Lab Mendelian Genetics-Exploring Genetic Probability
-Revisiting Mendel’s Observations
Purpose:
Students will
1. Learn that probability is strongly related to genetic outcomes.
2. Determine whether probability supports or does not support the 19th century data and conclusions of
Gregor Mendel.
3. Be introduced to the probability of crossing two traits at a time, known as dihybrid crosses.
Vocabulary:
Gene
Allele
Trait
Gamete
Sperm
Egg
Heterozygous
Homozygous
Probability
Law of Segregation
Law of Independent Assortment
Monohybrid
Dihybrid
Zygote
Fertilization
Punnett Square
Background:
When a coin is tossed, it can only land one of two ways -either heads up or tails up. This is similar to when an
organism makes gametes. Since an organism receives 2 alleles for a gene , one from the mother of the
gamete maker and one from the father of
the gamete maker, only one of these
alleles will be transferred to a single
gamete (sperm or egg)-not both.
If the gamete-making organism is a
heterozygous (ex. Aa), the resulting
gametes will either have the A
(dominant) or the a (recessive) allele.
Mathematically, this situation is
described as the probability of getting
one of the alleles (or side of the coin)
over the other is 50%, 0.5, ½ or a 1:2
ratio. Mendel summarized this finding in
his “Law of Segregation.”
Part 1: Monohybrid Crosses
In this Laboratory Activity, each side of a coin will represent an allele for a trait. Two coins will be tossed, one
representing the allele in the sperm and one representing the allele in the egg. The results of flipping both
coins will represent the resulting genotype of the zygote created by the union of the egg and sperm
(fertilization). Remember that the zygote is the first somatic cell of a new organism. From this genotype, the
phenotype can be determined.
The results of flipping these two coins will be used to mimic the genotypic and phenotypic outcomes recorded
by Mendel for a single train in a hybrid cross, specifically the mating of two heterozygous parents.
In pea-plant flowers, the color purple (P) is dominant over white (p). A heads-heads combination will
represent a homozygous dominant outcome (purple flowers). A heads-tails combination will stand for a
heterozygous outcome (purple flowers). A tails-tails combination will represent a homozygous-recessive
outcome (white flower).
Materials:
2 Pennies
2 Nickels
Prelab Questions: You may not begin the lab activity until these questions are completed in your lab
notebook and initialed by your teacher.
1. Use your notes to explain the Law of Segregation in your own words.
2. Draw and correctly label a Punnett Square for the following situation:
 One parent plant is homozygous for the dominant seed shape (round seeds-R).
 The other parent is heterozygous for the same trait.
 Give the phenotypes for all four possible outcomes.
3. It is possible and often happens that two parents with brown eyes produce a child with blue eyes.
Draw a Punnett Square illustrating this outcome. Compute the odds that it might happen.
Investigative Questions:
What is the effect of crossing 2 purple-flowered heterozygous parents on the resulting genotypes of the
offspring?
Hypothesis:
Write a hypothesis for the investigative question and use a Punnett Square and the genotypic ratios for this
cross as the “because” part of the hypothesis. Mendel observed this outcome many times during his testing.
The results were repeated with great precision.
Procedure:
1. Choose one person to be the Director and one person to be the data collector.
2. Copy the following data table into your lab notebook.
3. Label the top box in the left-most column of your data table as the phenotype that results from a PP
genotype
4. Label the second box in the left-most column of your data table as the phenotype that results from a
Pp genotype.
5. Label the third box in the left-most column of your data table as the phenotype that results from a pp
genotype.
6. Record the “expected outcome” in the appropriate column of the table. (Warning: remember that
there are two different ways for two pennies to land heads-tails (2 x 0.25 = 0.5)
7. Using 2 pennies, complete 100 simultaneous flips.
8. Record your results as appropriate hash-marks in the appropriate boxes under “Tally.”
9. When you finish, divide the results of each genotype (under “Tally”) by the total of tosses (100) to
obtain the “Experimental Outcome.”
Data Table A-Monohybrid Cross
Phenotype
Coin Combination
PENNY
PENNY
HEADS
HEADS
GENOTYPE(S)
PENNY
PENNY TAILS
HEADS
GENOTYPE(S)
PENNY TAILS
PENNY TAILS
GENOTYPE(S)
Tally
Expected
Probability
Experimental
Outcome
Part 2: Dihybrid Crosses
We will use four coins (two pennies and two nickels) to
test the genotype and phenotype ratio which would result
from a mating between parents who are heterozygous for
each of two different traits. We call this a dihybrid cross.
We will repeat the experiment in Part 1, this time
completing 200 coin flips. Because we are looking at 2
traits, we will need different coins for each parent. One
type of coin (penny) will represent one of the 2 traits
contributed by the parent. The other type of coin (nickel)
will represent the second trait contributed by that parent.
There are 2 sides to each coin and 2 possible outcomes for
each trait since each heterozygous parent carries 2
different alleles for each trait.
Do the results on the penny affect the results on the
nickel? No, they flip independently from each other. In
the same way, the genes for the two traits assort
independently from each other into gametes. Mendel called this the “Law of Independent Assortment. “
Any 1 of 4 possible combinations may therefore turn up in any single gamete (sperm cell or egg cell). There
are, therefore, 16 possible outcomes when the egg and sperm eventually meet.
