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INHERITANCE OF TWO OR MORE INDEPENDENT GENES
In Drosophila melanogaster, body colour is determined by the
e gene: the recessive allele is responsible for the black colour
of the body, the dominant allele e+
is responsible for the grey body. Vestigial wings are
determined by the recessive
allele vg, normal wings are determined by the dominant allele
vg+. These two genes are independent. If dihybrid flies for
these two genes are crossed and resulting progeny is
composed by 368 individuals, how many individuals are
present in every phenotypic class?
e+ e+
ee
vg+ vg+
vg vg
Parental Genotypes: e+e vg+vg X e+e vg+vg
vg+ (1/2)
e+ vg+ (1/4)
vg (1/2)
e+ vg (1/4)
e+ (1/2)
Gametes
e (1/2)
vg+ (1/2)
e vg+ (1/4)
vg (1/2)
e vg (1/4)
Phenotypes for gene e Phenotypes for gene vg Phenotipical classes
(cross e+e X e+e)
(cross vg+vg X vg+vg)
vg+ (3/4)
e+ (3/4)
vg (1/4)
e+ vg+
(¾ X ¾= 9/16)
e+ vg
(¾ X ¼ = 3/16)
vg+ (3/4)
e vg +
(¾ X ¼ = 3/16)
vg (1/4)
e vg
(¼ X ¼=1/16)
e (1/4)
9/16 di 368 = 207 individuals e+ vg+ grey body and normal wings
3/16 di 368 = 69 individuals e+ vg grey body and vestigial wings
3/16 di 368 = 69 individuals e vg+ black body and normal wings
1/16 di 368 = 23 individuals e vg black body and vestigial wings
Genotypes? - Punnet square
e+ vg+
1/4
e+ vg+
1/4
e+ vg
1/4
e vg+
1/4
e vg
1/4
e+ vg
1/4
e vg+
1/4
e vg
1/4
Genotypes? - Punnet square
e+ vg+
1/4
e+ vg
1/4
e vg+
1/4
e vg
1/4
e+ vg+
1/4
e+ e+ vg+ vg+ e+ e+ vg+ vg
1/16
1/16
e+ e vg+ vg+
1/16
e+ e vg+ vg
1/16
e+ vg
1/4
e+ e+ vg+ vg
1/16
e+ e+ vg vg
1/16
e+ e vg+ vg
1/16
e+ e vg vg
1/16
e vg+
1/4
e+ e vg+ vg+ e+ e vg+ vg
1/16
1/16
e e vg+ vg+
1/16
e e vg+ vg
1/16
e vg
1/4
e+ e vg+ vg e+ e vg vg
1/16
1/16
e e vg+ vg
1/16
e e vg vg
1/16
Genotypes
1/16 di 368 =
2/16 di 368 =
2/16 di 368 =
1/16 di 368 =
4/16 di 368 =
2/16 di 368 =
1/16 di 368 =
2/16 di 368 =
1/16 di 368 =
23 individuals
46 individuals
46 individuals
23 individuals
92 individuals
46 individuals
23 individuals
46 individuals
23 individuals
e+ e+ vg+ vg+
e+ e+ vg+ vg
e+ e vg+ vg +
e+ e+ vg vg
e+ e vg+ vg
e+ e vg vg
e e vg+ vg+
e e vg+ vg
e e vg vg
Phenotypes
9/16 di 368 = 207 individuals e+ vg+ grey body and normal wings
3/16 di 368 = 69 individuals e+ vg grey body and vestigial wings
3/16 di 368 = 69 individuals e vg+ black body and normal wings
1/16 di 368 = 23 individuals e vg black body and vestigial wings
RATIO 9:3:3:1
In guinea pigs, the R gene determinates fur rough or
smooth, while the B gene controls fur color. Determine the
genotype of individuals used for the crosses and verify your
hypothesis with χ2 test.
n° of individuals in the progeny (observed)
Phenotype of individuals used for the
cross
Black
Rough
Black
Smooth
Brown
Rough
Brown
Smooth
a) Black Rough X Brown Smooth
50
0
0
0
b) Black Rough X Black Rough
c) Black Rough X Brown Smooth
d) Brown Rough X Black Rough
a)The parents have different phenotypes and the progeny is all Black and
Rough:
- Black and Rough are dominant characters (B, black; R, rough)
- The parents are homozygous: BB RR X bb rr
We do not perfomed statistic test because we do not have differences between
expected progeny and observed progeny.
