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Experiment 1: Punnett square crosses

Materials
Red beads
Blue beads
Green beads
Yellow beads
2 100mL Beakers

Procedure
1. Set up and complete Punnett squares for each of the following crosses:
(remember Y = yellow, and y = blue)
Y Y and Y y

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Y Y and y y

?????

Y

Y

Y

YY

YY

y

Yy

Yy

?????

Y

Y

y

Yy

Yy

y

Yy

Yy

a) What are the resulting phenotypes?

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The punnet square shows all of the possible combinations that would result
from the mating of the given genotypes. In this example the dominant allele is
represented by Y, and determines the color of the corn to be yellow. The
results from the square is: 100 percent of the offspring will have a yellow
phenotype, because every combination expresses the dominant allele Y.

b) Are there any blue kernels? How can you tell?
For a recessive allele to be expressed there must be two recessive alleles
present.In this particular example, there is no combination that leads to a
homozygous recessive genotype or yy. Meaning there is no blue corn
expressed.

2. Set up and complete a Punnett squares for a cross of two of the F1 from 1b
above:
a) What are the genotypes of the F2 generation?
The results of the completed punnet square shows the probability of producing
a homozygous recessive offspring is 25%, a 50% chance of producing a
heterozygous offspring and a 25% chance the offspring will be homozygous
dominant.

b) What are their phenotypes?
The results show a 75% chance the offspring will be yellow and 25% chance
the offspring will be blue.

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c) Are there more or less blue kernels than in the F1 generation?
In F2 generation there is a 25% probability that a blue phenotype will be
expressed. However, all the genotypes in the F1 generation contain a
dominant allele supressing the recessive allele leaving a 0% chance of having
a blue phenotype. In other words, F1 has less blue kernels than the F2
generation.

3. Identify the four possible gametes produced by the following individuals:
a) YY Ss:

YS

YS

b) Yy Ss:

YS

yS

Ys
Ys

Ys
ys

c) Create a Punnett square using these gametes as P and determine the
genotypes of the F1:
There are 6 distinct genotypes expressed in F1. YYSS: expressed two times
YYSs: expressed four times YYss: expressed two times YySS: expressed two
times YysS: expressed four times Yyss: expressed two times.

d) What are the phenotypes? What is the ratio of those phenotypes?
When determining the phenotype of a gene the dominant allele is always
expressed as the phenotype. For Example in this punnet Square Ss=SS and
YY=Yy, so any time Y is expressed the phenotype is yellow and when S is
expressed the phenotype is smooth. There are two distinct phenotypes: 1.
yellow and smooth 2. yellow and wrinkled. Therefore, 75 percent of the
offspring will be yellow and smooth, and the other 25 percent will be yellow
and wrinkled. The ratio for the phenotype is 12:4. 12 combinations of Yellow
and smooth and 4 combinations of yellow and wrinkled.

4. You have been provided with 4 bags of different colored beads. Pour 50 of the
blue beads and yellow beads into beaker #1 and mix them around. Pour 50 of
the red beads and green beads in beaker #2 and mix them.
Attention! Do not pour the beakers together.
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?

#1 contains beads that are either yellow or blue.

?

#2 contains beads that are either green or red.

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Both contain approximately the same number of each colored bead.

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These colors correspond to the following traits (remember that Y/y is
for kernel color and S/s is for smooth/wrinkled):

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Yellow (Y) vs. Blue (y)

Green (S) vs. Red (s).

A. Monohybrid Cross: Randomly (without looking) take 2 beads out of #1.

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This is the genotype of individual #1, record this information. Do not
put those beads back into the beaker.

?

Repeat this for individual #2. These two genotypes are your parents
for the next generation. Set up a Punnett square and determine the
genotypes and phenotypes for this cross.

?

Repeat this process 4 times (5 total). Put the beads back in their
respective beakers when finished.

a) How much genotypic variation do you find in the randomly picked
parents of your crosses?
When I randomly picked the parents from the beaker I found six distinct
genotypes. Within all of the randomly selected parents, I chose a
heterozygous genotype five out of the ten times, homozygous dominant
genotype two out of the ten times, and a homozygous recessive genotype
three out of the ten times. The genotypes that were chosen: Yy, Ss, yy, SS,
ss, and YY.

b) How much in the offspring?
There was five homozygous dominant genotypes, seven homozygous
recessive genotypes, and eight heterozygous genotypes chosen out of the
twenty total offspring.

