2I. INCOMPLETE DOMINANCE Where a heterozygote has a phenotype intermediatebetween the corresponding homozygote phenotypes
3Incomplete or Partial Dominance involving Flower Color in Snapdragons R1R1 x R2R2Red WhiteR1R2 x R1R2Pink PinkR1R R1R2 R2R21/4 Red /2 Pink /4 Whitegenotypic and phenotypic ratios are the same (1:2:1)
4II. MULTIPLE ALLELES AND CODOMINANCE codominance - where the heterozygote shows thephenotype of both alleles equally
5Multiple Alleles & Codominance ABO Blood GroupsDiscoverer -- Landsteiner ( )Group A, B, O, ABGenetic Explanation -- Bernstein (1924)Series of multiple allelesIA, IB, IO (i)…O recessive, A and B alleles codominantthe A and B alleles make a different version of the same protein with a slightlydifferent function
6Molecular basis of ABO alleles fucose added to H-substance precursor product of A allele addsN-acetylgalactosamineproduct of B allele addsgalactoseboth molecules (A and B) canbe expressed on cell surfaceat the same timein OO individuals H substanceis not modified
7Red blood cell antigens type Atype ABA antigen A and B antigenB antigen H antigen(no A or B antigen)type Btype O
8IAIA IAIO Blood Group A 41% A Anti-B IBIB IBIO Blood Group B 10% B AntibodyinSerumGenotypePhenotypeFrequencyAntigenIAIAIAIOBlood GroupA41%AAnti-BIBIBIBIOBlood GroupB10%BAnti-AIOIOBlood GroupO45%NeitherAnti-AAnti-BIAIBBlood GroupAB4%BothNeitherNote: Caucasian frequencies given. Frequencies differ with population.
9Legal Applications of Blood Groups Hospital Cases -- Babies mixed upFather - O Baby 1 - BMother - A Baby 2 - OIAIA or IAIO x IOIOIBIB or IBIO impossible, but IOIO possible;therefore baby #2 belongs to these parents
10 Caroline is not Charlie Chaplin’s Blood tests can be used to rule out paternityCharlie Chaplin Joan BarryO=IOIO x A=IAIOCarolineB= IBIO Caroline is not Charlie Chaplin’sdaughter.
11Rh Blood Group Landsteiner and Weiner (1940) Rhesus Monkey BloodInjected into RabbitExtract serum from rabbit. Find that rabbit containsanti-Rh antibodies.Test in Man Phenotype Genotype1) 85% Agglutination (+) Rh+ RR or Rr2) 15% Non-Agglutination (-) Rh- rr
12Rh Blood Group Alleles Multiple Alleles Weiner R1 R2 R0 Rz r rl rll ry Closely Linked LociFisher-RaceCDecDEcDeCDEcdeCdecdECdEWeiner - One complex locus, 8 different allelesFisher-Race - Three separate loci; simple alleles, D allele most important; D = Rh+; dd = Rh-C and E loci may effect expression of Rh phenotypeRh+Rh-
13Rh Incompatibility Rh- x Rh+ rr RR all offspring Rr Affected (Erythroblastosis fetalis)
14Explanation for Rh Incompatibility Mother Rh-, lacks Rh antigen on blood cell surfacesSeepage of blood from 1st born into mother’s blood stream from trauma of birthMother’s immune system interprets entering Rh antigens as foreign protein to be eliminatedOffspring heterozygous Rr; produce Rh antigen on red blood cell surfaceMother builds up antibody against Rh antigen after birth of 1st child; takes time to build antibody; 1st child sparedRh antibody is IgG class and can pass across placenta and enter blood stream of fetus 2, 3 & 4Mother’s antibodies destroy fetus red blood cells; results in erythroblastosis fetalisExpected frequency is 10% of all pregnancies but observed at frequency of 1/200….C and E loci may modify immune response
16mice can have agouti (silver gray) or yellow coat coloryellow allele dominant to agoutiallelehomozygotes for the yellow allele(AY) do not survive embryonicdevelopmentF1 phenotypic ratio distorted (2:1)due to presence of lethal alleleyellow allele dominant with respectto coat color but recessive withrespect to embryonic developmentgenes can be multi-functional
17IV. EPISTASISAn interaction between genes such that a mutation in one gene masks the expression of the phenotype of another gene.
