Hereford cattle may be either Horned (A) or Polled (B) (without horns). This trait is controlled by a single gene pair where the polled gene is dominant. Polled cattle are either PP or Pp while horned cattle are homozygousrecessive (pp).

A B

Sample Problems:

Give genotypes of parents:

1. Round, yellow X wrinkled, yellow

Round, yellow

Round, green

wrinkled, yellow

wrinkled, green

2. Round, yellow X wrinkled, green

Round, yellow

Round, green

wrinkled, green

wrinkled, yellow

3. Round, yellow X Round, yellow

Round, yellow

Round, green

wrinkled, yellow

wrinkled, green

303 95 297 105

94 96 95 93

213 71 70 24

Sample Problems:

1. Round, yellow X wrinkled, yellow

Round, yellow

Round, green

wrinkled, yellow

wrinkled, green

2. Round, yellow X wrinkled, green

Round, yellow

Round, green

wrinkled, green

wrinkled, yellow

3. Round, yellow X Round, yellow

Round, yellow

Round, green

wrinkled, yellow

wrinkled, green

Ww , Gg ww , Gg

303 95 297 105(3) (1) (3) (1)

Ww , Gg ww , gg

94 96 95 93(1) (1) (1) (1)

Ww , Gg Ww , Gg

213 71 70 24(9) (3) (3) (1)

Using the information given, fill in all blanks below. Circle each gamete. A dihybrid cross for two autosomal traits in the guinea pig for hair type. L = short, l = long (complete dominance); W'W' = yellow, WW' = cream, WW = white (semidominance).

P Phenotypes Short, yellow x Long, white Genotypes LLW'W' x llWW .

Gametes x .

F1 Phenotypes x .

Genotypes x .

Gametes x .

F2 Genotypes .

F2 Phenotypes .

F2 Phenotypic ratio .

F2 Genotypic ratio .

USE THE INFORMATION ABOVE TO MAKE A TESTCROSS BELOW.

F1 Parent Testcross ParentPhenotypes x .

Genotypes x .

Gametes x .

Testcross ProgenyPhenotypes Genotypes

.

Testcross Phenotypic Ratio .

Testcross Genotypic Ratio .

41

Using the information given, fill in all blanks below. Circle each gamete. A dihybrid cross for two autosomal traits in the guinea pig for hair type. L = short, l = long (complete dominance); W'W' = yellow, WW' = cream, WW = white (semidominance).

P Phenotypes Short, yellow x Long, white .

Genotypes LLW'W' x llWW .

Gametes 1/1 LW’ x 1/1 lW .

F1 Phenotypes Short, cream x Short, cream .

Genotypes LlW’W x LlW’W .

Gametes 1/4 LW’ 1/4 LW 1/4 lW’ 1/4 lW x 1/4 LW’ 1/4 LW 1/4 lW’ 1/4 lW .

F2 Genotypes 32 = 9, (1+2+1)2.

F2 Phenotypes (21)(31) = 6, (3+1)(1+2+1) .

F2 Phenotypic ratio 6:3:3:2:1:1 .

F2 Genotypic ratio 4:2:2:2:2:1:1:1:1 .

USE THE INFORMATION ABOVE TO MAKE A TESTCROSS BELOW.

F1 Parent Testcross ParentPhenotypes Short x Long .

Genotypes Ll x ll .

Gametes 1/2 L 1/2 l x 1/1 l .

Testcross ProgenyPhenotypes Genotypes

Short Ll .

Long ll .

Testcross Phenotypic Ratio 1:1 .

Testcross Genotypic Ratio 1:1 .

