Mendel conducted both monohybrid and dihybrid crosses Dihybrid cross initial parents differ in two traits

Transcription

1 Mendel conducted both monohybrid and dihybrid crosses Dihybrid cross initial parents differ in two traits

2 Mendel s breeding scheme for a dihybrid cross Figure 3-3 step 3 R/R;y/y r/r;y/y : genes are in trans, as above R/R;Y/Y r/r;y/y : genes are in cis Both of these parental combinations = the same F 1

3 Mendel s dihybrid crosses 9 : 3 : 3 : 1 ratio round = = 423 wrinkled = = 133 Ratio = 3.2 : 1 yellow = = 416 green = = 140 Ratio = 3.0 : 1 Figure 3-3 step 5 dihybrid ratio = two superimposed monohybrid ratios gene pairs assort independently

4 How Two 3:1 Monohybrid Ratios Produce a 9:3:3:1 Dihybrid Ratio Monohybrid Ratios: Seed Shape: Round = 3/4, Wrinkled = 1/4 Seed Color: Yellow = 3/4, Green = 1/4 Resulting Dihybrid Phenotypic Ratios: Round Yellow 3/4 x 3/4 = 9/16 Round Green 3/4 x 1/4 = 3/16 Wrinkled Yellow 1/4 x 3/4 = 3/16 Wrinkled Green 1/4 x 1/4 = 1/16

5 Even segregation and independent assortment of alleles from two genes underlies a 9 : 3 : 3 : 1 ratio Figure 3-4 step 4 From each gene pair, 1/2 of gametes will receive a dominant allele and 1/2 a recessive allele 1/2 will be R (vs. r ) and 1/2 will be Y (vs. y ) 1/2 x 1/2 = 1/4 will be R;Y Other gametic combinations will be R;y, r;y and r;y, each 1/4 of the total

6 A punnett square can illustrate the genotypes underlying a 9:3:3:1 ratio Figure 3-4 step 4 1/4 male R;Y X 1/4 female R;Y = 1/16 R/R;Y/Y

7 Punnett square illustrating the genotypes underlying a 9 : 3 : 3 : 1 ratio Note: Figure 3-4 step 10 the ratio of yellow to green is 3:1 the ratio of round to wrinkled is 3:1

8 Chromosomal Basis for Independent Assortment: genes on different chromosomes assort independently

9

10 Dominant phenotypes are represented by partial genotypes (ex. A-, - indicates either dominant or recessive allele may be present) Yellow pea could be either Y/Y or Y/y Yellow can be designated as Y/- or Y- This is called a partial genotype Yellow Round could be any of four genotypes Y/Y ; R/R Y/y ; R/R Y/Y ; R/r Y/y ; R/r For simplicity the partial genotype is used: Y/- ; R/- or Y- ; R- or Y- R-

11 The ratio of partial genotypes is equal to the ratio of phenotypes A- ; B- 3/4 A- x 3/4 B- = 9/16 A- ; bb 3/4 A- x 1/4 bb = 3/16 aa ; B- 1/4 aa x 3/4 B- = 3/16 aa ; bb 1/4 aa x 1/4 bb = 1/16 When genes assort independently phenotypic and genotypic dihybrid ratios can be calculated as the product of their monohybrid ratios Also see:

12 Use of a Punnett square in dihybrid crosses is cumbersome Figure 3-4 step 10

13 What proportion would be RRyy? Figure 3-4 step 10

14 What proportion would be RRyy? 1/4 RR x 1/4 yy = 1/16 RRyy Figure 3-4 step 10

15 What proportion would be RrYy? Figure 3-4 step 10

16 What proportion would be RrYy? 1/2 Rr x 1/2 Yy = 1/4 RrYy Figure 3-4 step 10

17 What proportion would be R- yy? Figure 3-4 step 10

18 What proportion would be R- yy? 3/4 R- x 1/4 yy = 3/16 R- yy Figure 3-4 step 10

19 Calculation of Phenotypic Ratios for Crosses with Multiple Genes Purple, Yellow, Tall, Round white, green, short, wrinkled CC YY TT RR cc yy tt rr F 1 = all Purple, Yellow, Tall, Round Cc Yy Tt Rr F 1 shows the dominant characters, will be heterozygous for all

20 Calculation of Phenotypic Ratios for Crosses with Multiple Genes Purple, Yellow, Tall, Round white, green, short, wrinkled CC YY TT RR cc yy tt rr F 1 = all Purple, Yellow, Tall, Round Cc Yy Tt Rr What fraction of the F 2 will be Purple, green, short, Round? Answer = 3/4 x 1/4 x 1/4 x 3/4 = 9/256 What fraction will be CC Yy Tt rr? Answer = 1/4 x 1/2 x 1/2 x 1/4 = 1/64