18recessive epistasis mouse coat color A_ = agouti aa = black In a cross thought to be Aa x Aa see 9 agouti : 3 black : 4 albino.Modified dihybrid ratio implies 2 genes involved(9:4:3 is typical recessive epistatic ratio if 2 genes determine character)Can explain results if cross is actually AaBb x AaBb9 A_B_ agouti3 A_bb albino3 aaB_ black1 aabb albinob is epistatic to A or a.B allele required to depositpigment; A allele required tomodify pigment patternbb aaprecursor black agouti patterngene B gene A
19dominant epistasis squash color A_ = white aa = yellow In a cross thought to be Aa x Aa see 12 white : 3 yellow : 1 green.Modified dihybrid ratio implies 2 genes involved(12:3:1 is typical dominant epistatic ratio if 2 genes determine character)Can explain results if cross is actually AaBb x AaBb9 A_B_ white3 A_bb white3 aaB_ yellow1 aabb greenA is epistatic to B or b.a allele required to makeyellow pigment; b allelerequired to convert yellowto greenA_ B_white yellow greengene a (aa) gene b (bb)
20Blood Group Epistasis Bombay Phenotype O type woman found to beincompatible with type O bloodprior to transfusionbased on blood typing ofparents could not be type Oclearly passed on B alleles totwo of her children
21Red blood cell antigens type Atype ABA antigen A and B antigenB antigen H antigen(no A or B antigen)type Btype O
22Explanation hh IBIO x HH IAIO Hh IAIO or Hh IBIO or Hh IAIB If genotype is hh conversion does not occur and anti-H is madein serum….reacts with O-type blood cells which have H antigenh is epistatic to the A or B alleles
23V. PLEIOTROPYcondition in which a single mutation in one gene simultaneously affects several characters
24Examples of pleiotropy White eye gene in Drosophilaeye color whiteflight muscles defectiveAlbinism in humanslack of pigmenteye sight can be affectedhearing can be affectedHormones produced in pituitarygland required throughout body
25VI. POLYGENIC INHERITANCE interactions between two or more genes,each with an additive effect on the character
27Freq Height F1 F2 Tall Population Dwarf Population Intermediate Height Continuous VariationHeightFreq.F2Height
28Explanation of Kolreuter’s Results Residual = 2ftA = B = 1 ft addeda = b = 0 ft added(non-additivealleles)Explanation of Kolreuter’s ResultsParents AABB X aabb6 feet 2 feetF AaBb X AaBbF2AB Ab aB abABAbaBabAABB AABb AaBB AaBbAABb AAbb AaBb AabbAaBB AaBb aaBB aaBbAaBb Aabb aaBb aabb
30Calculating the number of genes involved in determining a polygenic character.The number of possible different F2 phenotypes = 2n+1n = the number of heterozygous gene pairs determiningthe characterWe know that all genes pairs in the F1 cross areheterozygous because we start the parental cross withpure breeding lines.
31VII. SEX-INFLUENCED TRAITS sex chromosome - a chromosome whose presence or absenceis correlated with the sex of the bearer,or,a chromosome that plays a role in sexdeterminationautosome - any chromosome that is not a sex chromosome
32I. Pattern BaldnessSex-influenced trait: Autosomal trait expressed as a dominant in one sex and as a recessive in the other.
37VIII. SEX-LIMITED TRAITS Sex-limited Trait: Autosomal trait expressed in one sex only.
38Cock and Hen Feathering in Chickens Hamburgh rooster (cock feathering)Sebright hen (hen feathering)cock feathers: curved and pointedhen feathers: shorter and roundedphenotype controlled by single gene, h
39h+h+ hh h+h h+h X cock feathers hen feathers X hen feathers
40Feathering in Chickens F2 Summary Genotype PhenotypeMales Females1 h+h hen hen2 h+h hen hen1 hh cock henLeghorn chickens all hh - males always differentSebright bantams all h+ h+ - males/females have samefeathers