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Where a heterozygote has a phenotype intermediate between the corresponding homozygote phenotypes.

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Presentation on theme: "Where a heterozygote has a phenotype intermediate between the corresponding homozygote phenotypes."— Presentation transcript:

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2 Where a heterozygote has a phenotype intermediate between the corresponding homozygote phenotypes

3 Incomplete or Partial Dominance involving Flower Color in Snapdragons R 1 R 1 xR 2 R 2 RedWhite R 1 R 2 xR 1 R 2Pink R 1 R 1 R 1 R 2 R 2 R 2 1/4 Red 1/2 Pink 1/4 White genotypic and phenotypic ratios are the same (1:2:1)

4 codominance - where the heterozygote shows the phenotype of both alleles equally

5 Multiple Alleles & Codominance ABO Blood Groups Discoverer -- Landsteiner ( ) Group A, B, O, AB Genetic Explanation -- Bernstein (1924) Series of multiple alleles I A, I B, I O (i) …O recessive, A and B alleles codominant the A and B alleles make a different version of the same protein with a slightly different function

6 Molecular basis of ABO alleles fucose added to H-substance precursor product of A allele adds N-acetylgalactosamine product of B allele adds galactose both molecules (A and B) can be expressed on cell surface at the same time in OO individuals H substance is not modified

7 type A type Btype O type AB Red blood cell antigens A antigenA and B antigen B antigen H antigen (no A or B antigen)

8 IAIAIAIOIAIAIAIO Blood Group A41%AAnti-B IBIBIBIOIBIBIBIO Blood Group B10%BAnti-A IOIOIOIO Blood Group O45% Neither Anti-A Anti-B IAIBIAIB Blood Group AB4%BothNeither GenotypePhenotypeFrequencyAntigen Antibody in Serum Note: Caucasian frequencies given. Frequencies differ with population.

9 Legal Applications of Blood Groups Hospital Cases -- Babies mixed up Father - OBaby 1 - B Mother - ABaby 2 - O I A I A or I A I O x I O I O I B I B or I B I O impossible, but I O I O possible; therefore baby #2 belongs to these parents

10 Blood tests can be used to rule out paternity Charlie ChaplinJoan Barry O=I O I O x A=I A I O Caroline B= I B I O Caroline is not Charlie Chaplins daughter.

11 Rh Blood Group Landsteiner and Weiner (1940) Rhesus Monkey Blood Injected into Rabbit Extract serum from rabbit. Find that rabbit contains anti-Rh antibodies. Test in ManPhenotypeGenotype 1) 85% Agglutination (+)Rh + RR or Rr 2) 15% Non-Agglutination (-)Rh - rr

12 Rh Blood Group Alleles Multiple Alleles Weiner R 1 R 2 R 0 R z r r l r ll r y Closely Linked Loci Fisher-Race CDe cDE cDe CDE cde Cde cdE CdE Weiner - One complex locus, 8 different alleles Fisher-Race - Three separate loci; simple alleles, D allele most important; D = Rh + ; dd = Rh - C and E loci may effect expression of Rh phenotype Rh + Rh -

13 Rh Incompatibility Rh - xRh + rrRR Affected (Erythroblastosis fetalis) all offspring Rr

14 Explanation for Rh Incompatibility Mother Rh -, lacks Rh antigen on blood cell surfaces Seepage of blood from 1st born into mothers blood stream from trauma of birth Mothers immune system interprets entering Rh antigens as foreign protein to be eliminated Offspring heterozygous Rr; produce Rh antigen on red blood cell surface Mother builds up antibody against Rh antigen after birth of 1st child; takes time to build antibody; 1st child spared Rh antibody is IgG class and can pass across placenta and enter blood stream of fetus 2, 3 & 4 Mothers antibodies destroy fetus red blood cells; results in erythroblastosis fetalis Expected frequency is 10% of all pregnancies but observed at frequency of 1/200….C and E loci may modify immune response

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16 mice can have agouti (silver gray) or yellow coat color yellow allele dominant to agouti allele homozygotes for the yellow allele (A Y ) do not survive embryonic development F1 phenotypic ratio distorted (2:1) due to presence of lethal allele yellow allele dominant with respect to coat color but recessive with respect to embryonic development genes can be multi-functional

