Genetics Review
Name:_______KEY___________
1. If both parents of a P generation are said to be heterozygous for height, with T representing
tall and t representing short, what will the F1 generation genotypic and phenotypic ratios be?
P cross ____Tt______ x ____Tt______
T
F1 genotypic ratio:
1:2:1 (TT:Tt:tt)
t
TT
Tt
Tt
tt
T
F1 phenotypic ratio:
3:1 (tall:short)
t
2. In a family of six, the children had the following genotypic and phenotypic ratios:
Genotypic ratio = 1:2:1
Phenotypic ratio = 1:2:1 (Blue skin: green skin: purple skin)
What are the genotypes and phenotypes of the parents?
Genotypes: ____Bb___ x ___Bb____
Phenotypes: _____green_____ x ___green___
B
BB
Bb
Bb
bb
B
What kind of dominance is this representing?
Incomplete dominance
b
b
3. Supposing curly hair, L, is dominant to straight hair, l, and green skin, G, is dominant to
orange skin, g, what would the genotypic and phenotypic ratios be for the following:
A parental generation of:
a heterozygous (both traits) male X a female who is homozygous dominant for the
hair trait and homozygous recessive for skin color LlGg x LLgg
What will the outcome of their F1 generation be?
LG
Lg
lG
lg
Lg
Genotypic ratio =
Lg
1:1:1:1 (LLGg:LLgg:LlGg:Llgg)
LLGg
LLgg
LlGg
Llgg
LLGg
LLgg
LlGg
Llgg
LLGg
LLgg
LlGg
Llgg
LLGg
LLgg
LlGg
Llgg
Lg
Phenotypic ratio =
1:1
Lg
(straight hair, green skin:straight hair,
orange skin)
4. Use this paragraph to answer the questions that follow.
Betsy the bumble bee was traveling all around Crystal Lake in search of her favorite
flower, a petunia. Although it took her hours to find this particular flower, as soon as she did, she
immediately landed on it to get a whif of its aroma. Whenever she was able to take in this particular
scent, she knew it was going to be a great start to the day. After basking in the petunia she had
found, she took off to meet up with Bud, her bee friend, at a flower he liked to call his own. As she
was flying to meet Bud, she noticed a strange yellow substance on her legs. She didn’t think much
of it, but as soon she landed on Bud’s flower, the yellow substance fell into a strange structure on
Bud’s flower. Betsy became worried, but Bud quickly reassured her, informing her that she had
simply helped continue her petunia’s legacy. Betsy wasn’t too sure what this meant but was too
enamored by Bud to care.
a. What was the strange, yellow substance on Betsy’s legs? ____pollen____________
What flower structure did this substance come from? _____anther__________
b. What flower structure did the substance fall into on Bud’s flower? ____stigma/pistil_____
Would you consider this structure male or female? ______female__________
c. This type of pollination is called ____cross____ pollination.
5. In squirrels, albinism is recessive. Crossing a pure strain normally pigmented squirrel (NN)
with an albino squirrel (nn) would yield offspring in the F1 generation that are
______100______% normally pigmented (Nn) and ____0_______% that are albino.
6. In incomplete dominance, what would the phenotypic ratio be if two heterozygotes are
crossed? Nn x Nn = 1NN : 2Nn : 1nn
1:2:1
7. If an organism has a genotype of Dd, which alleles do its gametes have?
A. All have D
B. all have d
C. three have D for each one that has d
D. 1/2 have D and 1/2 have d
8. State the probabilities of each of the following random events.
A. one die being rolled once and coming up a five __1/6_____
B. A pyramid, containing sides A-D, being tossed in the air and landing on side “A” five times
in a row. Show work.
________1/1024___________
¼x¼x¼x¼x¼
C. Having the following events all occur together. Drawing a spade from a deck of cards,
flipping a coin and landing on tails, rolling one die and getting an even number, and getting
three boys in a row in one family. Show work.
¼x½x½x½x½x½
__________1/128_________
9. Calculate the per cent deviation for the following random situation. Two coins are tossed in the
air randomly. The following data are obtained. Show all work and have answer to nearest .1%
HEAD/HEAD HEAD/TAIL
70
149
TAIL/TAIL
% DEV= ___4%______
81
l 75-70 l + l 150-149 l + l 75-81 l x 100%
300
= 4%
10. Using the information given in problem #2 above, calculate the Chi2 and “p” value for the same
situation. Use the Chi2 table provided on the back. Show all calculations, include all steps.
Chi2 ____.82_____
p value ____50% - 95%______
( 75-70 )2 + ( 150-149 )2 + ( 75-81 )2
75
= .33 + .007 + .48
= .82
150
75