In guinea pigs, the R gene determinates fur rough or smooth, while the B gene
controls fur color. Determine the genotype of individuals used for the crosses
and verify your hypothesis with χ2 test.
n° of individuals in the progeny (observed)
Phenotype of individuals used for the
cross
Black
Rough
Black
Smooth
Brown
Rough
Brown
Smooth
a) Black Rough X Brown Smooth
50
0
0
0
b) Black Rough X Black Rough
185
60
57
18
c) Black Rough X Brown Smooth
d) Brown Rough X Black Rough
b) We know that Black and Rough are dominant traits.
The crossed individuals have the same phenotypes but in the progeny we have
the recessive characters, thus both parents are double heterozygous: Bb Rr X
Bb Rr
In the progeny we excpect:
9/16 BR (Black rough): 3/16 Br (black smooth):
3/16 bR (brown rough): 1/16 br (brown smooth)
We want to compare experimentally observed numbers of items in several different categories
with numbers that are predicted on the basis of our hypothesis.
C2 (chi-square) test is used to help in making the decision to hold onto or reject the
hypothesis.
Phenotypes
Xo
(observed
individuals)
H
(hypothesi
s)
Black
Rough
185
9/16
Black
Smooth
60
3/16
Brown
Rough
57
3/16
Brown
Smooth
18
1/16
Total
320
16/16
Xa
Expected
individuals
(Xo –Xa)2
(Xo –Xa)2
Xa
We want to compare experimentally observed numbers of items in several different categories
with numbers that are predicted on the basis of our hypothesis.
C2 (chi-square) test is used to help in making the decision to hold onto or reject the
hypothesis.
Phenotypes
Xo
(observed
individuals)
H
(hypothesi
s)
Xa
Expected
individuals
Black
Rough
185
9/16
Black
Smooth
60
Brown
Rough
(Xo –Xa)2
(Xo –Xa)2
Xa
180
(185-180) 2=
25
25/180= 0,14
3/16
60
0
0/60= 0
57
3/16
60
(57-60) 2=
9
9/60=0,15
Brown
Smooth
18
1/16
20
(18-20) 2=
4
4/20= 0,20
Total
320
16/16
320
c2 = 0,49
Freedom Degree= 4 – 1 = 3 (number of phenotypes -1)
S =0,49
5% is the conventional decision line
Accepted
c2 = 0,49
P >90%, we accept the hypotesis
Rejected
In guinea pigs, the R gene determinates fur rough or smooth, while the B gene
controls fur color. Determine the genotype of individuals used for the crosses
and verify your hypothesis with χ2 test.
n° of individuals in the progeny
Phenotype of individuals used for the
cross
Black
Rough
Black
Smooth
Brown
Rough
Brown
Smooth
a) Black Rough X Brown Smooth
50
0
0
0
b) Black Rough X Black Rough
185
60
57
18
c) Black Rough X Brown Smooth
105
100
98
97
d) Brown Rough X Black Rough
c) The parents have different phenotypes. The second parent is homozygous
recessive for both genes (bb rr). The first parent has the dominant phenotypes
for both characters B - R -. In the progeny we observe classes of individuals with
recessive phenotypes thus the first indiviudual is double heterozygous: Bb Rr. In
the progeny we expect 4 phenotypic classes with ratio:
1:1:1:1
c2
Phenotypes
Xo
(observed
individuals)
H
(hypothes
is)
Black
Rough
105
¼
Black
Smooth
100
¼
Brown
Rough
98
¼
Brown
Smooth
97
¼
Total
400
4/4
Xa
Expected
individuals
(Xo –Xa)2
(Xo –Xa)2
Xa
c2
Phenotypes
Xo
(observed
individuals)
H
(hypothes
is)
Xa
Expected
individuals
Black
Rough
105
¼
Black
Smooth
100
Brown
Rough
(Xo –Xa)2
(Xo –Xa)2
Xa
100
(105-100) 2=
25
25/100= 0,25
¼
100
0
0/60= 0
98
¼
100
(98-100) 2=
4
4/100=0,04
Brown
Smooth
97
¼
100
(97-100) 2=
9
9/100= 0,09
Total
400
4/4
400
c2 = 0,38
Freedom Degree = 4 – 1 = 3
P =>90%, we accept the hypothesis
S =0,38
In guinea pigs, the R gene determinates fur rough or smooth, while the B gene
controls fur color. Determine the genotype of individuals used for the crosses
and verify your hypothesis with χ2 test.
