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IB/AP Biology Genetics Problems
1. In sweet peas (Lathyrus odoratus), the tall plant allele (T) is dominant to the short plant allele (t).
a. Complete a Punnett square to show the predicted ratios of the genotypes and phenotypes
expected for the following cross: heterozygous tall x heterozygous tall.
b. A cross of this type was performed to test for Mendelian inheritance. The result for the
next generation of plants was 65 tall plants and 23 short plants. Are these results
statistically indistinguishable from what you would expect?
2. In fruit flies (Drosophila melanogaster), wild type (+) is dominant to the apterous mutation (ap,
wingless).
a. Use Punnett squares to predict the F2 generation ratios of a cross between a homozygous
wild male parent (+ +) and a homozygous apterous female parent (ap ap). Use proper
symbols in this wild X mutant cross.
b. A student performed this cross and counted 824 flies with the wild phenotype and 297
flies with the apterous phenotype in the F2 generation. What is the null (H0) statistical
hypothesis? Are these results statistically indistinguishable from what you would expect?
3. In humans (Homo sapiens), the CFTR gene codes for a transmembrane protein that actively pumps
chloride ions from inside to outside of the cells that line the lungs. Water follows the ions by way of
osmosis and this keeps the mucous membranes of the lungs fluid. A mutant CFTR gene fails to
produce a functional protein which gives individuals who are homozygous for the mutant CFTR gene
a disease called cystic fibrosis. A man and a woman, both without cystic fibrosis, have a child with
the disease. What is the probability that if they have a second child, he or she will also have the
disease? Show your work.
4. A group of fish biologists were studying the genetics of eye color and body patterning in largemouth
bass (Micropterus salmoides). They crossed a largemouth bass with stripes and red eyes with another
largemouth bass that had black eyes and no stripes (plain). An F1 generation (brood) of 134 fish was
produced and all the offspring looked the same. From this F1 generation, 5 males and 5 females were
randomly selected, paired, and mated. Each breeding pair produced one brood (ranging between 45
and 187 little fish), thus creating the F2 generation. The results of the F1 cross (the F2 generation) are
summarized below in Table 1.
Table 1. Results of a dihybrid cross between five pairs of Largemouth Bass parents.
Breeding
Pair
1
2
3
4
5
Total
80
101
45
187
67
Stripes/Red Eyes
50
64
24
106
39
Phenotype
Stripes/Black Eyes
Plain/Red Eyes
15
12
21
12
10
9
31
36
15
9
Plain/Black Eyes
3
4
2
14
4
a. What are the dominant phenotypes and what are the recessive phenotypes?
b. What is the genotype for all 134 fish in the F1 Brood?
c. Construct a Punnett Square that shows the predicted outcome for the phenotypes of the F2
generation.
d. Can we accept these results as those from a true dihybrid cross? Provide statistical evidence and a
statement that argues in favor of the evidence.
5. In radishes (Raphanus sativus), two incompletely dominant genes control color and shape. Red and
white radishes are homozygous, whereas the hybrid is purple. Long and round radishes are
homozygous and if crossed will produce an oval hybrid. Complete a Punnett square to show the
genotypes and phenotypes produced by crossing pure breeding red, long radishes with purple, oval
radishes. What are the ratios of the phenotypes? Make sure you use the correct symbols.
6. A woman with type A blood has a child with type O blood. The father of the child has type B blood.
a. If the two parents have another child together, what are its possible blood types?
b. What is the probability that the parents have a child with type B blood?
c. What is the probability that the parents have a child who is homozygous for Type A
blood?
7. When two different pairs of genes control different steps in the same biochemical pathway, they will
interact with each other to produce modifications of typical Mendelian phenotypic ratios. In many
cases, however, a homozygous recessive pair of alleles is epistatic (“standing above”) to another pair
of alleles, regardless of whether they are homozygous dominant, heterozygous, or homozygous
recessive. The second pair of alleles is hypostatic (“standing below”) to the first pair. This
phenomenon is known as epistasis, and is defined as the masking of the phenotypic effects of allele
substitution at one locus by the presence of specific alleles at a second locus; perhaps even on a
different chromosome. A frequent result of epistasis is that typical Mendelian phenotypic ratios are
altered. For example, consider a portion of the pathway for anthocyanin flower pigment synthesis in
some species of cape primroses (Streptocarpus sp.). Gene pairs R,r and D,d show independent
assortment but control separate steps in a biochemical pathway producing salmon, rose and magenta
colored forms of anthocyanin pigment. See Figure 2.
In this example, epistasis occurs when rr masks the difference in phenotypic effects of D_ and dd. In
the dihybrid mating RrDd x RrDd, instead of getting a 9:3:3:1 ratio of phenotypes among the
offspring, what ratio occurs and why?
8. A human female (Homo sapiens) who is heterozygous (XCXc) for the recessive, X-linked trait causing
red-green color blindness marries a normal male (XCY). Determine what proportion of their male
progeny could have red-green color blindness and explain why.
9. The X-linked barred locus in fruit flies (Drosophila melanogaster) controls the pattern of the wings,
with the alleles (B) for barred pattern and (b) for no bars. A male with barred wings (XBY) is mated to
a barred female of unknown genotype. They produced 59 barred and 25 nonbarred offspring. What is
the gender of the nonbarred flies? What is the probability that this result occurred by chance? How do
you know if these results are in line with the predicted result of this cross?
10. Refer to the pedigree below. The shading indicates that the individual expresses full-blown
hemophilia.
a. Explain how individual III-2 inherited hemophilia.
b. What is the probability that individual II-5 is a carrier of hemophilia?
c. What is the probability that individual III-5 is a carrier of hemophilia?
11. The recessive gene p produces shrunken kernels in corn while the dominant gene P produces plump
kernels. The recessive gene y produces colorless kernels while Y produces yellow kernels. A fully
homozygous dominant plant is crossed with a shrunken, colorless plant. The F1 plants are all
phenotypically plump and yellow. A test cross is then performed with an F1 plant and a shrunken,
colorless plant. These are the offspring:
4035 plump, yellow
152 shrunken, yellow
149 plump, colorless
4032 shrunken, colorless
a) Identify the P (parental) cross.
b) Show the F1 results.
c) Show the F2 results if the genes are linked.
d) Show the F2 results if the genes are non-linked.
e) Are the traits linked or non-linked? How do you know?
f) How far apart in map units (centiMorgans) are the genes on the chromosome?
12. Suppose genes A and B are 14.5 map units apart. Another gene C is linked to A and B. Gene C is
found to cross over with gene B 7% of the time. Is this enough information to map the three genes?
Map as many possibilities as you can from this information showing the map distances between them.
13. In problem 12, it was later found that gene C crosses over with gene A 21% of the time. Make a
correct map of the genes showing the map distances between them.