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Transcript
GENETICS
Gregor Mendel – 1840’s
Conclusion:
Heredity material was packaged in discrete
transferable units; came up with law of
segregation and law of independent
assortment.
Thomas Morgan – early 1900’s
Discovered that
fruit flies’ genes
were associated
with chromosome
inheritance.
Conclusion:
Chromosomes were known to be composed of
proteins and DNA, so genes must be one of
these two macromolecules
How traits are inherited:
The debate in the 1800’s
The Blended Hypothesis: the traits from 2 parents
are mixed to become a third trait.
The Particulate Hypothesis: the traits from 2
parents are joined but remain discrete, and can be
separated again to their original forms.
Which hypothesis seems more logical to you?
-Come up with a response & give at least
one supporting argument.
In order to test these 2 hypotheses,
a scientist must have “true breeding
parents”
What does “true-breeding” mean?
The parent organism produces only one type of
offspring for a particular characteristic
The blended hypothesis
Parental generation
True breeding black coffee
F1 generation
True breeding white cream
The traits from two parents are mixed to become a
third trait.
F1 generation
F2 generation
What type of offspring can the parents in the above cross make?
Only mocha colored; one phenotype
What type of offspring can the F1 generation create?
Mocha-colored offspring; only one phenotype
Can the 2 traits for the characteristic of color be separated out again in this example?
No
Do we see examples of this type of
character blending in real life?
The particulate hypothesis
True breeding
yellow
True breeding
red
Parental generation
F1 generation
The traits from the two parents are joined but remain
discrete, and can be separated again to their original
forms.
F1 generation
What type of offspring can the parental generation make?
Only two-colored offspring; one phenotype possible
What type of offspring can the F1 generation create?
Yellow, red, or a mixture of red and yellow; 3 phenotypes are possible
Can the 2 traits for color be separated out again in this example?
Yes
Do we see examples of this type of
character mixing in real life?
Which hypothesis, do you
think, is more likely to be
accurate?
The particulate hypothesis could give just one
type of offspring, which could account for why we
see some blended traits, but with the blended
hypothesis, traits could never separate out to yield
variation in subsequent generations, as with
discrete traits.
MONOHYBRID CROSSES
PREDICTING PROBABILITY OF
POTENTIAL OFFSPRING
• All organisms pass on inherited information
using haploid gametes.
• Chromosomes are discrete packages of
genetic material that can be traced back to
the two parents that produced the zygote
The marbles in this example represent certain traits or
alleles. Where are these alleles located (in us)?
•They are located at specific places on paired chromosomes, called
genes.
• One chromosome in a pair has Mom’s allele at a gene locus & the
other chromosome in the pair has Dad’s allele at the same gene locus.
•Because the alleles come from different parents, they may not be the
same.
The zygote grows by mitosis to
become a multi-cellular organism
• How do inherited alleles from your parents
show themselves in physical traits all over
your body?
Because DNA replicates before cell division, the alleles
from each parent are passed on to every cell that makes
up your body and thus the chromosomes can express any
trait that is needed in any part of the body.
How do these inherited traits
get passed on to your offspring?
• One of these cells with all of Dad’s
chromosomes and all of Mom’s
chromosomes will become a gameteproducing cell in the gonads and begin
making sperm and eggs with half the
number of required chromosomes.
Which chromosomes will your
children get- the ones from your
dad or the ones from your mom?
• They will get a random mixture of
both as a consequence of crossingover and the law of segregation.
Do you know the following terms?
•
•
•
•
•
Genotype
Phenotype
Dominant
Recessive
Test cross
The alleles of a given gene locus; genetic make-up
The expression of gene locus; observable trait
An allele that is fully expressed in a heterozygote
An allele that isn’t expressed at all in a heterozygote
Breed an organism of unknown genotype with a
recessive homozygote
Crossing pea plants for the color of flower
Name the phenotypes of the
flowers in the P generation
Name the genotype of the
flower in the F1 generation
Which of the traits are
dominant and which are
recessive?
What are the possible gametes of
parents of each of the following
genotypes?
AA
Aa
A or A
A or a
aa
a or a
TEST CROSS
Let’s work out a problem with dogs…
Heterozygous parents for black hair are crossed and their
progeny are then test-crossed. Determine the expected
genotypic and phenotypic ratios among the test cross progeny.
Parental generation: Bb x Bb.
