GENETICS - Chandler Unified School District

GENETICS Gregor Mendel 1840s Conclusion: Heredity material was packaged in discrete transferable units; came up with law of segregation and law of independent assortment. Thomas Morgan early 1900s 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 1800s 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 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 Moms allele at a gene locus & the other chromosome in the pair has Dads 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 Dads chromosomes and all of Moms 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 isnt 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 Lets 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 b Bb Bb bb Test cross of the F1 is to a white haired dog (bb).

of the F1 will be black haired homozygous dominant. When crossed with bb the result will be all black (Bb). of the F1 will be black haired heterozygous (Bb). When crossed with bb the result will be black and white. of the F1 will be white haired (bb). When crossed with bb all offspring will be white haired So Bb and 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 x =

The rule of multiplication Used whenever there is a statement of one probability and another probability Lets practice: What is the probability that a couple would have 3 boys? x x = 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 and the same can be said for mom then x = Lets 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 black lambs. So 1/4 x 2/3 = 1/6 or 0.16 Lets 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. Lets 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? What type(s) of offspring will be produced? ig 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 dont. 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 doesnt Which genotype would make the darker skin? Lets 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 I i or AO A I I or BB

B B I i or BO B 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 IB i

IA IA IB IA i 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 Victorias 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 Huntingtons Disease: Dominant Trait Huntingtons 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

XH Y XhXh x XHY Xh Xh XHXh XH Xh Xh Y Xh Y 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 Linked Genes Linked genes tend to be inherited together because they are located near each other on the

same chromosome Each chromosome Has hundreds or thousands of genes Genes that are close together on the same chromosome are linked and do not assort independently Unlinked genes are either on separate chromosomes of are far apart on the same chromosome and assort independently with crossing over. The farther apart genes are on a chromosome The more likely they are to be

separated during crossing over If genes are located closer to one another, what happens to their recombination frequency? The percentage of recombination decreases toward 0% as distance between the genes decreases. A linkage map Is the actual map of a chromosome based on recombination frequencies APPLICATION A linkage map shows the relative locations of genes along a chromosome.

TECHNIQUE A linkage map is based on the assumption that the probability of a crossover between two genetic loci is proportional to the distance separating the loci. The recombination frequencies used to construct a linkage map for a particular chromosome are obtained from experimental crosses, such as the cross depicted in Figure 15.6. The distances between genes are expressed as map units (centimorgans), with one map unit equivalent to a 1% recombination frequency. Genes are arranged on the chromosome in the order that best fits the data. RESULTS In this example, the observed recombination frequencies between three Drosophila gene pairs (bcn 9%, cnvg 9.5%, and bvg 17%) best fit a linear order in which cn is positioned about halfway between the other two genes: Recombination frequencies 9.5% 9%

17% Chromosome b Figure 15.7 cn vg The bvg recombination frequency is slightly less than the sum of the bcn and cnvg frequencies because double crossovers are fairly likely to occur between b and vg in matings tracking these two genes. A second crossover would cancel out the first and thus reduce the observed bvg recombination frequency. If gene A recombines 12% of the time with gene

B but 16% of the time with gene C, which gene is closer to gene A ? B If gene B and C recombine with one another 28% of the time, what is the order of genes A, B, and C along the chromosome? B A C; B and A are 12 map units or centimorgans apart; A and C are 16 centimorgans apart; therefore, B and C are the furthest apart, with 28 centimorgans between them. Determine the sequence of genes along a chromosome based on the

following recombination frequencies: A B , 8 map units A C , 28 map units A D , 25 map units B C , 20 map units B D , 33 map units D A B C

A wild-type fruit fly (heterozygous for gray body color & normal wings) is mated with a black fly with vestigial wings. The offspring have the following phenotypic distribution: wild-type 778 black-vestigial 785 black-normal 158 gray-vestigial 162 What is the recombination frequency between these genes for body color and wing size? Take the total number of recombinant types & divide by the total number of offspring. 17%

A wild-type fruit fly (heterozygous for gray body color & red eyes) is mated with a black fruit fly with purple eyes. The following are their offspring: 721 wild-type 751 black-purple 49 gray-purple 45 black-red What is the recombination frequency between these genes for body color and eye color? 6% 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. When the genes that produce the dark coat are blocked by cold temperatures, a white coat results. In the summer these genes are not blocked, and a darker coat results.

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 Lebers 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 Its time for M & M CHI SQUARED ANALYSIS!!! Are the numbers of different colors of M & Ms 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. Mendels 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: 2 (o-e) X2 =

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. X = 2 (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

(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

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