Download Chi Square values have been published by statisticians in a table

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Mercer County Community College
Division of Science and Health Professions
BIO 208
Genetics
Applied Human Genetics
Single Gene (Monohybrid) Crosses
Probability
Chi Square Analysis
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Applied Human Genetics/Single Gene Traits
The classical study of single gene traits is by pedigree analysis. Over 7000 human single gene
traits have been recognized, many also assigned to the particular chromosome on which they
reside. The Human Genome Project will uncover many more genes and further study in
proteomics will assign function to these genes. Chromosome 22 contains over 500 genes and
the larger chromosomes contain thousands of genes. For more information on the
mechanism of inheritance of single gene traits, allelic variations, and chromosome mapping,
consult the Online Mendelian in Man web site at www.OMIM.gov.
The genotype of a homozygous recessive individual is usually known by the phenotype
exhibited. For example, individuals with blue eyes are bb. However, the dominant phenotype
could arise from two genotypes. A person with the dominant brown eye phenotype could have
the BB or Bb genotype. Therefore the genotype is represented as B-. The dash indicates that
while a second allele is present, the nature of that allele is unknown
BB = homozygous dominant
Bb = heterozygous
bb = homozygous recessive
B- indicates dominant phenotype
Single gene traits
Interlocking fingers (I)
Interlock your fingers and observe which thumb is on top (right or left). The tendency to place
the left thumb on top is due to a dominant allele (I). Those who place the left thumb on top
have the genotype I- (either II or I-). The right thumb on top is determined by the ii genotype
Ear lobes (E)
Examine your partner’s earlobes. The dominant allele (E) results in the phenotype of free
earlobes. The recessive allele (e) is for attached earlobes. Earlobes are attached if the bottom
lobe is attached directly to the head. Earlobes are free if the lobe hangs free.
Widow’s peak (W)
Widow’s peak occurs when the hairline forms a distinct point in the center of the forehead. Lift
up the hair of your forehead to score this trait. People without widow's peak have a smooth
hairline with no dip. Men starting to go bald (or already bald) may be unable to score this trait.
Widow's peak is controlled by a dominant allele (W)
Tongue curling (T)
A dominant allele (T) results in the ability to curl the tongue in a U-shape
Hitch hiker’s thumb (H)
A person homozygous recessive for this trait (hh) can bend the last (distal) thumb joint back to
about a 90 degree angle. Those with the H allele cannot
Pigmented iris (B)
A person with the dominant allele has brown or brownish eyes. The recessive allele encodes
blue eyes
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Polydactyly (PD)
A person with polydactyly has >5 fingers on 1 or more hands and feet. It is dominant (PD)
Mid-digital finger hair
Examine middle segment of your fingers. If hair is present, even one, you have mid-digital hair.
PTC tasting (P)
If you can taste the chemical PTC, you have a dominant allele for this trait (P). Place the strip
of PTC paper on your tongue for a few seconds. If you cannot taste anything, you do not
possess the dominant allele
Fill in the following table
Gene
Describe phenotype
Earlobes (E)
Genotype
Widows peak (W)
Tongue roll (R)
Hitchhikers thumb (H)
Pigmented iris (P)
Polydactyly (PD)
PTC taster (T)
Interlocking fingers (I)
Class phenotypic frequencies.
Gene
Tongue roll (R)
Pigmented iris (P)
Polydactyly (PD)
PTC taster (T)
Interlocking fingers (I)
f (E-) dominant phenotype
f(ee) recessive phenotype
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Maize Genetics: Single gene crosses
Zea mays is a model organism in genetics. A large number of progeny are produced from
each cross as each kernel is a single individual. All kernels on one ear represent the progeny
of a single cross. The effects of many genes can be observed in the kernels, eliminating the
need to grow plants. Many genes determine the phenotypes of the 3 tissues that control the
color of a corn kernel.
Endosperm (kernel) traits
Color gene
Kernel shape gene
P allele = purple (color)
p allele = yellow (no color)
R allele= smooth
r allele= wrinkled
STATION 1. Obtain an ear of corn from the instructor
1. What are the two color phenotypes observed?
2. Count 50 kernels. Compute the ratio of phenotypes:
3. Which phenotype is caused by a dominant allele?
4. Show the cross of the genotypes and phenotypes of the parental plants that led to this cob.
Parental genotypes  _____________
Parental phenotypes  _____________
X
_______________
_______________
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STATION 2: Obtain an ear of corn from the instructor.
1. What are the two kernel shape phenotypes observed?
2. Count 50 kernels. Compute the ratio of phenotypes:
3. Which phenotype is caused by a dominant allele?
4. Show the cross of the genotypes and phenotypes of the parental plants.
Parental genotypes  _____________
Parental phenotypes _____________
X
_______________
_______________
STATION 3: The ear of corn represents the progeny of a cross between two purple, smooth
parents.
Genotype:
PpRr
X
PpRr
Phenotype: Purple, smooth
Purple, smooth
1. What are the 4 allelic combinations of gametes (eggs and sperm) each parent can
produce? Remember, a gamete can only and must contain one allele of each gene.
2. Construct a Punnett square showing the expected genotypes and phenotypes of the
offspring
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3. What are the 4 phenotypes (with respect to both kernel shape and color) observed in
the offspring of this cross?
