Sample Problems: Give genotypes of parents: 1. Round, yellow X wrinkled, yellow

Sample Problems: Give genotypes of parents: 1. Round, yellow X wrinkled, yellow

Sample Problems:
Give genotypes of parents:
1. Round, yellow
Round,
yellow
303
Round,
green
95
2. Round, yellow
Round,
yellow
94
Round,
green
96
3. Round, yellow
Round,
yellow
213
Round,
green
71
X wrinkled, yellow
wrinkled,
yellow
297
wrinkled,
green
105
X wrinkled, green
wrinkled,
green
95
wrinkled,
yellow
93
X Round, yellow
wrinkled,
yellow
70
wrinkled,
green
24
Sample Problems:
1. Round, yellow
Ww , Gg
Round,
yellow
303
(3)
Round,
green
95
(1)
2. Round, yellow
Ww , Gg
Round,
yellow
94
(1)
213
(9)
Round,
green
71
(3)
wrinkled, yellow
ww , Gg
wrinkled,
yellow
297
(3)
X
Round,
green
96
(1)
3. Round, yellow
Ww , Gg
Round,
yellow
X
wrinkled, green
ww , gg
wrinkled,
green
95
(1)
X
wrinkled,
green
105
(1)
wrinkled,
yellow
93
(1)
Round, yellow
Ww , Gg
wrinkled,
yellow
70
(3)
wrinkled,
green
24
(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
Genotypes
Gametes
Short, yellow
LLW'W'
F1 Phenotypes
Genotypes
Gametes
F2
F2
F2
F2
x
x
x
Long, white
llWW
.
.
x
x
x
.
.
.
Genotypes
Phenotypes
Phenotypic ratio
Genotypic ratio
.
.
.
.
USE THE INFORMATION ABOVE TO MAKE A TESTCROSS BELOW.
F1 Parent
Phenotypes
Genotypes
Gametes
Testcross Parent
x
x
x
.
.
.
Testcross Progeny
Phenotypes
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
Genotypes
Gametes
Short, yellow
LLW'W'
1/1 LW’
F1 Phenotypes
Genotypes
Gametes
Short, cream
x
Short, cream
LlW’W
x
LlW’W
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
F2
F2
F2
x
x
x
Long, white
llWW
1/1 lW
.
.
.
.
Genotypes
32 = 9, (1+2+1)2
Phenotypes
(21)(31) = 6, (3+1)(1+2+1)
Phenotypic ratio
6:3:3:2:1:1
Genotypic ratio
4:2:2:2:2:1:1:1:1
.
.
.
.
USE THE INFORMATION ABOVE TO MAKE A TESTCROSS BELOW.
F1 Parent
Phenotypes Short
Genotypes
Ll
Gametes
1/2 L 1/2 l
.
Testcross Parent
x
Long
x
ll
x 1/1 l
Testcross Progeny
Phenotypes
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, round
3 yellow
1 wrinkled = 9 tall, yellow, wrinkled
3 tall
3 round = 9 tall, green, round
1 green
1 wrinkled = 3 tall, green, wrinkled
3 round = 9 dwarf, yellow, round
3 yellow
1 wrinkled = 3 dwarf, yellow, wrinkled
1 dwarf
3 round = 3 dwarf, green, round
1 green
1 wrinkled = 1 dwarf, green, wrinkled
44
P
DDGGWW x ddggww
F1 DdGgWw x DdGgWw
F2
GG
DD
2 Gg
gg
GG
2 Dd 2 Gg
gg
Genotypic
frequency
Phenotypic
ratio
Phenotypes
Genotypes
Tall, Yellow
round
DDGGWW
DDGGWw
DDGgWW
DDGgWw
DdGGWW
DdGGWw
DdGgWW
DdGgWw
1
2
2
4
2
4
4
8
27
DDGGww
DDGgww
DdGGww
1
2
2
9
2
WW = 1 DDGGWW
Ww = 2 DDGGWw
ww = 1 DDGGww
2
WW = 2 DDGgWW
Ww = 4 DDGgWw
ww = 2 DDGgww
2
WW = 1 DDggWW
Ww = 2 DDggWw
ww = 1 DDggww
2
WW = 2 DdGGWW
Ww = 4 DdGGWw
