A 2 /_ 3 a 1 /a 1 ; a 2 /a : 1 1 dominant allele  pigment

DUPLICATE RECESSIVE GENES (aka COMPLEMENTARY GENE ACTION) C/_; P/_ 9 C/_; p/p 3 c/c; P/_ 3 c/c; p/p 1 9 : 7 1 recessive allele  pigment

REGULATING GENE & TARGET INTERACTION r + /_; a + /_ 9 r/r; a + /_ 3 r + /_; a/a 3 r/r; a/a 1 9 : 7

(A POSSIBLE) SUPPRESSION MECHANISM

RECESSIVE SUPPRESSION pd + /_; su + /_ 9 pd + /_; su/su 3 pd/pd; su + /_ 3 pd/pd; su/su 1 13 : 3 pd + su + pd supd mechanism for genotype  not explained by model

GENE & PROTEIN INTERACTIONS   Mendelian dihybrid ratio parallel pathways dominant epistasis recessive epistasis duplicate additive genes duplicate dominant genes duplicate recessive genes recessive suppression (1) recessive suppression (2)

MODIFIERS Fmodifier mutations influence expression of a mutation at a second locus Fbroadly used term (text gives a few examples) Fgene activity can be modified by... Fup- or down-regulation Fenhancement or suppression

SYNTHETIC LETHALS

PENETRANCE & EXPRESSIVITY Fpenetrance: the % of individuals of given genotype that exhibit a phenotype... a population measurement Fexpressivity: the extent to which a given genotype is expressed at the phenotypic level... in each individual

PENETRANCE & EXPRESSIVITY Fcan be due to environmental factors 2. genetic factors 3. unknown / difficult to measure (text) ??? F e.g., the rover / sitter foraging polymorphism

POLYGENY GENE  GENE  PHENOTYPE GENE      BIOCHEMICAL PATHWAY:       PHENOTYPE   COMPETITION / INHIBITION:   PHENOTYPE     ENZYME CATALYSIS:     PHENOTYPE

PLEIOTROPY PHENOTYPE  GENE PHENOTYPE  PHENOTYPE    SEQUENTIAL PHENOTYPES:GENE         RELATED PHENOTYPES:GENE       UNRELATED PHENOTYPES:GENE    

GENE & PROTEIN INTERACTIONS        

CHI-SQUARE APPLICATIONS Fexample question on p. 209 Ftrue-breeding P yellow x red petal plants all orange petal F 1. F 1 are bred & F 2 are... red 77 orange182 yellow 61 total320 Fwith 3 phenotypes, reasonable hypotheses are... H 1 = incomplete dominance... 1:2:1 ratio E = 80 G 1 /G G 1 /G G 2 /G 2  2 = 7.6, P < (2.5%),  H 1 

H 2 = recessive epistasis of r on Y & y... 9:3:4 ratio E = 180 Y/_; R/_ + 60 y/y; R/_ + 80 [ Y/_; r/r + y/y; r/r ]  2 = 0.15, P > 0.9 (90%),  H 2  CHI-SQUARE APPLICATIONS Fexample question on p. 209 Ftrue-breeding P yellow x red petal plants all orange petal F 1. F 1 are bred & F 2 are... red 77 orange182 yellow 61 total320 Fwith 3 phenotypes, reasonable hypotheses are...

PRACTICE PROBLEMS you should be able to do CH6, p. 214 #