n° of individuals in the progeny
Phenotype of individuals used for the
cross
Black
Rough
Black
Smooth
Brown
Rough
Brown
Smooth
a) Black Rough X Brown Smooth
50
0
0
0
b) Black Rough X Black Rough
185
60
57
18
c) Black Rough X Brown Smooth
105
100
98
97
d) Brown Rough X Black Rough
63
17
58
22
d) The first individual is homozygous recessive for color (bb) and he has the
dominant phenotype for Rough gene (R -). The second individual is dominant
for both characters (B - R -).
In the progeny we observe the classes with recessive phenotypes for brown and
smooth hair, thus the original individuals have a recessive allele to donate to
the progeny: b b R r X B b R r
With branch diagram determine expected phenotypic classes:
Gene B
bb X Bb
Gene R
Rr x Rr
phenotypic Classes
R (3/4)
B R (1/2 x 3/4 = 3/8)
black rough
r (1/4)
B r (1/2 x 1/4 = 1/8)
black smooth
R (3/4)
b R (1/2 x 3/4 = 3/8)
brown rough
r (1/4)
b r (1/2 x 1/4 = 1/8)
brown smooth
B (1/2)
b (1/2)
c2
Phenotypes
Xo
H
(observed (hypothe
individual
sis)
s)
Black
Rough
63
3/8
Black
smooth
17
1/8
Brown
Rough
58
3/8
Brown
smooth
22
1/8
Totale
160
8/8
Xa
Expected
individuals
(Xo –Xa)2
(Xo –Xa)2
Xa
c2
Phenotyp
Xo
es
(observed
individual
s)
H
(hypoth
esis)
Xa
Expected
individuals
(Xo –Xa)2
(Xo –Xa)2
Xa
Black
Rough
63
3/8
60
(63-60) 2=
9
9/60= 0,15
Black
smooth
17
1/8
20
(17-20) 2=
9
9/20= 0,45
Brown
Rough
58
3/8
60
(58-60) 2=
4
4/60=0,07
Brown
smooth
22
1/8
20
(22-20) 2=
4
4/20= 0,2
Totale
160
8/8
160
c2 = 0,87
Degrees of freedom = 4 – 1 = 3
P =80-90%, we accept the hypothesis
S =0,87
Determine the genotype of individuals used for the crosses and verify
your hypothesis with χ2 test. For each cross create a table, according to
the scheme reported below. Gene W determines Red color, gene D
determines the plant height
Numero di individui della progenie
Phenotype of individuals used for
Red
Red
White
White
the cross
tall
short
tall
short
120
0
45
0
a) red tall x red tall
b) red tall
x
white short
c) red tall
x
white short
d) white tall x
e) red tall
red tall
x
red tall
a) Color: the crossed individuals have the same phenotypes but in the progeny
we have also recessive phenotype individuals: both parents are heterozygous
(Ww) and Red (W) is dominant over white (w).
b) Height: we cannot establish wich is the dominant allele since the character
does not segregate. All the progeny is tall. Both parents could be recessive
(dd) or D D x D - o or D - x D D
c2
Phenotypes
Xo
H
(observed (hypothes
individuals)
is)
Xa
Expected
individuals
(Xo –Xa)2
(Xo –Xa)2
Xa
Red tall
120
3/4
123,75
(120-123,75) 2=
14
14/123,75=
0,11
White tall
45
1/4
41,25
(45-41,25) 2=
14
14/41,25= 0,33
Total
165
4/4
165
c2 = 0,44
Degrees of freedom= 2 – 1 = 1
P =50-70%, we can accept the hypothesis
S =0,44
Determine the genotype of individuals used for the crosses and verify
your hypothesis with χ2 test. For each cross create a table, according to
the scheme reported below. Gene W determines Red color, gene D
determines the plant height
n° of individuals in the progeny
Phenotype of individuals used for
Red
Red
White
the cross
tall
short
tall
120
0
45
a) red tall x red tall
b) red tall
x
white short
c) red tall
x
white short
d) white tall x
e) red tall
100
0
105
White
short
0
0
red tall
x
red tall
b) The crossing plants have two different phenotypes.