ww = 2 DdGGww
Tall, yellow
wrinkled
DdGgww
2
WW = 4 DdGgWW
Ww = 8 DdGgWw
ww = 4 DdGgww
Tall, green,
round
DDggWW
DDggWw
DdggWW
DdggWw
2
WW = 2 DdggWW
Ww = 4 DdggWw
ww = 2 Ddggww
Tall, green
wrinkled
DDggww
Ddggww
45
4
1
2
2
4
9
1
2
3
GG
dd
2 Gg
gg
2
WW = 1 ddGGWW
Ww = 2 ddGGWw
ww = 1 ddGGww
2
WW = 2 ddGgWW
Ww = 4 ddGgWw
ww = 2 ddGgww
2
WW = 1 ddggWW
Ww = 2 ddggWw
ww = 1 ddggww
Genotypic
frequency
1
2
2
4
Phenotypic
ratio
9
Phenotypes
Dwarf, yellow,
round
Genotypes
ddGGWW
ddGGWw
ddGgWW
ddGgWw
Dwarf, yellow
wrinkled
ddGGww
ddGgww
1
2
3
Dwarf, green
round
ddggWW
ddggWw
1
2
3
Dwarf, green
wrinkled
ddggww
1
1
(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 similar
to 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, AND
THE FULLY RECESSIVE PARENTAL TYPE WITH DWARF VINES, GEEEN AND WRINKLED SEEDS.
F1 Parent
Tall, yellow, round
Seed parent
DdGgWw
x
DGW DGw DgW Dgw dGW dGw dgW dgw
Testcross Parent
Dwarf, green, wrinkled
Pollen parent
ddggww
dgw Gametes
DGW DGw DgW Dgw dGW dGw dgW dgw
dgw
Phenotypes
Tall, yellow, round
Tall, yellow, wrinkled
Tall, green, round
Tall, green, wrinkled
Dwarf, yellow, round
Dwarf, yellow, wrinkled
Dwarf, green, round
Dwarf, green, wrinkled
DGW DGw DgW Dgw dGW dGw dgW dgw
dgw dgw dgw dgw dgw dgw dgw dgw
Genotypic
Genotypes
frequency
DdGgWw
1
DdGgww
1
DdggWw
1
Ddggww
1
ddGgWw
1
ddGgww
1
ddggWw
1
ddggww
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
Phenotypic
ratio
1
1
1
1
1
1
1
1
P
DD x dd
D – Tall d - Dwarf
F1 Dd
Number of different heterozygous gene pairs in F1 genotype: n
Number of different F1 gametes = 2n: D d
Number of different F2 genotypes = 3n: DD, Dd, dd
Number of different F2 phenotypes = 2n: Tall, dwarf
Number of F2 progeny required to get all possible combinations in the correct proportions = 4n:
DD Dd
Dd 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 specific
F2
F2
combination of
heterozygous
kinds of
F2
progeny
alleles in a gamete
pairs
F1 gametes genotypes phenotypes
1
2
3
2
4
1/2
2
4
9
4
16
1/4
3
8
27
8
64
1/8
4
16
81
16
256
1/16
10
1024
59049
1024
1,048,576
1/1024
n
n
n
n
3
2
4
(1/2)n
n
2
_______________________________________________________________________________
F2 phenotypic ratio (1 pair): 3:1
(2 pairs): (3:1)2
(3 pairs): (3:1)3
F2 genotypic ratio (1 pair): 1:2:1
(2 pairs): (1:2:1)2
(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 PROBLEM
F1
Aa
n
No. diff.
Gametes
No. diff.
F2 Geno.
No. diff.
F2 Pheno.
No. F2
Progeny
BbCc
DdEeFf
GgHhMmNn
Example problem:
P
F1
AAbbCCddEE x aaBBccDDee
n = _____
AaBbCcDdEe
49
No. diff. F1 Gametes
No. diff. F2 Geno.
No. diff. F2 Pheno.
No. F2 Progeny
______
______
______
______
EXAMPLE PROBLEM
F1
Aa
n
1
No. diff.
Gametes 2n
=2
No. diff.
F2 Geno. 3n
=3
No. diff.