B
b
B
BB
Bb
b
Bb
bb
Test cross of the F1 is to a white
haired dog (bb).
¼ of the F1 will be black haired homozygous dominant. When crossed with bb the result will
be all black (Bb).
½ of the F1 will be black haired heterozygous (Bb). When crossed with bb the result will be ½
black and ½ white.
¼ of the F1 will be white haired (bb). When crossed with bb all offspring will be white haired
So… ½ Bb and ½ bb ratio= 1:1 for phenotype and 0:1:1 for genotype
Recall the Law of Independent Assortment
Probability
• If a coin is flipped once, what is the
probability it will land on heads?
• If it lands on heads, what is the
probability it will land on heads again?
• If I keep flipping the coin and it keeps
coming up heads, will that affect the
probability of the outcome of my next
flip?
What is the probability that I will get
heads on 2 coin tosses in a row?
• This changes the probability because you are
now asking for the probability of flipping
heads and flipping heads again.
• When you have 2 separate probabilities that
both must occur, the 2 separate probabilities
must be multiplied to find the overall
probability.
So the probability of heads and heads occurring is ½ x ½ = ¼
The rule of multiplication
• Used whenever there is a statement of one
probability “and” another probability
Let’s practice:
• What is the probability that a couple would
have 3 boys? ½ x ½ x ½ = 1/8
• What is the probability that parents with the
genotypes Aa and Aa would have a child with
the genotype AA?
Probability of dad giving “A” is ½ and the same can be
said for mom then… ½ x ½ = ¼
Let’s try another!!!
Black wool in sheep is recessive (w) to white wool which is dominant
(W). A white ram is crossed with a white ewe. Both parents carry the
allele for black (w).
They produce a white ram lamb which is then back crossed to the
female parent.
Determine the probability of the back cross offspring being black.
The white ram lamb (F1) is either WW or Ww.
W
w
W
WW
Ww
w
Ww
ww
1/3 of the white offspring (WW) crossed with
parent Ww will result in zero black lambs
2/3 of the white offspring (Ww) crossed with
parent Ww will result in ¼ black lambs.
So… 1/4 x 2/3 = 1/6 or 0.16
Let’s use a deck of cards (52 cards)
• If the probability of getting dealt an ace from a
deck of cards is 4 out of 52, what is the
probability of being dealt an ace or a king?
• The probability of getting an ace is 4/52 and the
probability of getting a king is 4/52. You must add
the probabilities together to get the total
probability.
• 4/52 + 4/52 = 8/52
Rule of addition
• Used whenever there is a statement of one
probability “or” another probability.
Let’s practice:
• What is the probability of being dealt an ace
or a face card from one deck of cards?
• The probability of getting an ace is 4/52 & the
probability of getting a face card is 12/52
So… 4/52 + 12/52 = 16/52 = 4/13
Dihybrid crosses
IIGG x iigg
I = inflated
i= constricted
G= green
g= yellow
How many traits or genes are represented by these
genotypes? 2 – pod texture & pod color
When gametes are made in meiosis, how many alleles does
each gamete get of a single gene? One allele of each gene
What gametes can the 1st parent make? IG
What gametes can the 2nd parent make? ig
What type(s) of offspring will be produced? IiGg
= F1
generation
If the F1 generation was allowed to cross, what types of
gametes would result?
“IG” “Ig” “iG” and “ig”
If both parents are heterozygous
(IiGg x IiGg)
• How can we predict the outcome?
Creating a dihybrid cross!!!
Recall the law of segregation:
• Do both dominant alleles have to go to the same
gamete? No
• Can a gamete have a mixture of dominant and
recessive alleles? Yes
• Do you have a mixture of dominant and recessive
traits from your parents? Yes
• How is it that each gene can segregate
independently when they are all on a limited
number of chromosomes?
Crossing over during prophase 1 allows alleles to mix randomly as the
gene loci are not located too near one another on the chromosome
GENE EXPRESSION
How does a genotype result in a
phenotype?
• The genotype is an abbreviation for the allele
found at a specific gene.
• The gene is a sequence of DNA that codes for
the “recipe” to make a particular protein.
• For example” melanin gene locus can have
different sequences of DNA (different alleles)
that either make melanin – a dark skin
pigment that absorbs UV or don’t.
• If the genotype is “MM” then the person has 2
alleles, or strands of DNA, that make the protein
melanin.
