Dihybrid Genetics
Dihybrid Genetics
Dihybrid Genetics
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Page 127, <strong>Dihybrid</strong> Problems #1<br />
1<br />
In rabbits, black coat color is due to a dominant<br />
gene B, brown coat color is due to a recessive<br />
gene b. Short hair is due to a dominant gene S,<br />
long hair to its recessive allele s. In a cross<br />
between 2 heterozygous black, short haired<br />
rabbits, what would be the phenotypic ratio of<br />
the offspring?
2<br />
First, write the genotypes of the parents.<br />
Then, combine each of the alleles to find<br />
the gametes formed by each parent.<br />
Cross between 2 heterozygous<br />
black, short haired rabbits<br />
BbSs b s x BbSs<br />
BS Bs bS bs
#1 Results<br />
3<br />
BS<br />
Bs<br />
bS<br />
bs<br />
BS<br />
BBSS BBSs BbSS BbSs<br />
Bs<br />
bS<br />
BBSs BBss BbSs Bbss<br />
BbSS BbSs bbSS bbSs<br />
bs<br />
BbSs Bbss bbSs bbss
4<br />
B – black; b – brown; S – short; s - long<br />
BS<br />
Bs<br />
bS<br />
BS Bs bS bs<br />
BBSS BBSs BbSS BbSs<br />
BBSs BBss BbSs Bbss<br />
BbSS BbSs bbSS bbSs<br />
Four combinations<br />
for phenotypes:<br />
Black (B_), short (S_)<br />
Black (B_), long (ss)<br />
Brown (bb), short (S_)<br />
Brown (bb), long (ss)<br />
bs<br />
BbSs<br />
Bbss<br />
bbSs<br />
bbss<br />
9 : 3 : 3 : 1
Page 127, <strong>Dihybrid</strong> Problems #2<br />
5<br />
In horses, black color is dependent upon a<br />
dominant gene B, and chestnut upon its<br />
recessive allele b. The trotting gait is due to a<br />
dominant gene T and the pacing gait to its<br />
recessive allele t. if a homozygous black pacer<br />
is mated to a homozygous chestnut trotter,<br />
what will the genotype and phenotype of the F 1<br />
generation be?
6<br />
B – black<br />
T - trotting<br />
b – chestnut<br />
t – pacing<br />
Homozygous x Homozygous<br />
black pacer chestnut trotter<br />
BBtt x bbTT<br />
Bt Bt Bt Bt<br />
bT bT bT bT
7<br />
B – black; b – chestnut; T – trotting; t - pacing<br />
Bt<br />
Bt<br />
Bt<br />
Bt<br />
bT<br />
bT<br />
bT<br />
bT<br />
BbTt BbTt BbTt BbTt<br />
BbTt BbTt BbTt BbTt<br />
BbTt BbTt BbTt BbTt<br />
BbTt BbTt BbTt BbTt<br />
All offspring<br />
BbTt<br />
—<br />
Black Trotters
Page 127, <strong>Dihybrid</strong> Problems #3<br />
8<br />
If two F 1 individuals from #2 were mated, what<br />
would be the phenotypic ratio of the offspring?<br />
BbTt x BbTt<br />
BT Bt bT bt<br />
BT Bt bT bt
9<br />
B – black; b – chestnut; T – trotting; t - pacing<br />
BT Bt bT bt<br />
BT<br />
Bt<br />
bT<br />
BBTT BBTt BbTT BbTt<br />
BBTt BBtt BbTt Bbtt<br />
BbTT BbTt bbTT bbTt<br />
9:3:3:1<br />
bt<br />
BbTt Bbtt bbTt bbtt
Page 127, <strong>Dihybrid</strong> Problems #4<br />
10<br />
An organisms has the genotype AABb. It is<br />
crossed with an individual with the genotype<br />
Aabb. What type of gametes can be produced<br />
by these two individuals?<br />
AABb x Aabb<br />
AB Ab AB Ab<br />
Ab Ab ab ab
Page 128 Codominance #1<br />