41

FORKED LINE METHOD FOR SOLVING CROSSES

P DDGGWW x ddggww

F1 DdGgWw x DdGgWw

F2 3 round = 27 tall, yellow, round3 yellow

1 wrinkled = 9 tall, yellow, wrinkled3 tall

3 round = 9 tall, green, round1 green

1 wrinkled = 3 tall, green, wrinkled

3 round = 9 dwarf, yellow, round3 yellow

1 wrinkled = 3 dwarf, yellow, wrinkled1 dwarf

3 round = 3 dwarf, green, round1 green

1 wrinkled = 1 dwarf, green, wrinkled

44

P DDGGWW x ddggww

Genotypic PhenotypicF1 DdGgWw x DdGgWw Phenotypes Genotypes frequency ratio

F2 WW = 1 DDGGWW Tall, Yellow DDGGWW 1 27GG 2 Ww = 2 DDGGWw round DDGGWw 2

ww = 1 DDGGww DDGgWW 2 DDGgWw 4

WW = 2 DDGgWW DdGGWW 2DD 2 Gg 2 Ww = 4 DDGgWw DdGGWw 4

ww = 2 DDGgww DdGgWW 4DdGgWw 8

WW = 1 DDggWWgg 2 Ww = 2 DDggWw Tall, yellow DDGGww 1 9

ww = 1 DDggww wrinkled DDGgww 2DdGGww 2

WW = 2 DdGGWW DdGgww 4GG 2 Ww = 4 DdGGWw

ww = 2 DdGGwwDDggWW 1 9

WW = 4 DdGgWW Tall, green, DDggWw 22 Dd 2 Gg 2 Ww = 8 DdGgWw round DdggWW 2

ww = 4 DdGgww DdggWw 4

WW = 2 DdggWW Tall, green DDggww 1 3gg 2 Ww = 4 DdggWw wrinkled Ddggww 2

ww = 2 Ddggww

45

Genotypic PhenotypicPhenotypes Genotypes frequency ratio

WW = 1 ddGGWW Dwarf, yellow, ddGGWW 1 9GG 2 Ww = 2 ddGGWw round ddGGWw 2

ww = 1 ddGGww ddGgWW 2ddGgWw 4

WW = 2 ddGgWW Dwarf, yellow ddGGww 1 3dd 2 Gg 2 Ww = 4 ddGgWw wrinkled ddGgww 2

ww = 2 ddGgwwDwarf, green ddggWW 1 3

WW = 1 ddggWW round ddggWw 2gg 2 Ww = 2 ddggWw

ww = 1 ddggww Dwarf, green ddggww 1 1wrinkled

(a) (b)

(a) Cross between two F1 garden peas of the genotype DdGgWw. The forked-line method is employed and the genotypes are illustrated. These results represent the F2 of a cross similarto those obtained from the Punnett square method involving 64 squares.

(b) Summary of F2 from trihybrid cross resulting in a 27:9:9:9:3:3:3:1 phenotypic ratio.

45 cont.

METHOD FOR SOLVING TESTCROSS TYPE PROBLEMS INVOLVING THREE GENE PAIRS. THIS CROSS IS BETWEEN AN F1 GARDEN PEA WITH TALL VINES, YELLOW AND ROUND SEEDS, ANDTHE FULLY RECESSIVE PARENTAL TYPE WITH DWARF VINES, GEEEN AND WRINKLED SEEDS.

F1 Parent Testcross ParentTall, yellow, round Dwarf, green, wrinkled

Seed parent Pollen parentDdGgWw x ddggww

DGW DGw DgW Dgw dGW dGw dgW dgw dgw Gametes

DGW DGw DgW Dgw dGW dGw dgW dgw

dgw DGW DGw DgW Dgw dGW dGw dgW dgwdgw dgw dgw dgw dgw dgw dgw dgw

Genotypic PhenotypicPhenotypes Genotypes frequency ratioTall, yellow, round DdGgWw 1 1Tall, yellow, wrinkled DdGgww 1 1Tall, green, round DdggWw 1 1Tall, green, wrinkled Ddggww 1 1Dwarf, yellow, round ddGgWw 1 1Dwarf, yellow, wrinkled ddGgww 1 1Dwarf, green, round ddggWw 1 1Dwarf, green, wrinkled ddggww 1 1

1. A 1:1 phenotypic ratio from a testcross indicates a monohybrid.2. A 1:1:1:1 phenotypic ratio from a testcross indicates a dihybrid.3. A 1:1:1:1:1:1:1:1 phenotypic ratio from a testcross indicates a trihybrid.

46

P DD x dd D – Tall d - DwarfF1 DdNumber of different heterozygous gene pairs in F1 genotype: nNumber of different F1 gametes = 2n: D dNumber of different F2 genotypes = 3n: DD, Dd, ddNumber of different F2 phenotypes = 2n: Tall, dwarfNumber of F2 progeny required to get all possible combinations in the correct proportions = 4n:

DD DdDd dd

Probability of getting one specific combination of alleles in a gamete = 1/2n

Relations among pairs of independent alleles, gametes, F2 genotypes and F2 phenotypes when dominance is present:_______________________________________________________________________________Number of Number of Number of Number of Number of One specificheterozygous kinds of F2 F2 F2 combination ofpairs F1 gametes genotypes phenotypes progeny alleles in a gamete