• If the genotype is “Mm” then they have one
allele that makes the protein & one that doesn’t
• Which genotype would make the darker skin?
Let’s look at Snap Dragons
CR CR
CW CW
What color would a snap dragon be
with a genotype of CR CW ?
THIS IS AN EXAMPLE OF
INCOMPLETE DOMINANCE
Gene expression
• Alleles do not have to be completely
expressed or unexpressed.
• An allele can make small amounts of a protein
in its recessive form and greater amounts in
its dominant form.
• There is a wide range of expression for any
allele termed dominant or recessive.
• CODOMINANCE- neither of the 2
alleles of the same gene totally
masks the other. The result is a
combination of both dominant
traits.
CODOMINANCE
R = red colored coat
W= white colored coat
RR x WW
R
R
All offspring will be …..?
W
RW
RW
RED & WHITE !!
W
RW
RW
Blood Types
• There are 6 genotypes.
• They make up 4 phenotypes (blood types).
• A and B are codominant, and O is recessive.
Genotype
IAIA or AA
IAi or AO
IBIB or BB
IBi or BO
IAIB or AB
ii or OO
Phenotype
(Blood Type)
A
A
B
B
AB
O
We all inherit a set of three Rhesus (Rh) genes from
each parent called a haplotype. They are referred to
as the c, d, e, C, D and E genes. The upper case
letters denote Rh positive genes and the lower case,
negative and we inherit either a positive or negative
of each gene from each parent (eg. CDe/cde, cdE/cDe
etc.). This means that we then possess two of each
gene and can pass either to our offspring.
If a person is tested Rh positive, their blood is said to
contain the Rhesus factor - if they are tested negative
it does not. A person possessing one or more positive
Rh genes (C, D or E), anywhere in their inherited
haplotypes, has inherited the Rh factor (eg. cdE/De,
cde/cDe etc.) and they are tested Rh positive - only a
person with a genotype of cde/cde is truly Rh
negative.
The baby of an Rh negative woman may inherit the Rh
positive factor from his/her father. This would result in the
mother and baby having different blood types. During
pregnancy, some of the baby's Rh positive red blood cells
may enter the mother's circulation. The cells are
recognized as being "foreign" by the mother's immune
system, and she may produce antibodies. These
antibodies can be permanent, and are capable of crossing
over into the baby's blood and break down his/her Rh
positive red blood cells; they will not harm the mother.
Antibodies are usually produced too late in the first
pregnancy to affect the baby being carried. Future babies
are at risk since the antibodies are already present when
pregnancy occurs.
What are the possible blood types of the
potential children of an AB (IAIB) male
and an B (IB i) female?
What % chance will the offspring be type B?
Hint: use a punnett square
IAIB x IBi
IA
IB
i
I AI B
IAi
IB
IBIB
IBi
What are the possible blood
types of the potential children
of an AB (IAIB) male and an B
(IB i) female?
TYPES: A, B, or AB
What % chance will the
offspring be type B?
50%
PEDIGREES
Collecting information about a family ‘s history
for a particular trait and creating a family tree
that displays how the trait is passed down from
generation to generation.
• One type of pedigree shows only the
phenotypes of individuals.
• Another type of pedigree chart gives
information about all of the individuals’
genotypes for a trait.
Addams Family
Bb
bb
Bb
bb
bb
What is the percent chance the new
Addams’ baby will have web feet?
Bb
bb
Bb
B_
Using pedigrees to trace
inheritance patterns WS
When you are done with the WS, get
together with your lab groups and come
up with 3-4 rules that apply to recessive
allele inheritance patterns and 3-4 rules
that apply to dominant allele
inheritance patterns.
Possible Rules for Inheritance Patterns
Recessive
Dominant
• Unaffected parents can
have affected offspring
• The phenotype can skip a
generation
• Individuals with no sign of
the allele can be carriers
• Affected offspring must
have at least one affected
parent
• The phenotype appears in
every generation without
skipping
• Two unaffected parents
have no affected offspring
Genetic Testing
Chromosomal Basis of Sex
• Short segments at either end of the Y
chromosome are homologous with the same
regions on the X chromsome
– Allows the X and Y chromosomes to pair up during
meiosis.
• XX= female and XY= male
– Anatomical distinction of sex in an embryo occurs
at about 2 months
– SRY gene on the Y chromosome is a trigger for
testes to form.
• If not present, ovaries form.