11<br />
Indian corn is multicolored due to the fact that yellow<br />
and purple are codominant. When a yellow corn (C Y C Y )<br />
is crossed with a purple corn (C P C P ) the heterozygous<br />
condition results in multicolored corn (C Y C P ). Show the<br />
Punnett Square for this cross.<br />
C Y<br />
C Y<br />
C Y C Y x C P C P<br />
yellow purple<br />
C P C Y C P C Y C P<br />
C P C Y C P C Y C P All offspring C Y C P<br />
All offspring multicolored
Page 128 Codominance #1 (cont)<br />
12<br />
What would be the result of two of the F 1 individuals?<br />
C Y C Y C P C Y C Y<br />
C P<br />
C P<br />
C P C P C Y C P<br />
multicolored<br />
C Y<br />
C Y C P<br />
x C Y C P<br />
multicolored<br />
Genotypic Ratio<br />
1 C P C P : 2 C Y C P : 1 C Y C Y<br />
Phenotypic Ratio<br />
1 purple : 2 multicolored : 1 yellow
Page 128 Codominance #2<br />
13<br />
A white chicken was mated with a black chicken. The<br />
resulting offspring was a black and white chicken.<br />
A. What is the name of this genetic pattern of inheritance?<br />
Codominance<br />
B. If one chicken’s genotype was F W F W and the other<br />
chicken’s genotype was F B F B , what was the black and<br />
white offspring’s genotype?<br />
F W F B heterozygous
Page 128 Codominance #2 (cont)<br />
14<br />
C. Complete a Punnett Square and show the genotypic and<br />
phenotypic ratios for a cross between a black and white<br />
chicken and a black chicken.<br />
F B F W<br />
Black and white<br />
x F B F B<br />
black<br />
F B<br />
F W<br />
Genotypic Ratio<br />
F B F B F B F B F W<br />
2 F B F B : 2 F B F W<br />
Phenotypic Ratio<br />
2 black : 2 black and white
Page 128 Incomplete Dominance #1<br />
15<br />
In Andalusian fowl, the gene for black plumage, B,<br />
is incompletely dominant to<br />
the gene for white plumage, W.<br />
The heterozygous condition results in blue plumage.<br />
List the genotypic and phenotypic ratios expected<br />
from the following crosses.<br />
BB – black plumage<br />
WW – white plumage<br />
BW – blue plumage
16<br />
A. black x blue<br />
BB x BW<br />
black blue<br />
B<br />
B<br />
Genotypic Ratio<br />
B<br />
BB<br />
BB<br />
2 BB : 2 BW<br />
W<br />
BW<br />
BW<br />
Phenotypic Ratio<br />
2 black : 2 blue
17<br />
B. blue x blue<br />
BW x BW<br />
blue blue<br />
B<br />
W<br />
Genotypic Ratio<br />
B<br />
BB<br />
BW<br />
1 BB : 2 BW : 1 white<br />
W<br />
BW<br />
WW<br />
Phenotypic Ratio<br />
1 black : 2 blue : 1 white
18<br />
C. blue x white<br />
BW x WW<br />
blue white<br />
B<br />
W<br />
Genotypic Ratio<br />
W<br />
BW<br />
WW<br />
2 BW : 2 WW<br />
W<br />
BW<br />
WW<br />
Phenotypic Ratio<br />
2 blue : 2 white
Page 128 Incomplete Dominance #2<br />
19<br />
In short-horned beef cattle, coat color shows<br />
incomplete dominance. A reddish-brown or roan<br />
calf is the offspring of homozygous white (W) and<br />
homozygous red (R) parents. Give the genotypic<br />
and phenotypic ratios of the following crosses.