In the progeny we observe plants red and white: we know that red is dominant over
white thus the genotypes are W w X w w
Height: since we have only tall plants we estabish that tall is dominant over short. In the
progeny there isn’t segregation of character then the tall plant is homozygous. The
genotypes are: D D X d d
Ww DD X ww dd
We expect: ½ WD (red and tall) and ½ wD (white and tall).
c2
Phenotypes
Xo
H
(observed (hypothes
individuals)
is)
Xa
Expected
individuals
(Xo –Xa)2
(Xo –Xa)2
Xa
Red tall
100
½
102,5
(100-102,5) 2=
6,25
6,25/102,5=
0,06
White tall
105
½
102,5
(105-102,5) 2=
6,25
6,25/102,5=
0,06
Total
205
2/2
205
c2 = 0,12
Degrees of freedom = 2 – 1 = 1
P = 70-80%, we accept the hypothesis
S =0,12
Determine the genotype of individuals used for the crosses and verify
your hypothesis with χ2 test. For each cross create a table, according to
the scheme reported below. Gene W determines Red color, gene D
determines the plant height
n° of individuals in the progeny
Phenotype of individuals used for
Red
Red
White
the cross
tall
short
tall
120
0
45
a) red tall x red tall
White
short
0
b) red tall
x
white short
100
0
105
0
c) red tall
x
white short
45
43
48
44
d) white tall x
e) red tall
red tall
x
red tall
c) The crossed plants have different phenotypes. In the progeny we observed segregation
of the character: we know that red is dominant over white and we assess that the cross is
between a heterozygous and a homozygous recessive: W w X w w
Height: Tall is dominant over short thus also for height the cross is between a
heterozygous and a homozygous recessive D d X d d
WwDd X wwdd
We expect 4 phenotypic classes with ratio 1 : 1 : 1 : 1.
test c2
Phenotypes
Xo
(observed
individuals)
H
(hypothes
is)
Red tall
45
¼
Red short
43
¼
White tall
48
¼
White short
44
¼
Total
180
4/4
Xa
Expected
individuals
(Xo –Xa)2
(Xo –Xa)2
Xa
WwDd X wwdd
We expect 4 phenotypic classes with ratio 1 : 1 : 1 : 1.
test c2
Phenotypes
Xo
(observed
individuals)
H
(hypothes
is)
Xa
Expected
individuals
Red tall
45
¼
Red short
43
White tall
(Xo –Xa)2
(Xo –Xa)2
Xa
45
(45-45) 2=
0
0/45= 0
¼
45
(43-45) 2=
4
4/45= 0,09
48
¼
45
(48-45) 2=
9
9/45=0,2
White short
44
¼
45
(44-45) 2=
1
1/45= 0,02
Total
180
4/4
180
c2 = 0,31
Degrees of freedom = 4 – 1 = 3
P =>90%, we can accept the hypothesis
S =0,31
Determine the genotype of individuals used for the crosses and verify
your hypothesis with χ2 test. For each cross create a table, according to
the scheme reported below. Gene W determines Red color, gene D
determines the plant height
n° of individuals in the progeny
Phenotype of individuals used for
Red
Red
White
the cross
tall
short
tall
120
0
45
a) red tall x red tall
White
short
0
b) red tall
x
white short
100
0
105
0
c) red tall
x
white short
45
43
48
44
red tall
175
67
182
58
d) white tall x
e) red tall
x
red tall
d) In the progeny we have recessive phenotypic classes.
We can suppose that the cross is between: w w D d X W w D d
What are the expected phenotypic classes?