21 A dihybrid testcross = F 1 crossed to double recessive. It produces a 1:1:1:1 phenotypic ratio The parental combinations (round, green and wrinkled, yellow) equal the recombinants (round, yellow and wrinkled, green) Gametes = R;Y R;y r;y r;y Gametes = r;y

22 Genetic and Chromosomal Basis for 1:1:1:1 Testcross Ratio Testcross: Phenotypes of testcross progeny reflect the genotypes of gametes from heterozygous parent Figure 3-13 step 4 Dom/Dom, rec/rec = Parental Dom/rec, rec/dom = Recombinant

23 Independent assortment cis or trans Parental genes in cis Parental genes in trans p 2 F 1 produces 4 different gametic genotypes in equal frequency for cis or trans parental combinations: AB, ab (cis) = Parental ab, Ab (trans) = Recombinant ab, Ab (trans) = Parental AB, ab (cis) = Recombinant

24 Linked genes do not assort independently Linkage: genes on same chromosome reduces the frequency of recombinants Figure 4-8 step 2 Linked gene combinations will tend to be inherited together. F 1 will produce 4 different gametic genotypes in unequal frequency: mostly parental combinations = AB, ab fewer recombinants = Ab, ab

25 Linkage Reduces the Frequency of Recombinants Parentals in cis Phenotypes of testcross progeny reflect the genotypes of the F 1 gametes cis combinations trans combinations

26 Linkage Reduces the Frequency of Recombinants Parentals in trans Phenotypes of testcross progeny reflect the genotypes of the F 1 gametes trans combinations cis combinations

27 Introduction to Polygenic Inheritance Some traits show intermediate inheritance in heterozygotes due to incomplete dominance This is incomplete dominance More than one incompletely dominant trait may control a trait. This in combination with environmental effects on gene expression can lead to continuous phenotypic distribution.

28 Inheritance of incompletely dominant traits follows Mendelian principles Petal color in snapdragons: Red White F 1 Pink Pink F 1 consistent with blending - F 2 1/4 Red 1/2 Pink These results are not consistent 1/4 White with blending

29 Inheritance of incompletely dominant traits follows Mendelian principles Red RR White rr F 1 Pink Rr F 2 1/4 Red RR 1/2 Pink Rr 1/4 White rr Incomplete dominance: Mendelian F 2 genotypic ratios, Heterozygote phenotype intermediate between homozygotes

30 A Common Mistake (I found this on the internet): Different letters may not be used for different alleles of the same gene Also: RR, RR, R R Archaic, not commonly used because: R and R both imply active pigment alleles, white results from a null allele

31 Pink appearance is the result of less red pigment on an unpigmented background a dilution effect It is not due to modification of red pigment to pink

32 Polygenic Inheritance Polygenes active alleles of two or more incompletely dominant genes all have the same effect on phenotype This is also called quantitative inheritance Each active allele produces one dose of phenotypic determinant (ex. = pigment) with a dihybrid cross, gene A and gene B: AABB = 4 doses, AaBb or AAbb = 2 doses Since both genes have the same effect they are designated by subscripts: R 1 or r 1, R 2 or r 2 Dihybrid cross = R 1 R 1 R 2 R 2 r 1 r 1 r 2 r 2 R 1 r 1 R 2 r 2 same as: AABB aabb AaBb

33 Polygenes in progeny of a dihybrid self 2 doses = R 1 R 2 Figure 3-15 step 1 1 dose = R 1 r 2 or r 1 R 2 0 doses = r 1 r 2

34 Polygenes in progeny of a dihybrid self Figure 3-15 step 5

35 Polygenes in progeny of a dihybrid self

36 Explanation for combinations of R 1 or R 2 alleles: R 1 or R 2 = active allele = + r 1 or r 2 = inactive allele = - All R 1, R 2 = = 1/16 of the total /16 each one r 1 or r 2 4 combinations any position /16 of the total /16 each two r 1 or r 2 6 combinations any position /16 of the total

37 Phenotypic Distribution For polygenic traits, the frequency of a phenotype depends on how many combinations of alleles will result in that phenotype 16ths with "+" Alleles Stronger phenotypic expression Number of "+" Alleles

38 64ths with + alleles For polygenic traits, the frequency of a phenotype depends on how many combinations of alleles will result in that phenotype Additive polygenic model for three genes with two alleles each = 6 max active alleles Number of + alleles

39 For polygenic traits, genotypic expression may be altered by environmental effects that smooth the phenotypic distribution into a continuum.