20<br />
Page 128 Incomplete Dominance #2 (cont)<br />
A. A cross between a roan and white animal<br />
RW x WW<br />
roan white<br />
W<br />
R<br />
RW<br />
W<br />
WW<br />
Genotypic Ratio<br />
2 RW : 2 WW<br />
W<br />
RW<br />
WW<br />
Phenotypic Ratio<br />
2 roan : 2 white
21<br />
Page 128 Incomplete Dominance #2 (cont)<br />
B. A cross between a roan and red animal<br />
RW x RR<br />
roan red<br />
R<br />
R<br />
RR<br />
W<br />
RW<br />
Genotypic Ratio<br />
2 RW : 2 RR<br />
R<br />
RR<br />
RW<br />
Phenotypic Ratio<br />
2 roan : 2 red
Page 128 Multiple Alleles #2<br />
22<br />
One parent has type A blood and the other parent has type<br />
B blood. What are their genotypes if they produce a<br />
large number of children whose blood types are:<br />
A. all AB<br />
I A I A<br />
I B I A I B I A I B<br />
I A I A x I B I B<br />
B. ½ AB and ½ B<br />
I B I A I B I A I B<br />
I A i<br />
I B I A I B I B i<br />
I A i x I B I B<br />
I B I A I B I B i
Page 128 Multiple Alleles #2 (cont)<br />
23<br />
One parent has type A blood and the other parent has type<br />
B blood. What are their genotypes if they produce a<br />
large number of children whose blood types are:<br />
I B i<br />
I A I A I B I A i<br />
I B i x I A I A<br />
C. ½ AB and ½ A<br />
I A I A I B I A i<br />
I B<br />
i<br />
D. Type O<br />
I B i x I A i<br />
I A I A I B I A i<br />
i I B i ii
24<br />
Page 128 Sex-Linked Traits #1<br />
In fruit flies, eye color is a sex-linked trait. Red eye<br />
color is determined by the dominant gene R, white eye<br />
color is determined by the recessive allele r.<br />
Diagram a cross between a<br />
carrier female and a white eye male.<br />
Remember: This sex-linked trait is found on<br />
the X chromosome, not the Y.<br />
Females have XX, but males have XY<br />
X R X r x X r Y<br />
red eyed white eyed<br />
female male
Page 128 Sex-Linked Traits #1 (cont)<br />
25<br />
X r<br />
X R X r X r X r<br />
X R X r x X r Y<br />
red eyed<br />
female<br />
X R X r<br />
white eyed<br />
male<br />
Female Phenotypic Ratio<br />
1 red eyed : 1 white eyed<br />
Y<br />
X R Y<br />
X r Y<br />
Male Phenotypic Ratio<br />
1 red eyed : 1 white eyed
Page 128 Sex-Linked Traits #2<br />
26<br />
In humans, the gene for normal<br />
blood clotting, H, is dominant to the gene for<br />
hemophilia, h. The trait is sex-linked. This means<br />
it is located on the X chromosome.<br />
Complete the following crosses.<br />
Give the genotypes of the parents.<br />
Give the phenotypic ratio of the offspring.<br />
Be sure to include the words male and female.
Page 128 Sex-Linked Traits #2 (cont)<br />
27<br />
A. A heterozygous female and a hemophiliac male<br />
X h<br />
X H X h X h X h<br />
X H X h x X h Y<br />
heterozygous<br />
female<br />
X H X h<br />
hemophiliac<br />
male<br />
Female Phenotypic Ratio<br />
1 normal : 1 hemophilac<br />
Y<br />
X H Y<br />
X h Y<br />
Male Phenotypic Ratio<br />
1 normal : 1 hemophilac
Page 128 Sex-Linked Traits #2 (cont)<br />
28<br />
B. A heterozygous female and a normal male<br />
X H<br />
X H X H X H X h<br />
X H X h x X H Y<br />
heterozygous<br />
female<br />
X H X h<br />
normal<br />
male<br />
Female Phenotypic Ratio<br />
All normal<br />
Y<br />
X H Y<br />
X h Y<br />
Male Phenotypic Ratio<br />
1 normal : 1 hemophilac
Page 128 Sex-Linked Traits #2 (cont)<br />
29<br />
C. A homozygous dominant female and a normal male<br />
X H<br />
X H X H X H X H<br />
X H X H x X H Y<br />
homozygous<br />
dominant female<br />
X H X H<br />
normal<br />
male<br />
Female Phenotypic Ratio<br />
All normal<br />
Y<br />
X H Y<br />
X H Y<br />
Male Phenotypic Ratio<br />
All normal
Page 128 Sex-Linked Traits #2 (cont)<br />
30<br />
D. A homozygous dominant female and a hemophiliac male<br />
X h<br />
X H X h X H X h<br />
X H X H x X h Y<br />
homozygous<br />
dominant female<br />
X H X H<br />
hemophiliac<br />
male<br />
Female Phenotypic Ratio<br />
All normal<br />
Y<br />
X H Y<br />
X H Y<br />
Male Phenotypic Ratio<br />
All normal
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