Gene W
ww X Ww
Gene D
Dd x Dd
Phenotypic classes
D (3/4)
W D (1/2 x 3/4 = 3/8)
Red Tall
d(1/4)
W d (1/2 x 1/4 = 1/8)
Red short
D (3/4)
w D (1/2 x 3/4 = 3/8)
White Tall
d (1/4)
w d (1/2 x 1/4 = 1/8)
White short
W (1/2)
w (1/2)
We expect 4 phenotypic classes: 3/8 : 1/8 : 3/8 : 1/8
Phenotype
s
Xo
(observed
individuals)
H
(hypothes
is)
Red tall
175
3/8
Red short
67
1/8
White tall
182
3/8
White
short
58
1/8
Total
482
8/8
Xa
Expected
individuals
(Xo –Xa)2
(Xo –Xa)2
Xa
Phenotype
s
Xo
(observed
individuals)
H
(hypothes
is)
Xa
Expected
individuals
Red tall
175
3/8
Red short
67
White tall
(Xo –Xa)2
(Xo –Xa)2
Xa
180,75
(175-180,75) 2=
33,06
33,06
/180,75= 0,18
1/8
60,25
(67-60,25) 2=
45,56
45,56
/60,25= 0,75
182
3/8
180,75
(182-180,75) 2=
1,56
1,56/180,75=0,
1
White
short
58
1/8
60,25
(58-60,25) 2=
5,06
5,06/60,25=
0,08
Total
482
8/8
482
c2 = 0,31
Degrees of freedom = 4 – 1 = 3
P =70-80%, Hypothesis accepted
S =1,11
Determine the genotype of individuals used for the crosses and verify
your hypothesis with χ2 test. For each cross create a table, according to
the scheme reported below. Gene W determines Red color, gene D
determines the plant height
n° of individuals in the progeny
Phenotype of individuals used for
Red
Red
White
the cross
tall
short
tall
120
0
45
a) red tall x red tall
White
short
0
b) red tall
x
white short
100
0
105
0
c) red tall
x
white short
45
43
48
44
red tall
175
67
182
58
265
92
93
28
d) white tall x
e) red tall
x
red tall
e)In the progeny we observe recessive phenotypes. The individuals must be double
heterozygous: W w D d X W w D d
From a dihybrid cross we expect 4 phenotipic classes:
9/16 WD (Red tall) : 3/16 Wd (Red short) : 3/16 wD (white tall) : 1/16 wd (white short).
Phenotype
s
Xo
(observed
individuals)
H
(hypothes
is)
Red tall
265
9/16
Red short
92
3/16
White tall
93
3/16
White
short
28
1/16
Total
478
16/16
Xa
Expected
individuals
(Xo –Xa)2
(Xo –Xa)2
Xa
Phenotype
s
Xo
(observed
individuals)
H
(hypothes
is)
Xa
Expected
individuals
Red tall
265
9/16
Red short
92
White tall
(Xo –Xa)2
(Xo –Xa)2
Xa
268,87
(265-268,87) 2=
14,97
14,97
/268,87= 0,05
3/16
89,62
(92-89,62) 2=
5,66
5,66
/89,62= 0,06
93
3/16
89,62
(93-89,62) 2=
11,42
11,42/89,62=0,
12
White
short
28
1/16
29,87
(28-29,87) 2=
3,49
3,49/29,87=
0,11
Total
478
16/16
478
c2 = 0,34
Degrees of freedom = 4 – 1 = 3
P >90%, hypothesis accepted
S =0,34
Crossing a pure line of melons whit white and spherical fruits with an other pure
line of melons with yellow and flat fruits the f1 progeny is melons with white and
spherical fruits.
P1
RR GG
X
P2
rr gg
Rr Gg
F1
Crossing two plants of F1 we have:
148 Plants with white and spherical fruits
52 Plants with yellow and spherical fruits
F2
49 Plants with white and flat fruits
23 Plants with yellow and flat fruits
Determine the genotypes of individuals of P1, P2 e F1, make the hypothesis of traits
segregation and verify the results with X2
From first cross we know that spherical is dominant over flat and white is dominant over
yellow. WE decide to name R the gene that control the shape melon and G the gene that
control the color melon: the first parent is homozygous dominant and the second parent
is homozygous recessive. The progeny will be heterozygous for both genes.
In F2 we have 4 phenotipic classes. This is a dihybrid cross and we expect
frequencies 9 : 3 : 3 : 1.
test c2.