Prelab Questions: You may not begin Part 2 of this lab activity until these questions are completed in your lab
notebook and initialed by your teacher.
4. Use your notes to explain the Law of Independent Assortment in your own words.
5. Following is a partially completed Punnett Square for a dihybrid cross. In it, both parents are
heterozygous for brown eyes and carry the recessive trait for spinal muscular atrophy (SMA-a rare
genetic disorder).
 Write the four possible gamete combinations. Note: Always keep the alleles for the same trait
together and put the dominant (capital letter) allele and the recessive (lower case) allele
second.
 Complete the Punnett Square.
 Compute the odds of having a child with blue eyes.
 Compute the odds of having a child with blue eyes and SMA.
 Compute the odds of having a child with brown eyes and SMA.
B = brown eyes;
N = normal Central Nervous System development;
BN
b = blue eyes
n = Spinal Muscular Atrophy
Bn
FATHER
bN
bn
MOTHER
BN
Bn
bN
bn
In Pea Plants,
 Rough seed shape (R) is dominant over smooth seed shape (r).
 Yellow seeds (Y) are dominant over white seeds (y).
Investigative Questions:
What is the effect of crossing 2 rough seed shape, yellow seeded heterozygous parents on the resulting
genotypes of the offspring?
Hypothesis:
Write a hypothesis for the investigative question and use a Punnett Square and the genotypic ratios for this
cross as the “because” part of the hypothesis. Mendel observed this outcome many times during his testing.
The results were repeated with great precision.
Procedure:
6. Choose one person to be the Director and one person to be the data collector.
7. Determine the 4 possible gametes produced by either parent.
8. Create a Punnett square to show the possible offspring from such a cross.
9. Determine the possible seed shape and color of all offspring whose parents are each heterozygous for
the two traits.
10. Copy the following data table into your lab notebook.
11. Label the heads side of both pennies “R” and the tails “r.”
12. Label the heads side of both nickels “Y” and the tails “y.”
13. Label the second column of the table (below coin combination) with the correct genotypes.
14. Label each box in the first column of the data table “Phenotype” with the appropriate phenotype.
15. From the outcomes in your Punnett Square, enter the “expected probability” for each genotype. This
is done by counting up the boxes in the Punnett Square that match each genotype. Remember that
some of the alleles in your data table are produced in more than one way (ex. Heads-tails is the same
as tails-heads). Then take the number of Punnett square boxes that match one genotype and divide by
16 to get the probability of that outcome. (Number of Squares for a Genotype/16 = Probability)
16. Toss both coins 200 times.
17. Record your results as appropriate hash-marks in the appropriate boxes under “Tally.”
18. When you finish, divide the results of each genotype (under “Tally”) by the total of tosses (200) to
obtain the “Experimental Outcome.”
Data Table B-Dihybrid Cross
Phenotype
Coin Combination
Rough,
yellow
seeds
PENNY PENNY
HEADS HEADS
GENOTYPE(S)
RR only
PENNY PENNY
HEADS HEADS
GENOTYPE(S)
RR only
PENNY PENNY
HEADS TAILS
GENOTYPE(S)
NICKEL NICKEL
HEADS HEADS
GENOTYPE(S)
YY only
NICKEL NICKEL
HEADS TAILS
GENOTYPE(S)
Yy or yY
NICKEL NICKEL
HEADS TAILS
GENOTYPE(S)
PENNY PENNY
HEADS TAILS
GENOTYPE(S)
NICKEL NICKEL
HEADS HEADS
GENOTYPE(S)
PENNY PENNY
HEADS HEADS
GENOTYPE(S)
NICKEL NICKEL
TAILS
TAILS
GENOTYPE(S)
PENNY PENNY
HEADS TAILS
GENOTYPE(S)
NICKEL NICKEL
TAILS
TAILS
GENOTYPE(S)
PENNY PENNY
TAILS
TAILS
GENOTYPE(S)
NICKEL NICKEL
HEADS HEADS
GENOTYPE(S)
PENNY PENNY
TAILS
TAILS
GENOTYPE(S)
NICKEL NICKEL
HEADS TAILS
GENOTYPE(S)
PENNY PENNY
TAILS
TAILS
GENOTYPE(S)
NICKEL NICKEL
TAILS
TAILS
GENOTYPE(S)
Tally
Expected
Probability
Experimental
Probability
(outcome)
1 ÷ 16 =
0.0625
(tally) ÷ 200 =
Analysis Questions:
1. How closely did your data resemble the expected probability for the monohybrid cross? Does your data support the
data and conclusions published by Mendel?
2. Does your data more closely match Mendel’s for the dihybrid cross than they did in Part A (a monohybrid cross)?
3. Whatever your answer, explain why you obtained the outcome you did. (Why do the data from Part B match
Mendel’s more closely- or less closely- than the monohybrid cross in Part A?)
4. Students almost never obtain a perfect match for the expected outcomes. Why do you think this is so?
5. If we combined the data from the whole class, do you think your data would more closely or less closely match
Mendel’s results? Why?
Extra credit: Complete a Punnett Square for any tri-hybrid cross.