Genetics Practice Problems for AP Exam
1.
Fill in the following table, assuming complete dominance in each situation.
CROSS
PHENOTYPIC RATIO
GENOTYPIC RATIO
Tt x tt
1:1
1:1
50% Tt 50% tt
TT x Tt
100% dominant
1:1
50% TT 50% Tt
Tt x tt
Same as first
Same as first
Tt x Tt
3:1
1:2:1 25% TT, 50% Tt, 25% tt
tt x Tt
Same as first
Same as first
TT x tt
100% dom
100% Tt
TT x TT
100% dom
100% TT
Tt x TT
100% dom
1:1
50% dom 50% rec
75% dom 25% rec
50% TT, 50% Tt
2.
In horses, black coat color is influenced by the dominant allele (B), while chestnut coat color is influenced by the recessive
allele (b). Trotting gait is due to a dominant genet (T), pacing gait to the recessive allele (t). A homozygous black trotter is
crossed to a chestnut pacer.
What will be the appearance and gait of the F1 and F2 generations?
F1 generations will be all black and all trotting; F2 generation will be 9/16 black & trotting; 3/16 black & pacing; 3/16
chestnut & trotting; and 1/16 chestnut & pacing
Which phenotype will be most common?
Black and trotting
Which genotype will be most common?
BbTt
Which of the potential offspring will be certain to breed true?
Homozygous dominant for black and/or trotting (BB or TT)
3.
In four o-clock plants, red flower color is determined by gene R and white flower color by R’, while the heterozygous
condition is pink. Complete the table below by giving the genotypic & phenotypic ratios.
CROSS
RR x R’R’
PHENOTYPIC RATIO
100% pink
GENOTYPIC RATIO
100% R R’
R’R’ x R’R’
100% white
100% R’ R’
RR x RR
100% red
100% RR
4.
Sickle cell anemia is a genetic disease in which children who are homozygous for a defective gene produce defective
hemoglobin (HbS/HbS). The genotypes of normal persons are Hb A/ HbA. If the level of blood oxygen drops below a certain
level in a person with the Hb S/HbS genotype, the hemoglobin chains stiffen and cause the red blood cells to form sickle, or
crescent, shapes. These cells clog and rupture capillaries, resulting in oxygen-deficient tissues where metabolic wastes collect.
Several body functions are body damaged. Severe anemia and other symptoms develop, and death nearly always occurs
before adulthood. The sickle-cell gene is considered PLEIOTROPIC. Persons who are heterozygous (HbS/HbA) are said to
possess sickle-cell trait. They are able to produce enough normal hemoglobin molecules to appear normal, but their red blood
cells will sickle if they encounter oxygen tension (such as at high altitudes).
A. A man whose sister died of sickle-cell anemia married a woman whose blood is found to be normal. What
advice would you give this couple about the inheritance of this disease as they plan their family?
If the father’s sister had sickle cell anemia there is a 50% chance (assuming their parents were carriers) that he is
a carrier. If the woman has normal blood, she is homozygous dominant. If they have a child, it will not have
sickle cell anemia, however, if the father is a carrier there is a 50% he will pass the allele to the child and it will
be a carrier. In other words there is a 25% chance they will have a carrier child (1/2 chance he is a carrier X ½
chance the child will inherit the recessive allele = ¼ or 25%)
B. If a man and a woman, each with sickle-cell trait, were planning to marry, what information could you provide
them regarding the genotypes and phenotypes of their future children?
If they are both carriers, then they have a 25% chance of having a child with sickle cell disease. Additionally,
they have a 25% chance of having a child that is a carrier. They only have a 25% chance of having a normal
child who is not a carrier.
In one example of multiple alleles with codominance, the three genes I A , IB, and I produce proteins found on the surface of
red blood cells that determine the four blood types in the ABO system. Give the genotypic and phenotypic ratios of the
following crosses:
CROSS
PHENOTYPIC RATIO
GENOTYPIC RATIO
IAi x IAIB
50% type A; 25% type B; 25% type AB
25% IAIA; 25% IAi; 25% IBi; 25% IAIB
5.
IBi x IAi
25% type A; 25% type B; 25% type AB;
25% IAi; 25% IBi; 25% IAIB; 25% ii
25% type O
IAi x IBi
Same as #2
Same as #2
IAIB x ii
50% type A; 50% type B
50% IAi; 50% IBi
6.
In epistasis, two alleles of a gene mask the expression of alleles of another gene, so that some expected phenotypes never
appear. In sweet peas, genes C and P are necessary for colored flowers. In the absence of either of these dominant alleles
(__pp or cc__) or both of them (ccpp), the flowers are white. What will be the color of the offspring of the following crosses
and in what proportions will they appear?