Phenotype
s
Xo
(observed
individuals)
H
(hypothes
is)
White
Spherical
148
9/16
Yellow
Spherical
52
3/16
White
Flat
49
3/16
Yellow
Flat
23
1/16
Total
272
16/16
Xa
Expected
individuals
(Xo –Xa)2
(Xo –Xa)2
Xa
In F2 we have 4 phenotipic classes. This is a dihybrid cross and we expect
frequencies 9 : 3 : 3 : 1.
test c2.
Phenotype
s
Xo
(observed
individuals)
H
(hypothes
is)
Xa
Expected
individuals
White
Spherical
148
9/16
Yellow
Spherical
52
White
Flat
(Xo –Xa)2
(Xo –Xa)2
Xa
153
(148-153) 2=
25
25
/153= 0,16
3/16
51
(52-51) 2=
1
1
/51= 0,02
49
3/16
51
(49-51) 2=
4
4/51=0,08
Yellow
Flat
23
1/16
17
(23-17) 2=
36
36/17= 2,12
Total
272
16/16
272
c2 = 2,38
Degrees fo freedom = 4 – 1 = 3
P=30-50%, hypothesis accepted
S =2,38
Using the branch diagram, determine the gametes produced by
individuals with the following genotype (A, B and C genes are
independent):
a) Aa Bb CC
Gene A
Aa
Gene B
Bb
Gene C
CC
Gametes
B (1/2)
C (1)
ABC (½ X ½ X 1= 1/4)
b (1/2)
B (1/2)
C (1)
C (1)
AbC (½ X ½ X 1= 1/4)
aBC (½ X ½ X 1= 1/4)
b (1/2)
C (1)
abC (½ X ½ X 1= 1/4)
A (1/2)
a (1/2)
b) Aa bb Cc Dd
Gene A
Aa
Gene B
bb
Gene C
Cc
Gene D
Dd
Gametes
D (1/2)
AbCD (½ X 1 X ½ X ½ = 1/8)
d (1/2)
D (1/2)
AbCd (½ X 1 X ½ X ½ = 1/8)
AbcD (½ X 1 X ½ X ½ = 1/8)
d (1/2)
D (1/2)
Abcd (½ X 1 X ½ X ½ = 1/8)
abCD (½ X 1 X ½ X ½ = 1/8)
d (1/2)
D (1/2)
abCd (½ X 1 X ½ X ½ = 1/8)
abcD (½ X 1 X ½ X ½ = 1/8)
d (1/2)
abcd (½ X 1 X ½ X ½ = 1/8)
C (1/2)
A (1/2)
b (1)
c (1/2)
C (1/2)
a (1/2)
b (1)
c (1/2)
c) Aa Bb Cc
Gene A
Aa
Gene B
Bb
Gene C
Cc
Gametes
C (1/2)
ABC (½ X ½ X ½ = 1/8)
c (1/2)
C (1/2)
ABc (½ X ½ X ½ = 1/8)
AbC (½ X ½ X ½ = 1/8)
c (1/2)
C (1/2)
Abc (½ X ½ X ½ = 1/8)
aBC (½ X ½ X ½ = 1/8)
c (1/2)
C (1/2)
aBc (½ X ½ X ½ = 1/8)
abC (½ X ½ X ½ = 1/8)
c (1/2)
abc (½ X ½ X ½ = 1/8)
B (1/2)
A (1/2)
b (1/2)
B (1/2)
a (1/2)
b (1/2)
For the following cross, determine the phenotypic classes expected
in the progeny and their frequencies (A, B and C genes are
independent)
AaBbCc x aabbcc
Phenotypes for
Phenotypes for Phenotypes for Gene C
Gene A
Gene B
Cc X cc
Aa X aa
Bb X bb
C (1/2)
B (1/2)
c (1/2)
A (1/2)
C (1/2)
b (1/2)
c (1/2)
C (1/2)
B (1/2)
a (1/2)
c (1/2)
C (1/2)
b (1/2)
c (1/2)
Final Phenotypes
ABC (½ X ½ X ½ = 1/8)
ABc (½ X ½ X ½ = 1/8)
AbC (½ X ½ X ½ = 1/8)
Abc (½ X ½ X ½ = 1/8)
aBC (½ X ½ X ½ = 1/8)
aBc (½ X ½ X ½ = 1/8)
AbC (½ X ½ X ½ = 1/8)
Abc (½ X ½ X ½ = 1/8)
Which are the gametes produced by the first individual and their frequencies?