Hemophilia: A Sex-Linked
Trait
• Hemophilia is an inherited condition in which the blood
clots slowly or not at all
• Two genes that encode blood-clotting proteins reside on
the X chromosome
• Hemophilia is an X-linked recessive disorder
– Males develop hemophilia if they inherit one mutant
allele from their mother
– For females to develop hemophilia, they have to
inherit two mutant alleles, one from each parent
• Royal hemophilia
– Started by a mutant allele in Queen Victoria of England
– Three of her nine children received the defective allele
• They transferred it by marriage to other royal families
Queen
Victoria
Fig. 8.28
•
In all, 10 of Victoria’s male descendants had hemophilia
Fig. 8.28
Escaped the disorder
Sickle-Cell Anemia: Recessive Trait
• Sickle-cell anemia is an autosomal recessive trait in which
the protein hemoglobin is defective
– Affected individuals cannot properly transport oxygen to their tissues
Fig. 8.29
Huntington’s Disease: Dominant Trait
• Huntington’s disease is an autosomal dominant trait that
causes progressive deterioration of brain cells
• It is a fatal disease
– However, it persists
in human
populations
because it has a late
onset
Fig. 8.33
SEX LINKED TRAITS
• THESE ARE TRAITS LOCATED ON THE SEX
CHROMOSOMES.
• MALES PASS GENES LOCATED ON THE “X” CHROMOSOME
TO ALL OF THEIR DAUGHTERS & NONE OF THEIR SONS.
• WHATEVER MOM HAS ON HER “X” CHROMOSOME WILL
BE EXPRESSED IN HER SONS EVEN IF THE TRAIT IS
RECESSIVE.
• Examples of sex linked traits: hemophilia and
colorblindness
SEX LINKED ( X-LINKED)
H= non hemophilia
h= hemophilia
XhXh x XHY
Xh
XH
Y
Xh
XHXh
XHXh
XhY
XhY
IF THESE PARENTS HAVE A BOY, WHAT ARE HIS
CHANCES THAT HE WILL HAVE HEMOPHILIA?
100%
X inactivation in Female Mammals
• One of the two X chromosomes
in each cell is randomly
inactivated during embryonic
development
• If a female is heterozygous for a particular gene
located on the X chromosome
– She will be a mosaic for that character due to the
turning off or on of the gene in a cell
Two cell populations
in adult cat:
Active X
Early embryo:
X chromosomes
Cell division
Inactive X
and X
chromosome Inactive X
inactivation
Black
fur
Allele for
orange fur
Allele for
black fur
Figure 15.11
Orange
fur
Active X
Interactions between genes
• Some genes may not just control a single
characteristic or trait in the phenotype of an
organism.
• Most genes probably have an effect on one or
more phenotypic traits
Pleiotropy
Interactions between genes
• In some cases a single characteristic may be
controlled by more than one gene. -Polygeny
EPISTASIS
• INTERACTION BETWEEN THE PRODUCTS
OF TWO GENES IN WHICH ONE OF THE
GENES MODIFIES THE PHENOTYPIC
EXPRESSION PRODUCED BY THE OTHER.
–EX: COAT COLOR FOR LABRADOR
RETRIEVERS.
• The E gene determines if dark pigment will be
deposited in the fur or not.
• If the dog has ee there is no pigment & dog
will be yellow.
• The B gene determines how dark the pigment
will be.
• In yellow labs the B gene indicates the color
on their nose, lips, & eye rims.
E_B_=black
E_bb=chocolate/brown
B gene=black
eeB_
b= brown
eebb
SEX INFLUENCED TRAITS
• THE PRESENCE OF MALE OR FEMALE
HORMONES INFLUENCES THE EXPRESSION
OF CERTAIN TRAITS.
–EX: PATTERN BALDNESS
• IF FEMALE IS HETEROZYGOUS SHE WILL NOT
BE BALD.
-the gene is recessive in females
• IF A MALE IS HETEROZYGOUS HE WILL BE
BALD.
-the gene is dominant in males
ENVIRONMENTAL
EFFECTS
• Many alleles are expressed
depending on the environment.
–Some are heat sensitive
• Ex:Arctic foxes make fur pigment
only when the weather is warm.
CAN YOU SEE WHY THIS TRAIT
WOULD BE AN ADVANTAGE?