A. CcPp x ccpp
25% colored; 75% white
B. CcPP x Ccpp
75% colored; 25% white
C. Ccpp X ccPp
50% colored; 50% white
7.
In poultry, an epistatic interaction occurs in which two genes produce a phenotype that neither gene can produce alone. The
two interacting genes (R and P) produce comb shape in chickens. The possible genotypes and phenotypes are as follows:
R_P_
R_pp
rrP_
rrpp
walnut comb
rose comb
pea comb
single comb
What are the genotypic and phenotypic ratios of the offspring of a heterozygous walnut-combed male and a single-combed
female?
25% RrPp; 25% Rrpp; 25% rrPp; 25% rrpp
25% walnut comb; 25% rose comb; 25% pea comb; 25% single comb
Cross a homozygous rose-combed rooster with a homozygous single-combed hen, and list the possible genotypes and
phenotypes for their offspring. 100% Rrpp; 100% rose comb
8.
A color-blind man and a woman with normal vision whose father was color blind have a son. Color blindness, in this case, is
caused by an X-linked recessive gene. If only the male offspring are considered, what is the probability that their son is color
blind?
Color Blind Man—XeY;
Normal Woman—XEX?
Man’s Father-- XeY
If the woman is a carrier there is a 50% chance their son will be color blind. If she is normal, there is a 0%
chance their son will be color blind. X-linked traits are passed from mother to son.
9.
Hemophilia A is caused by an X-linked recessive gene. A woman who is seemingly normal but whose father was a
hemophiliac marries a normal man. What proportion of their sons will have hemophilia? What proportion of their daughters
will have hemophilia? What proportion of their daughters will be carriers?
Woman—MUST BE A CARRIER IF HER FATHER HAD HEMOPHILIA—XNXn
Normal Man--XNY
Each of their sons has a 50% chance of being hemophiliac (50% will inherit the normal X from mom; 50% will inherit the
mutated X from mom)
50% of their daughters will be carriers (50% will inherit the normal X from mom; 50% will inherit the mutated X from mom)
10. If genes A and B are twice as far apart on a chromosome as genes C and D, how often would you expect that crossing over
occurs between genes A and B as compared to between genes C and D?
REMEMBER: The further apart 2 genes are the more likely they are to cross over. This is how linkage maps are constructed.
Therefore, you would expect genes A and B to cross over twice as often as genes C and D.
11. In a heterozygous, heterozygous dihybrid cross, the following data was obtained:
# of offspring dominant for both traits
570
# of offspring dominant for trait 1 and recessive for trait 2
185
# of offspring dominant for trait 2 and recessive for trait 1
190
# of offspring recessive for both traits
55
Perform a chi-square analysis to see if the data above agrees with the predicted outcome of this cross.
NOTE: To get the Expected Number: Total up the offspring (570 + 185 + 190 + 55 = 1000), and then multiply each group by its
expected outcome. In a dihybrid heterozygous, heterozygous cross, you would expect 9:3:3:1 ratio (9/16 dominant, dominant,
etc.)
Group
Dominant for both
Dom for trait 1;
rec for trait 2
Dom for trait 2;
Rec for trait 1
Recessive for both
Observed
570
185
Expected
563 (9/16 X 1000)
188 (3/16 X 1000)
O-E
7
-3
(O-E)2
49
9
(O-E)2/E
.007
.048
190
188 (3/16 X 1000)
2
4
.021
55
63 (1/16 X 1000)
-8
64
Chi Square Value
1.016
1.092
Degrees of Freedom = Number of Groups – 1 = 4 groups – 1 = 3
Look up critical value in a critical values chart; always use probability = 0.05
Probability (p)
0.05
0.01
0.001
1
3.84
6.64
10.8
2
5.99
9.21
13.8
Degrees of Freedom
3
7.82
11.3
16.3
4
9.49
13.2
18.5
5
11.1
15.2
20.5
Since there are 3 degrees of freedom and we want 0.05 probability, the critical value for our Chi-square analysis is 7.82.
Once you determine the critical value, compare it to your computed Chi-square value (which is 1.092). If the Chi Square value is
equal to or less than the critical value the null hypothesis is accepted (in other words, chance did not produce the results, the results
actually followed the expected 9:3:3:1 ratio. If the Chi Square value is more than the critical value, the null hypothesis is rejected
and alternative explanations have to be developed to explain the data—the hypothesis does not explain it.
In this experiment, the null hypothesis is supported because 1.092 is less than 7.82.