F2 Pheno. 2n = 2
No. F2
Progeny 4n = 4
BbCc
2
DdEeFf
3
GgHhMmNn
4
4
8
16
9
27
81
4
8
16
16
64
256
Example problem:
P
F1
AAbbCCddEE x aaBBccDDee
n=
AaBbCcDdEe
49
5
No. diff. F1 Gametes
32
No. diff. F2 Geno.
243
No. diff. F2 Pheno.
32
No. F2 Progeny
1,024
.
.
.
.
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 cross
AaBbccddEe x AabbccDdEe
having 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.
Aa BB cc Dd EE × AA Bb Cc dd Ee
Number of possible genotypes in progeny = 32
(2) (2) (2) (2) (2) = (2)5 = 32
Genotypic 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
.
.
.
.
.
.
Testcross Progeny Genotypic/Phenotypic ratio of :
1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1
Give genotype of unknown dominant phenotype and
TP (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
and
ratios
Phenotypic
Genotypic
Testcross results
. Number of F gamete types?
. Use 3 to predict number of F
. Number of F phenotypes?
1
n
2
2
genotypes.
Ll WW′ XX ZZ × Ll WW′ XX ′ ZZ ′
. Give Genotypic and Phenotypic results
-ratios
. Total number of genotypes and phenotypes
in progeny?
. Perform testcross using each parent separatelygive testcross genotypic/phenotypic ratios.
A = Red
aa = White
B = Long
bb = Short
Red, Long × Red, Short
A_B_ × A_ bb
3
:
1
:
Red,
White,
Long
Long
Genotypes of Parents?
Answer: AaBb × Aabb
3
Red,
Short
:
1
White,
Short
Gene Interaction
• Although genes may affect phenotypes
differently, the expression of a gene will
sometimes modify the expression of another
non-allelic gene.
GENE INTERACTION
A
B
C
D
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)
62
rrP- = pea (3)
rrpp = single (1)
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
Genotypes
RRPP
Gametes
1/1 RP
x
x
x
single
rrpp
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
F2 Phenotypes
F2 Phenotypic ratio
F2 Genotypic ratio
Working space
Working space
9:3:3:1
4:2:2:2:2:1:1:1:1
.
.
.:
.:
.
.
.
.
USE THE INFORMATION ABOVE TO MAKE A TESTCROSS BELOW
F1 Parent
Testcross Parent
Phenotypes walnut
Genotypes
RrPp
Gametes 1/4 RP 1/4 Rp 1/4 rP 1/4 rp
x
x
x
single
rrpp
1/1 rp
Testcross Progeny
Phenotypes
1/4 walnut
1/4 rose
.
Genotypes
1/4 pea
1/4 single
Testcross Phenotypic Ratio
Testcross Genotypic Ratio
1/4 RrPp
1/4 Rrpp
1:1:1:1
1:1:1:1
63
1/4 rrPp
1/4 rrpp
.
.
.
.
.
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
Genotypes
RRpp
Gametes
1/1 Rp
x
x
x
pea
rrPP
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
F2 Phenotypes
F2 Phenotypic ratio
F2 Genotypic ratio
Working space
Working space
9:3:3:1
4:2:2:2:2:1:1:1:1
.
.
.:
.:
.
.
.
.
USE THE INFORMATION ABOVE TO MAKE A TESTCROSS BELOW
F1 Parent
Testcross Parent
Phenotypes walnut
Genotypes
RrPp
Gametes 1/4 RP 1/4 Rp 1/4 rP 1/4 rp
x
x
x
single
rrpp
1/1 rp
Testcross Progeny
Phenotypes
1/4 walnut
1/4 rose
.
Genotypes
1/4 pea
1/4 single
Testcross Phenotypic Ratio
Testcross Genotypic Ratio
1/4 RrPp
1/4 Rrpp
1:1:1:1
1:1:1:1
64
1/4 rrPp
1/4 rrpp
.
.
.
.
.
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 phenotypic
ratios.
Ex.
Classical
F2 ratio
9
3
3
1
Genotype
Phenotype
A_B_
A_bb
aaB_
aabb
disc
sphere
sphere
long
65
Final F2 Phenotypic ratio
9
disc
6
sphere
1
long