Inheritance of Organelle Genes
Mitochondria and chloroplasts have small
circular DNA (aka extranuclear genes)
Nonnuclear inheritance
• Chloroplasts & mitochondria are randomly
assorted to gametes & daughter cells
– Traits determined by them do not follow simple
Mendelian rules.
– Does not distribute genes to offspring through the
same process as nuclear DNA by way of meiosis.
• In animals, mitochondrial DNA is transmitted by
the egg & not the sperm
– Mitochondrial determined traits are maternally
inherited.
• Help make up protein complexes of the ETC
• ATP synthase
Most susceptible to energy deprivation
• Nervous system and muscles.
– Mitochondrial myopathy = weakness, intolerance
of exercise, and muscle deterioration
– Leber’s hereditary optic neuropathy= produces
sudden blindness in people as young as 20.
Chi Square Test
a statistical test commonly used to compare
observed data with data we would expect to
obtain according to a specific hypothesis
It’s time for …
M & M CHI SQUARED ANALYSIS!!!
Are the numbers of different colors of M & M’s
in a package really different from one package
to the next? Or does the Mars Company do
something to ensure that each package gets a
certain number of each M & M color?
Chi-Square Test:
• A fundamental problem in genetics is determining
whether the experimentally determined data fits the
results expected from theory (i.e. Mendel’s laws as
expressed in the Punnett square).
• How can you tell if an observed set of offspring counts
is legitimately the result of a given underlying simple
ratio?
– For example, you do a cross and see 290 purple flowers
and 110 white flowers in the offspring.
– This is pretty close to a 3/4 : 1/4 ratio, but how do you
formally define "pretty close"? What about 250:150?
“Goodness of fit”
• The chi-square test is a “goodness of fit” test:
it answers the question of how well do
experimental data fit expectations.
• We start with a theory for how the offspring
will be distributed: the “null hypothesis”.
• We will discuss the offspring of a selfpollination of a heterozygote.
– The null hypothesis is that the offspring will
appear in a ratio of 3/4 dominant to 1/4 recessive
An Example
• If, according to Mendel's laws, you expected
10 of 20 offspring from a cross to be male and
the actual observed number was 8 males,
then you might want to know about the
"goodness to fit" between the observed and
expected.
The Question is…
• Were the deviations (differences between
observed and expected) the result of chance,
or were they due to other factors?
• How much deviation can occur before you, the
investigator, must conclude that something
other than chance is at work, causing the
observed to differ from the expected?
The answer is…
• The chi-square test is always testing what scientists
call the null hypothesis, which states that there is
no significant difference between the expected and
observed result.
• The formula for calculating chi square is:
(o-e)2
2
X =Ʃ e
• That is, chi-square is the sum of the squared
difference between observed (o) and the expected
(e) data (or the deviation, d), divided by the
expected data in all possible categories.
More examples…
• Suppose that a cross between two pea plants
yields a population of 880 plants, 639 with
green seeds and 241 with yellow seeds. You are
asked to propose the genotypes of the parents.
• Your hypothesis is that the allele for green is
dominant to the allele for yellow and that the
parent plants were both heterozygous for this
trait because of the observed outcomes.
• If your hypothesis is true, then the predicted
ratio of offspring from this cross would be 3:1
(based on Mendel's laws) as predicted from
the results of the Punnett square
Predicted offspring from
cross between green and
yellow-seeded plants.
Green (G) is dominant (3/4
green; 1/4 yellow).
• To calculate X2 , first determine the number
expected in each category. If the ratio is 3:1
and the total number of observed individuals
is 880, then the expected numerical values
should be 660 green and 220 yellow.
X2
=Ʃ
(o-e)2
e
Calculating Chi-Square
Green
Yellow
Observed (o)
639
241
Expected (e)
660
220
Deviation (o - e)
-21
21
Deviation2 (d2)
(o-e)2
441
441
d2/e
0.668
2
Ʃ (d2/e) = X2 Your final answer should be: 2.668
So what does this answer mean?
• Here's how to interpret the X2 value:
• Determine degrees of freedom (df). Degrees
of freedom can be calculated as the number
of categories in the problem minus 1. In our
example, there are two categories (green and
yellow); therefore, there is I degree of
freedom.
• Determine a relative standard to serve as the
basis for accepting or rejecting the hypothesis.
• The relative standard commonly used in
biological research is p > 0.05.
• The p value is the probability that the
deviation of the observed from that expected
is due to chance alone (no other forces
acting).