1 2 3 2 4 1/22 4 9 4 16 1/43 8 27 8 64 1/84 16 81 16 256 1/16

10 1024 59049 1024 1,048,576 1/1024n 2n 3n 2n 4n (1/2)n

_______________________________________________________________________________

F2 phenotypic ratio (1 pair): 3:1 F2 genotypic ratio (1 pair): 1:2:1(2 pairs): (3:1)2 (2 pairs): (1:2:1)2

(3 pairs): (3:1)3 (3 pairs): (1:2:1)3

48

List the number of ways the following exponents can be used:

2n

3n

4n

(1/2)n

EXAMPLE PROBLEMF1 Aa BbCc DdEeFf GgHhMmNnnNo. diff.GametesNo. diff.F2 Geno.No. diff.F2 Pheno.No. F2Progeny

Example problem:

P AAbbCCddEE x aaBBccDDee n = _____ No. diff. F1 Gametes ______No. diff. F2 Geno. ______

F1 AaBbCcDdEe No. diff. F2 Pheno. ______No. F2 Progeny ______

49

EXAMPLE PROBLEMF1 Aa BbCc DdEeFf GgHhMmNnn 1 2 3 4No. diff.Gametes 2n = 2 4 8 16No. diff.F2 Geno. 3n = 3 9 27 81 No. diff. F2 Pheno. 2n = 2 4 8 16No. F2Progeny 4n = 4 16 64 256

Example problem:

P AAbbCCddEE x aaBBccDDee n = 5 No. diff. F1 Gametes 32 .

No. diff. F2 Geno. 243 .

F1 AaBbCcDdEe No. diff. F2 Pheno. 32 .

No. F2 Progeny 1,024 .

49

The following cross is made:AaBbccDDEe x AaBbCcddEE

• 1) How many different phenotypes might you expect to get in the offspring?

• 2) How many different genotypes might you expect to get in the offspring?

AaBbccDDEe x AaBbCcddEE

• 1) How many different phenotypes might you expect to get in the offspring?

22 x 2 x 1 x 1 = 8• 2) How many different genotypes might

you expect to get in the offspring?32 x 2 x 1 x 2 = 36

Assuming independent assortment, what is the probability of the offspring of the crossAaBbccddEe x AabbccDdEehaving a aabbccddee genotype?

Answer: 1/4 x 1/2 x 1/1 x 1/2 x 1/4 = 1/64

Aa BB cc Dd EE × AA Bb Cc dd Ee• Number of gamete types produced by each parent?• Probability for progeny with Aa BB cc dd Ee genotype?• Probability for progeny with A_ B_ C_ D_ E_

phenotype?• Number of possible genotypes in progeny?• Number of possible phenotypes in progeny?• Genotypic ratio of progeny?• Phenotypic ratio of progeny?• Genotypic ratio of progeny?• Perform testcross on each parent separately and give

testcross results phenotypically and genotypically- give testcross genotypic and phenotypic ratios.

Number of possible genotypes in progeny = 32 (2) (2) (2) (2) (2) = (2)5 = 32Genotypic ratio of progeny = (1 + 1) (1 + 1) (1 + 1) (1 + 1) (1 + 1) = (1 + 1) 5

= (1 + 1 + 1 + 1) (1 + 1 + 1 + 1) (1 + 1)= (1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1) (1 + 1)= 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1

+ 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1

Aa BB cc Dd EE × AA Bb Cc dd Ee

Testcross Progeny Genotypic/Phenotypic ratio of : 1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1Give genotype of unknown dominant phenotype andTP (testcross parent)Number of F2 genotypes possible?Number of F2 phenotypes possible?Number of F1 gamete types?Common denominator of proportions for each genotypic and phenotypic category?Probability for one specific combination of alleles in one gamete type?

.

.

.

.

.

.

Ll WW′ × Ll WW′

F2 results ratiosPhenotypicGenotypic

Testcross results

Number of F1 gamete types?Use 3n to predict number of F2 genotypes.Number of F2 phenotypes?

..

and

.

Ll WW′ XX ZZ × Ll WW′ XX ′ ZZ ′

Give Genotypic and Phenotypic results-ratiosTotal number of genotypes and phenotypes in progeny?Perform testcross using each parent separately-give testcross genotypic/phenotypic ratios.

.

.