© 2010 eScience Labs, Inc.
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© 2010 eScience Labs, Inc.
All Rights Reserved

c) How much phenotypic variation?
There were four different phenotypes for this exercise: 1. Yellow and Smooth
2. Yellow and Wrinkled 3. Blue and Smooth 4. Blue and Wrinkled. Five were
yellow and smooth phenotype, three were yellow and wrinkled phenotype, five
were blue and smooth phenotype, seven were blue and wrinkled phenotype of
the twenty total offspring.

d) Is the ratio of observed phenotypes the same as the ratio of predicted
phenotypes? Why or why not?
?????

e) Pool all of the offspring from your 5 replicates. How much phenotypic
variation do you find?
?????

f)

What is the difference between genes and alleles?

A gene is a section of DNA that determines a specific trait such as hair color or
blood type. Genes are functional units of heredity. Alleles are variations of the
same gene. For Example: One single gene determines a persons height and
whether or not you are tall or short would be determined through the
combinations of alleles within the height gene. Genes determine the genetic
traits a person will have and the different sequences of alleles within the gene
will determine a single characterisitc of the individual.
g) How might protein synthesis execute differently if there a mutation
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occurs?
A gene mutation is an alteration in the sequence of nucleotides in DNA. DNA
consists of nucleotides. During protein synthesis, DNA is transcribed into RNA
and then translated to produce proteins. Altering nucleotide sequences most
likely will result in nonfunctioning proteins. For example: if a nonsense
mutation occurs it will alter the nucleotide sequence coding a stopcodon in
place of a amino acid, signaling the end of the translation process, stopping
protein production.

h) Organisms heterozygous for a recessive trait are often called carriers
of that trait. What does that mean?
Carriers contain a recessive allele and are capable of passing on the
recessive allele to their offspring, even though they do not express the
recessive allele, because of the presence of a dominant allele. A person that
inherits a genetic trait but does not display or show symptoms of that trait, but
are able to pass the gene on to their offspring.

i)

In peas, green pods (G) are dominant over yellow pods. If a
homozygous dominant plant is crossed with a homozygous recessive
plant, what will be the phenotype of the F1 generation? If two plants
from the F1 generation are crossed, what will the phenotype of their
offspring be?

If a homozygous dominant plant is crossed with a homozygous recessive
plant, 100% of the offspring will be green. If two plant from the F1 generation
are crossed,since they are all heterozygous plants, 3/4 (75%) of the plant’s
offspring will be green and 1/4 (25%) of the plant’s offspring will be yellow.

© 2010 eScience Labs, Inc.
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B. Dihybrid Cross: Randomly (without looking) take 2 beads out of beaker #1 AND 2
beads out of beaker #2.

?
?

These four beads represent the genotype of individual #1, record this
information.
Repeat this process to obtain the genotype of individual #2.
a) What are their phenotypes?

?????

b) What is the genotype of the gametes they can produce?
?????

?

Set up a Punnett square and determine the genotypes and
phenotypes for this cross.
c) What is your predicted ratio of genotypes? Hint: think back to our
example dihybrid cross

?????

?

Repeat this process 4 times (for a total of 5 trials).
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d) How similar are the observed phenotypes in each replicate?
?????

e) How similar are they if you pool your data from each of the 5
replicates?
?????

f)

Is it closer or further from your prediction?

?????

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g) Did the results from the monohybrid or dihybrid cross most closely
match your predicted ratio of phenotypes?
?????

h) Based on these results; what would you expect if you were looking
at a cross of 5, 10, 20 independently sorted genes?
?????

i)

Why is it so expensive to produce a hybrid plant seed?

?????

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j)

In certain bacteria, an oval shape (S) is dominant over round and
thick cell walls (T) are dominant over thin. Show a cross between
a heterozygous oval, thick cell walled bacteria with a round, thin
cell walled bacteria. What are the phenotype of the F1 and F2
offspring?

?????

5. The law of independent assortment allows for genetic recombination. The
following equation can be used to determine the total number of possible
genotype combinations for any particular number of genes:
2g= Number of possible genotype combinations (where “g” is the number of
genes)
1 gene:
21= 2 genotypes
2 genes: 22= 4 genotypes
3 genes: 23 = 8 genotypes
Consider the following genotype:
Yy Ss Tt
We have now added the gene for height: Tall (T) or Short (t).
a) How many different gamete combinations can be produced?
?????

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b) Many traits (phenotypes), like eye color, are controlled by multiple
genes. If eye color were controlled by the number of genes
indicated below, how many possible genotype combinations would
there be?
5:
10:
20:

?????
?????

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