• In this case, using p > 0.05, you would expect
any deviation to be due to chance alone 5% of
the time or less.
Using the appropriate degrees of
'freedom, locate the value closest to
your calculated chi-square in the table.
Determine the closest p (probability) value
associated with your chi-square and degrees
of freedom.
Chi-Square Distribution
Degrees
of
Freedom
(df)
Probability (p)
0.95
0.90
0.80
0.70
0.50
0.30
0.20
0.10
0.05
0.01
0.001
1
0.004
0.02
0.06
0.15
0.46
1.07
1.64
2.71
3.84
6.64
10.83
2
0.10
0.21
0.45
0.71
1.39
2.41
3.22
4.60
5.99
9.21
13.82
3
0.35
0.58
1.01
1.42
2.37
3.66
4.64
6.25
7.82
11.34
16.27
4
0.71
1.06
1.65
2.20
3.36
4.88
5.99
7.78
9.49
13.28
18.47
5
1.14
1.61
2.34
3.00
4.35
6.06
7.29
9.24
11.07
15.09
20.52
Nonsignificant
Significant
• In this case (X2=2.668), the p value is about 0.10.
• This means that there is a 10% probability that
any deviation from expected results is due to
chance only.
• Based on our standard p > 0.05, this is within the
range of acceptable deviation.
• In terms of your hypothesis for this example, the
observed chi-square is not significantly different
from expected.
• The observed numbers are consistent with those
expected under Mendel's law.
• Our hypothesis is accepted.
Step-by-Step Procedure for Testing Your
Hypothesis and Calculating Chi-Square
• State the hypothesis being tested and the
predicted results.
• Gather the data by conducting the proper
experiment (or, if working genetics problems,
use the data provided in the problem).
• Determine the expected numbers for each
observational class.
• Remember to use numbers, not percentages.
• Calculate X2 using the formula.
• Complete all calculations to three significant
digits.
• Round off your answer to two significant digits.
• Use the chi-square distribution table to
determine significance of the value.
– Determine degrees of freedom and locate the value in
the appropriate column.
– Locate the value closest to your calculated 2 on that
degrees of freedom df row.
– Move up the column to determine the p value.
•
• State your conclusion in terms of your hypothesis.
– If the p value for the calculated X2 is p > 0.05, accept
your hypothesis. 'The deviation is small enough that
chance alone accounts for it. A p value of 0.6, for
example, means that there is a 60% probability that any
deviation from expected is due to chance only. This is
within the range of acceptable deviation.
– If the p value for the calculated X2 is p < 0.05, reject your
hypothesis, and conclude that some factor other than
chance is operating for the deviation to be so great. For
example, a p value of 0.01 means that there is only a 1%
chance that this deviation is due to chance alone.
Therefore, other factors must be involved.
Statistical analysis Section- Homework Questions:
A Punnett square of the F1 cross Gg x Gg would predict that the
expected proportion of green: albino seedlings would be 3:1.
The following information is what was observed from this cross:
72 green seedlings & 12 albino seedlings. Determine if these
results are due to chance alone or some other circumstance.
Null hypothesis: there is no statistically significant difference between the observed and
expected data
Phenotype # observed (o) # expected (e) (o-e) (o-e)2 (o-e)2 /e
Green
72
63
9
81
1.29
Albino
12
21
-9
81
3.85
5.14
If the calculated chi-square value is greater than or equal to the critical value then the null
hypothesis is rejected. Since 5.14 > 3.84, we reject our null hypothesis. Chance alone cannot
explain the deviations observed & there is reason to doubt our original hypothesis.
Chi-Square Distribution
Degrees
of
Freedom
(df)
Probability (p)
0.95
0.90
0.80
0.70
0.50
0.30
0.20
0.10
0.05
0.01
0.001
1
0.004
0.02
0.06
0.15
0.46
1.07
1.64
2.71
3.84
6.64
10.83
2
0.10
0.21
0.45
0.71
1.39
2.41
3.22
4.60
5.99
9.21
13.82
3
0.35
0.58
1.01
1.42
2.37
3.66
4.64
6.25
7.82
11.34
16.27
4
0.71
1.06
1.65
2.20
3.36
4.88
5.99
7.78
9.49
13.28
18.47
5
1.14
1.61
2.34
3.00
4.35
6.06
7.29
9.24
11.07
15.09
20.52
Nonsignificant
Significant
Reject the null
hypothesis