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Chi-Square and Degree of Freedom Material

by: Notetaker

Chi-Square and Degree of Freedom Material BIOL 01104 5 (Mcardle, BIOLOGY 1: DIVERS/EVOL/ ADAP

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Daniel McArdle

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These are the notes for Chi square and Degree of freedom. I added a few more notes about Chi-square and Degrees of freedom from prior knowledge in Statistics class. Check it out!
Daniel McArdle
Biology, Chi-square, DOF, Degree of Dreedom
75 ?





Popular in Biological Sciences

This 9 page Bundle was uploaded by Notetaker on Monday October 19, 2015. The Bundle belongs to BIOL 01104 5 (Mcardle, BIOLOGY 1: DIVERS/EVOL/ ADAP at Rowan University taught by Daniel McArdle in Fall 2015. Since its upload, it has received 31 views. For similar materials see BIOLOGY 1: DIVERS/EVOL/ ADAP in Biological Sciences at Rowan University.

Similar to BIOL 01104 5 (Mcardle, BIOLOGY 1: DIVERS/EVOL/ ADAP at RU

Popular in Biological Sciences


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Date Created: 10/19/15
Basic Genetic Terminology Gene 0 Allele Genotype vs Phenotype Allele Frequency Homozygous vs heterozygous Dominant vs Recessive Adaptive Melanism in rockpocket mice lnhabitis lightcoloured sandy substrate In most of its range Mice generally have light hair D dominant d recessive zygotes information that is passed on from parents to offspring forms a new zygote Gametes zygote juveniles l adults gene pool gametes and so on Chi square calculation Xquot2 Sum of observedexpectedquot2expected Hardy Weinberg predicts probability 0 Alleles if so with what frequency Punnett Square Dominant always a capital Letter Recessive always a lower cased letter ie AA or Aa or Aa Frequency is a percentage Chisquare statistic Calculate degrees of freedom number of classes minus of independent variables Compare the calculated statistic to the appropriate entry on the table using the degrees of freedom and the P value Natural selection is not just about survival 0 Individuals with genes that increase the number of successful offspring will pass on more copies of those genes to the next generation 0 Individuals may have equal survival probabilities but differ in their chances of producing offspring quotSuccessquot is often very different for males an females in a species What does the graph show 0 quotNatural selection that arises from differences in mating successquot Arnold 1994 0 Originally hypothesized by Darwin 1871 Why should sexual selection exist 0 Females of most species invest a large amount of energy in a few large gametes o Males typically invest energy in many small gametes Expected behaviors Females should resist mating unless presented with the quotrightquot stimulus o Males will tend to compete for mating opportunities Male competition 0 Traits that increase male success in gaining matings will be selected for 0 Successful males will tend to pass on more copied of the genes for the successful trait than unsuccessful males Female choice 0 Females control male access to mating Choice between males based on differences in a particular traits Females randomly prefer one form of a trait over others eg brighter coloration Trait will increase in frequency simply because females prefer it 0 May not be any other advantage Hardy Weinberg Practice Problems 0 P q 1 relative frequency proportion of 2 different alleles in a population Pquot2 2pq qquot2 1 relative frequency proportion of given genotypes in a population P frequency of the dominant allele in the population Q frequency of the recessive allele in the population Pquot2 of homozygous dominant individuals Qquot2 of homozygous recessive individuals 2pq of heterozygous individuals Examples 1 You have a sampled a population of 10599 individuals that you know is in hardy Weinberg equilibrium HWE The percentage of the homozygous recessive genotype aa is 36 Using that 36 calculate the following o A the frequency of the quotaaquot genotype 0 B the frequency of the quotaquot allele 0 a 6 which is qquot2 sqrd rt of qquot2sqrd rt or 36 o C the frequency of the A allele o a 6 p 1 o 16 D The frequencies of the genotypes quotAAquot and quotAaquot o For AA 44 16 o 4 24q qquot21 o quotAaquot 48 o E The frequencies of the two possible phenotypes if A is completely dominant over quotaquot o Pquot2 2p36 36quot2 o Pq1 o 136 64 o A 64 and a 36 o F How many of the individuals in this population show the dominant phenotype o 6410599 678336 2 Sickle cell anemia is an interesting genetic disease Normal homozygous individuals SS have normal blood cells that are easily infected with the malarial parasite Thus many of these individuals become very ill from the parasite and may die Individuals homozygous for the sickle cell trait 55 have red blood cells that readily coapse when deoxygenated Although malaria cannot grow in these red blood cells individuals often die because of this genetic defect causes deformation of red blood cells However individuals with the heterozygous condition 55 have some sickling of red blood cells it generally not enough to cause mortality In addition malaria cannot survive well within these quotpartially defectivequot red blood cells Thus Heterozygotes tend to survive better than either of the dominant homozygous conditions Assuming HWE if 9 of an African population is born with a severe form of sicklecell anemia 55 what percentage of the population will be more resistant to malaria because they are heterozygous 5 for the sicklecell gene HWE if 9 is 55 o What is the of SS 0 00000 Pquot2 2pq qquot2 1 qquot209 sqrd rt of 09 3 s q 1 3 7 7 S p 2 7 342 42 Ss 3 Within a population of butter ies the color brown B is dominant over the color white b and 40 of all butter ies are white Given this information and HWE calculate the following o A the percentage of butter ies in the population that are heterozygous 0 000000 HWE B brown b white40 or 4 q value Pquot2 2pq qquot21 Pquot2 2p4 4quot2 To nd q vaue Sqrd rt of 4 63 Q 63 16327 p Heterozygous 46 or 46 o B the frequency of the homozygous dominant individuals 0 Dominant is 137 or 137 The following genotypes were observed in a population Genotypes of individuals HH 40 Hh 45 Hh 50 o A Calculate the observed genotypic and allele frequencies for this population 0 HH 401353 o Hh 45135 33 o Hh 50135 37 0 Total of dominant allele 40 x 2 45 80 45 12527046 0 Total number of recessive alleles 50 x 2 45 10045 145270 54 o B calculate the number of individuals of each genotypic expected if this population is in HardyWeinberg equilibrium 0 H46 h54 o HH 2857 Hh 6706 hh 3937 0 C using a chisquare test determine whether the population is in HardWeinberg equilibrium A table of critical values for the chisquare distribution at p 05 is directly below Degrees of freedom Value 1 3841 2 5991 3 7815 4 9488 Category Observed Expected OE OEquot2 OEquot2 E count count 40 2857 1143 13056 457 HH 45 6707 2207 48708 726 Hh 50 3937 1063 113 287 hh Column Totals 135 13501 147 NA NA Easier explanation of Hardy Weinberg Equation Think of p as the number of successes within the problem The number of successes is how successful the Dominant trait is expressed within the genotypes and phenotypes So therefore p Dominance Think of q as the number of failures within the problem The number of failures is how unsuccessful the Dominant trait is expressed within the genotypes and phenotypes The failures of the dominant traits are also known as the recessive traits So therefore q Recessive The Relative Frequencies of Successes p of failures q1 therefore pq1 pquot2 2pq qquot2 1 The of successes sqrd pquot2 2 times the of the heterozygous traits or successes and failures2pq the of failures sqrd qquot2 1 When asked about the degrees of freedom the formula would be 3 x genotypes 2 x successes 1 If the degree of freedom number comes out to be more or less than the Chi square number than it would be considered signi cantly different


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