17 An interaction between genes such that a mutation in one gene masks the expression of the phenotype of another gene.

18 recessive 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 AaBb 9 A_B_agouti 3 A_bb albino 3 aaB_ black 1 aabb albino gene B gene A b is epistatic to A or a. bb aa B allele required to deposit pigment; A allele required to modify pigment pattern precursor black agouti pattern

19 white yellow green dominant epistasis squash colorA_ = 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 AaBb 9 A_B_white 3 A_bb white 3 aaB_ yellow 1 aabb green gene a (aa) gene b (bb) A is epistatic to B or b. A_ B_ a allele required to make yellow pigment; b allele required to convert yellow to green

20 O type woman found to be incompatible with type O blood prior to transfusion based on blood typing of parents could not be type O clearly passed on B alleles to two of her children

21 type A type Btype O type AB Red blood cell antigens A antigenA and B antigen B antigen H antigen (no A or B antigen)

22 hh I B I O x HH I A I O Hh I A I O or Hh I B I O or Hh I A I B Explanation If genotype is hh conversion does not occur and anti-H is made in serum….reacts with O-type blood cells which have H antigen h is epistatic to the A or B alleles

23 condition in which a single mutation in one gene simultaneously affects several characters

24 Examples of pleiotropy White eye gene in Drosophila eye color white flight muscles defective Albinism in humans lack of pigment eye sight can be affected hearing can be affected Hormones produced in pituitary gland required throughout body

25 interactions between two or more genes, each with an additive effect on the character

26 Kolreuters Tobacco Plants

27 Dwarf Population Tall Population Height Freq Intermediate Height F1F1 Height Freq. Height F2F2 Continuous Variation

28 Explanation of Kolreuters Results ParentsAABB Xaabb 6 feet2 feet F 1 AaBbXAaBb AABB AABb AaBB AaBb AABb AAbb AaBb Aabb AaBB AaBb aaBB aaBb AaBb Aabb aaBb aabb AB Ab aB ab F2F2 AB Ab aB ab Residual = 2ft A = B = 1 ft added a = b = 0 ft added (non-additive alleles)

29 Polygenic Inheritance F 2 Results Genotypes Phenotypes Frequency AABB 6 feet tall 1 AABb 5 feet tall 2 AaBB 5 feet tall 2 AAbb 4 feet tall 1 AaBb 4 feet tall 4 aaBB 4 feet tall 1 Aabb 3 feet tall 2 aaBb 3 feet tall 2 aabb 2 feet tall

30 Calculating the number of genes involved in determining a polygenic character. The number of possible different F2 phenotypes = 2n+1 n = the number of heterozygous gene pairs determining the character We know that all genes pairs in the F1 cross are heterozygous because we start the parental cross with pure breeding lines.

31 sex chromosome - a chromosome whose presence or absence is correlated with the sex of the bearer, or, a chromosome that plays a role in sex determination autosome - any chromosome that is not a sex chromosome

32 I. Pattern Baldness Sex-influenced trait: Autosomal trait expressed as a dominant in one sex and as a recessive in the other.

33 Sex-Influenced Trait Inheritance Pattern B B B B pattern baldness X B N B N non- bald B B B N pattern baldness B B B N non- bald X B allele is dominant in males and recessive in females. N allele is dominant in females and recessive in males.

34 Pattern Baldness F2 Summary Genotype Phenotype Males Females 1 B B B B Bald Bald 2 B B B N Bald Non-Bald 1 B N B N Non-Bald Non-Bald

35 Sex Influenced Trait II. 2nd Finger Shorter than 4th S S S S shorter X S L S L longer S S S L shorter S S S L longer X

36 2nd Finger Shorter than 4th F2 Summary Genotypes Phenotypes Males Females 1 S S S S Shorter Shorter 2 S S S L Shorter Longer 1 S L S L Longer Longer

37 Sex-limited Trait: Autosomal trait expressed in one sex only.

38 Cock and Hen Feathering in Chickens Hamburgh rooster (cock feathering) Sebright hen (hen feathering) cock feathers: curved and pointed hen feathers: shorter and rounded phenotype controlled by single gene, h

39 hh X h+h+h+h+ h+hh+h h+hh+h X cock feathers hen feathers

40 Feathering in Chickens F2 Summary Genotype Phenotype Males Females 1 h + h + hen hen 2 h + h hen hen 1 hh cock hen Leghorn chickens all hh - males always different Sebright bantams all h + h + - males/females have same feathers


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