4. Count 100 kernels. Compute the ratio of phenotypes (divide each number by the
smallest number)
5. Does this observed ratio approach the expected ratio?
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Chi Square
The purpose of the Chi Square (X2) test is to determine if data obtained experimentally
(observed data) fit into a model of expected data. If data are shown to fit into the model, then
any variations from the model are due to chance. If the observed data do not approximate the
model, then variations are said to be significant, or not due to chance. If deviations are small
they are more easily attributed to chance than are large ones.
X2 =

( O – E )2
E
O observed data
E expected data
Chi Square values have been published by statisticians in a table (below). Across the top of
the table are probability values (P). A P value >0.05 indicates that in over 5% of trials, similar
data would be obtained. A p >0.05 leads the researcher to accept the null hypothesis, that the
differences between observed and expected data are due to chance. In other words, the
observed data fits the expected data.
A P value <0.05 corresponds to a 1 chance in 20 (5%) that deviations are due to significant
variations. A p < 0.05 indicates that observed variations cannot be explained by the
expected model. In other words differences in the observed data are not due to chance.
The null hypothesis states that any variations between observed and expected data are only
due to chance. Rejecting the null hypothesis indicates that observed and expected data are
significantly different.
To determine which row of Chi Square values to use, first determine the degrees of freedom.
The degrees of freedom are one less than the number of phenotypic classes. For example if
white, yellow, purple, and red corn kernels are observed, the degrees of freedom is 3.
Critical Values of the Chi Square Distribution
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I. Application of data: Maize
A. Enter the data obtained from ear 1
Phenotype
Observed O
Expected E
(O-E)2
E
Deviations
(O-E)2
TOTAL
X2 = ______
1. How many degrees of freedom (df) are there?
2. Locate the two Chi Square values on either side of the calculated X 2. What is the lowest
probability value assigned to these X2s?
p=
3. Do you accept the null hypothesis (variations between observed and expected data are
only due to chance) or do you reject the null hypothesis (variations between observed
and expected data are significantly different? Explain
4.
In general, does a low Chi Square value support the null hypothesis?
B. Enter the data obtained from ear 3 (2 gene cross)
Phenotype
Observed
O
Expected
E
Deviations
(O-E)2
(O-E)2
E
TOTAL
X2 = ______
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1. How many degrees of freedom (df) are there?
2. What is the lowest probability value assigned to these X2s?
p=
3. Do you accept the null hypothesis (variations between observed and expected data are
only due to chance) or do you reject the null hypothesis (variations between observed
and expected data are significantly different? Explain.
II. Application of data: Beans
Obtain a handful of beans from a jar in which two different colors of beans have been mixed in
equal amounts. Count each color and enter in the table below
Phenotype
Observed O
Expected E
Deviations
(O-E)2
(O-E)2
E
TOTAL
X2 = ______
1. How many degrees of freedom (df) are there?
2. Locate the two Chi Square values on either side of the calculated X2. What is the lowest
probability value assigned to these X2s?
p=
3. Do you accept the null hypothesis (variations between observed and expected data are only
due to chance) or do you reject the null hypothesis (variations between observed and
expected data are significantly different? Explain
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Probability
The coin toss is a convenient tool to demonstrate principles of probability. Probability in coin
tossing involves chance. There is a ½ chance that a head will face up and a ½ chance that
tails will occur. No matter how many times a coin is tossed, the probability of heads or tails
remains the same.
If two coins are tossed at the same time, there are 4 combinations that can occur:
2 heads
1 head, one tail
2 tails
HH
HT, TH
TT
The chance of obtaining two heads is determined by employing the product rule: the
probability of two events occurring simultaneously is the product of their individual
probabilities.
1. What is the probability of obtaining two heads? ___________
The chance of obtaining a two heads OR a tail and a head is determined by using the sum
rule: The chance of either of two events occurring is the sum of their individual probabilities.
First use the product rule to determine the chance of HT and TH, then the sum rule to
determine the chance of HT OR TH.
2. The probability of obtaining HH or TH is
_________
3. Data collection
Obtain 3 coins.
a. What are the possible combinations of each class? Enter this information in the table
b. What is the probability of each class? Enter this information in the table
c. Toss the 3 coins 12 times. Combine your data with that of another group so that you
have a total of 24 throws. Enter in the observed numbers column
Classes
Combinations
Probability___________
3 heads
____________
__________________
____________
2 heads, 1 tail
HHT, HTH, THH
1/8+ 1/8+ 1/8 = 3/8
____________
1 head, 2 tails
____________
__________________
____________
3 tails
____________
__________________
____________
Totals
____________
8/8
_____
Observed numbers
24_____
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Binomial expansion
Expectations for combinations of outcomes in groups can be obtained mathematically by using
the following binomial formula:
n!
___________ p x q n-x
P=
x! (n-x)!
Note: 0! = 1
1. 5 babies are born. What is the probability that 2 are boys and 3 are girls?
n! = 5! =
5 X 4X 3X 2X1
x! = 2! =
2X1
(n-x)! =
3 X 2 X 1
p = probability of a boy = ½
q = probability of a girl = ½
n = number of events
x = number of event 1
n-x = number of event 2
Determine P (2 boys, 3 girls)
2. Albinism is a recessive trait. Those with the albino phenotype are aa. Heterozygotes are Aa,
they are not albino, but are carriers for the trait. Two black mice, heterozygous for albinism
were mated.
Aa
X
Aa
1. What is the probability of obtaining a white mouse?
2. A black mouse?
3. In a litter of 6 mice, determine P (3 white, 3 black)?
4. In a litter of 5 mice, determine P (all black)?