POLYGENIC INHERITANCE PROBLEMS
1 A farmer notices that the length of his watermelons ranges from very long (dominant) to very short
(recessive) with many watermelons falling in-between at average lengths.
a.
The trait of watermelon length is called polygenic inheritance? What does this term mean? Write the definition.
Let’s say this trait is controlled by 4 different genes. Write the genotype (letter combinations) of a very long
watermelon.
b.
c.
Write the genotype (letter combinations) of a very short watermelon.
d.
Write the genotype (letter combinations) of an average length watermelon.
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A geneticist studying the inheritance of color in flowers crossed a pure white flower (aabb) with a
dark red flower (AABB) and got all pink offspring. Interbreeding the pink flowers produced the following
data:
43
162
245
166
41
dark red
red
pink
light pink
white
a)
Why is this particular example a case of polygenic inheritance?
b)
What are the genotypes of…

The dark red flower _______

The red flower _______

The pink flower _______

The light pink flower _______

The white flower _______
c)
Cross a red flower (AaBB) with a pink flower (AaBb). What are the expected phenotypic ratios of
offspring?
d)
Cross a light pink flower (aaBb) with a pink flower (AaBb). What are the expected phenotypic ratios
of offspring?
This is controlled by as many as 4 genes each with its own alleles.
As the number of genes increases the amount of phenotypic variation increases.
The alleles control the production of melanin which is a pigment that colours skin.
In this example the calculation is performed with 2 genes each with 2 alleles. The cross is between
two individuals heterozygous at both alleles
Allele Key


A= add melanin (with a nominal value of 1 unit)
a= no melanin added (with a nominal value of 0 units)


B= adds melanin (with a nominal value of 1 unit)
b= no melanin added (with a nominal value of 0 units)
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Human skin color is a good example of polygenic (multiple gene) inheritance. Assume that
three genes control skin color. The capital letter alleles (A, B and C) control dark pigmentation because
more melanin is produced. The lower case alleles of these three genes (a, b & c) control light pigmentation
because lower amounts of melanin are produced. A genotype with all capital genes (AABBCC) has the
maximum amount of melanin and very dark skin. Another way to think about it is to imagine that each
capital letter allele makes one unit of melanin…by that logic, a skin cell with the genotype AABBCC would
make 6 units of melanin and be dark. A genotype with all lower case allele (aabbcc) has no capital letters,
and would very light skin. Remember, each capital allele produces one unit of color, so that a wide range of
intermediate skin colors are produced, depending on the number of capital alleles in the genotype. For
example, a genotype with three capital alleles and three lower case alleles (AaBbCc) has a medium amount
of melanin and an intermediate skin color.
In the following cross (AABbCc x AaBbCC), how many kids would you expect to have 5 or more units of
melanin?
3. Skin color in humans is determined by a polygenic inheritance system, possibly involving involving
as many as 9 genes. For simplicity let’s consider the influence of 3 genes: A, B, and C, where the
dominant allele darkens skin color. Suppose a women who is AABbCc mates with a man who is
AaBbcc.
A. List all of the possible genotypes of the gametes that could be produced by each the parents?
: ______________ : _______________
B. Draw a Punnett square that shows the genotypes possible, and number each genotype from lightest
(1) to darkest skin coloration.
C. In this cross, how many dominant alleles will children with the darkest skin coloration possess, and
what theoretical fraction of the children will have this coloration?
# of alleles: ____
fraction: _____
In the next two problems please include the following:
B. include a key showing all possible genotypes and phenotypes.
E. perform a Punnett Square.
F. include the genotypic ratio of the offspring.
G. include the phenotypic ratio of the offspring.
H. perform all necessary crosses.
3. Show types of eye color a couple could expect in their children if their genotypes are
AaBb and aaBb. What are the phenotypes of the parents?
Polygenic Inheritance – Eye Color
Key:
AABB = Black
A-B- = Brown
A-bb = Hazel
aaB- = Green
aabb = blue
5. What eye colors could be expected from a couple who both have green eyes?
Choose two different green genotypes.
Polygenic Inheritance – Eye Color
AABB = Black
A-B- = Brown
A-bb = Hazel
aaB- = Green
aabb = blue
TRY THIS ONE IF YOU HAVE TIME…IF YOU GET IT RIGHT, YOU GET 1 BONUS POINT ON
THE NEXT TEST!
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Erma and Harvey are a compatible barnyard pair, but a curious sight. Harvey’s tail was only
6 cm, while Erma’s was 30 cm. Their F1 piglet offspring all grew tails that were 18 cm. When interbred the
F2 generation resulted in many piglets (Erma and Harvey’s grandpigs) whose tails ranged in 4 cm intervals
from 6 cm to 30 cm (6, 10, 14, 18, 22, 26, and 30). Most had 18 cm tails, while 1/64 had 6 cm and 1/64 had
30 cm tails. If one of the 18 cm F1 pigs were mated with the 6 cm F2 pigs, what phenotypic ratio would be
predicted in the offspring?
Study Hints:
--------------------------------------------------------------POLYGENIC PROBLEMS
When a problem says “many different genes” it is polygenic. REMEMBER:


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Upper case letters = Dominant
Lower case letters = Recessive
Look for the # of genes. For example: If height is controlled by 4 genes. . .
A person who is extremely tall would have the genotype AABBCCDD.
--------------------------------------------------------------------------------------------MENDEL PROBLEMS
FLASHBACK! When solving problems that follow Mendel’s Pattern REMEMBER: You only
need to use uppercase (dominant) or Lowercase (recessive) letters.
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