.

A = Red aa = WhiteB = Long bb = Short

Red, Long × Red, ShortA_B_ × A_ bb

3 : 1 : 3 : 1Red, White, Red, White,Long Long Short Short

Genotypes of Parents?Answer: AaBb × Aabb

Gene Interaction

• Although genes may affect phenotypes differently, the expression of a gene will sometimes modify the expression of another gene.

• Gene interaction pertains to relationships between alleles belonging to different gene pairs.

GENE INTERACTION

A B

Two different allelic pairs interact in affecting comb shape in the fowl. A. rose (R-pp) B. pea (rrP-) C. walnut (R-P-) D. single (rrpp)

R-P- = walnut (9) R-pp = rose (3) rrP- = pea (3) rrpp = single (1)

C D

62

Chicken Comb-type Gene Interaction example:

• 1) Phenotypic ratio in F2 remains unaltered from classical ratio.

• 2) One distinct phenotype in each phenotypic class-rather than two, although two separate genes are being considered.

• 3) The appearance of “novel” or new phenotypes in the F1 and/or F2 generations that had not appeared in the P generation.

Illustrate gene interaction for comb-type in chickens by filling in all blanks below. Circle all gametes. R-P- = walnut; R-pp = rose; rrP- = pea; rrpp = single

P Phenotypes walnut x single .

Genotypes RRPP x rrpp .

Gametes 1/1 RP x 1/1 rp .

F1 Phenotypes walnut x walnut .

Genotypes RrPp x RrPp .

Gametes 1/4 RP 1/4 Rp 1/4 rP 1/4 rp x 1/4 RP 1/4 Rp 1/4 rP 1/4 rp

F2 Genotypes Working space .

F2 Phenotypes Working space .

F2 Phenotypic ratio 9:3:3:1 .:

F2 Genotypic ratio 4:2:2:2:2:1:1:1:1 .:

USE THE INFORMATION ABOVE TO MAKE A TESTCROSS BELOW

F1 Parent Testcross Parent

Phenotypes walnut x single .

Genotypes RrPp x rrpp .

Gametes 1/4 RP 1/4 Rp 1/4 rP 1/4 rp x 1/1 rp .

Testcross Progeny

Phenotypes Genotypes

1/4 walnut. 1/4 pea 1/4 RrPp 1/4 rrPp1/4 rose 1/4 single 1/4 Rrpp 1/4 rrpp

Testcross Phenotypic Ratio 1:1:1:1 .

Testcross Genotypic Ratio 1:1:1:1 .

63

Illustrate gene interaction for comb-type in chickens by filling in all blanks below. Circle all gametes. R-P- = walnut; R-pp = rose; rrP- = pea; rrpp = single

P Phenotypes rose x pea .

Genotypes RRpp x rrPP .

Gametes 1/1 Rp x 1/1 rP .

F1 Phenotypes walnut x walnut .

Genotypes RrPp x RrPp .

Gametes 1/4 RP 1/4 Rp 1/4 rP 1/4 rp x 1/4 RP 1/4 Rp 1/4 rP 1/4 rp

F2 Genotypes Working space .

F2 Phenotypes Working space .

F2 Phenotypic ratio 9:3:3:1 .:

F2 Genotypic ratio 4:2:2:2:2:1:1:1:1 .:

USE THE INFORMATION ABOVE TO MAKE A TESTCROSS BELOW

F1 Parent Testcross Parent

Phenotypes walnut x single .

Genotypes RrPp x rrpp .

Gametes 1/4 RP 1/4 Rp 1/4 rP 1/4 rp x 1/1 rp .

Testcross Progeny

Phenotypes Genotypes

1/4 walnut. 1/4 pea 1/4 RrPp 1/4 rrPp1/4 rose 1/4 single 1/4 Rrpp 1/4 rrpp

Testcross Phenotypic Ratio 1:1:1:1 .

Testcross Genotypic Ratio 1:1:1:1 .

64

Novel phenotypes in the F2 as well as modified F2 dihybrid phenotypic ratios can be seen in summer squash (Cucurbita pepo) fruit shape. This is in contrast to chicken comb-type, which only produces novel phenotypes and not modified F2 phenotypicratios.

Ex.

Classical Genotype Phenotype Final F2 Phenotypic ratioF2 ratio9 A_B_ disc 9 disc3 A_bb sphere3 aaB_ sphere1 aabb long 1 long

6 sphere

65