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Presentation on theme: "EXTENSIONS OF MENDELIAN GENETICS."— Presentation transcript:


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

3 Incomplete or Partial Dominance involving Flower Color in Snapdragons
R1R1 x R2R2 Red White R1R2 x R1R2 Pink Pink R1R R1R2 R2R2 1/4 Red /2 Pink /4 White genotypic and phenotypic ratios are the same (1:2:1)

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 IA, IB, IO (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 Red blood cell antigens
type A type AB A antigen A and B antigen B antigen H antigen (no A or B antigen) type B type O

8 IAIA IAIO Blood Group A 41% A Anti-B IBIB IBIO Blood Group B 10% B
Antibody in Serum Genotype Phenotype Frequency Antigen IAIA IAIO Blood Group A 41% A Anti-B IBIB IBIO Blood Group B 10% B Anti-A IOIO Blood Group O 45% Neither Anti-A Anti-B IAIB Blood Group AB 4% Both Neither Note: Caucasian frequencies given. Frequencies differ with population.

9 Legal Applications of Blood Groups
Hospital Cases -- Babies mixed up Father - O Baby 1 - B Mother - A Baby 2 - O IAIA or IAIO x IOIO IBIB 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 paternity Charlie Chaplin Joan Barry O=IOIO x A=IAIO Caroline B= IBIO  Caroline is not Charlie Chaplin’s 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 Man Phenotype Genotype 1) 85% Agglutination (+) Rh+ RR or Rr 2) 15% Non-Agglutination (-) Rh- rr

12 Rh Blood Group Alleles Multiple Alleles Weiner R1 R2 R0 Rz r rl rll ry
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- x Rh+ rr RR all offspring Rr Affected
(Erythroblastosis fetalis)

14 Explanation for Rh Incompatibility
Mother Rh-, lacks Rh antigen on blood cell surfaces Seepage of blood from 1st born into mother’s blood stream from trauma of birth Mother’s 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 Mother’s 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

15 III. Lethal Alleles

16 mice can have agouti (silver gray)
or yellow coat color yellow allele dominant to agouti allele homozygotes for the yellow allele (AY) 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 IV. EPISTASIS 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 b is epistatic to A or a. B allele required to deposit pigment; A allele required to modify pigment pattern bb aa precursor black agouti pattern gene B gene A

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

20 Blood Group Epistasis Bombay Phenotype
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 Red blood cell antigens
type A type AB A antigen A and B antigen B antigen H antigen (no A or B antigen) type B type O

22 Explanation 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 made in serum….reacts with O-type blood cells which have H antigen h is epistatic to the A or B alleles

23 V. PLEIOTROPY 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

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

26 Kolreuter’s Tobacco Plants

27 Freq Height F1 F2 Tall Population Dwarf Population Intermediate Height
Continuous Variation Height Freq. F2 Height

28 Explanation of Kolreuter’s Results
Residual = 2ft A = B = 1 ft added a = b = 0 ft added (non-additive alleles) Explanation of Kolreuter’s Results Parents AABB X aabb 6 feet 2 feet F AaBb X AaBb F2 AB Ab aB ab AB Ab aB ab AABB AABb AaBB AaBb AABb AAbb AaBb Aabb AaBB AaBb aaBB aaBb AaBb Aabb aaBb aabb

29 Polygenic Inheritance F2 Results
Genotypes Phenotypes Frequency AABB 6 feet tall AABb feet tall AaBB feet tall AAbb feet tall AaBb feet tall aaBB feet tall Aabb feet tall aaBb feet tall aabb feet tall 1 4 6 4 1

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.

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
BBBB pattern baldness X BNBNnon- bald BBBN pattern baldness BBBN non-bald B allele is dominant in males and recessive in females. N allele is females and in males.

34 Pattern Baldness F2 Summary
Genotype Phenotype Males Females 1 BBBB Bald Bald 2 BBBN Bald Non-Bald 1 BNBN Non-Bald Non-Bald

35 II. 2nd Finger Shorter than 4th
Sex Influenced Trait II. 2nd Finger Shorter than 4th SSSS shorter X SLSL longer SSSL shorter X SSSL longer

36 2nd Finger Shorter than 4th F2 Summary
Genotypes Phenotypes Males Females 1 SSSS Shorter Shorter 2 SSSL Shorter Longer 1 SLSL Longer Longer

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