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Gordon B. Hinckley Standing for Something pg.92

hyacinth-everett
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Gordon B. Hinckley Standing for Something pg.92

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  1. Contemplate the wonders of the age in which we live, this greatest of all ages in the history of humanity. More inventions and scientific discoveries have been made during my lifetime than in all the previous centuries of human history combined. This is the remarkable function of the efforts of thinking men and women who have applied their inquisitive and dedicated thought processes in the fields of medicine, industrial safety, hygiene and sanitary measures, chemistry, and research in genetics, microbiology, the environment, and other disciplines, all involving the processes of the human mind. How can we help but be grateful for such miracles? Gordon B. Hinckley Standing for Something pg.92

  2. Review of Hardy Weinberg Here is what you need to know already.

  3. When working with one polymorphic gene.... • There are only two alleles, A and a, possible for a gene locus. • Let the original A allelefrequency be represented by pand the original aallelefrequency be represented by q • Possible genotypes are: AA; Aa; aa • AA = p2Aa= 2pqaa= q2 • In anypopulation • the frequencies of p + q =1.0 • Genotype frequencies can be represented as p2+2pq+q2=1.0 • Allele frequencies can be calculated from genotype frequencies as follow: • A = p = p2+ ½ (2pq) (see page 178-79) • a = q = q2+ ½ (2pq) • Only in Hardy Weinberg equilibrium we can also calculate genotype frequencies from allele frequencies • Genotypic frequencies = AAor p2= (p x p); Aaor 2pq= (2 x p x q); and aaor q2 = (q x q)

  4. A gene pool • The sum total of all alleles for all genes which exist in a population. • This means that genotype numbers can be used to calculate allele frequencies. • Example:ina population of 100 people (with 200 alleles):36 are AA; 48are Aa;16 are aa. • How many A alleles are in this population’s gene pool? _____ • How many a alleles?_____ (36*2)+48)= 120 Aallelesso A frequency =120A/200 total alleles= .6 (16*2) +48 = 80 a alleles so a frequency =80a/200 total alleles= .4

  5. Hardy Weinberg Conclusions • The allele frequencies in a population will not change from generation to generation. • The genotype frequencies can be calculated from the allele frequencies as follows: AA= p x p; Aa= 2pq ; aa=q x q. Alternatively if AAcannot be calculated by p2 (pxp), Aacannot be calculated by 2pq and aacannot be calculated by q2 (qxq) then the population is not in equilibrium

  6. What is necessary for a population to be in Hardy-Weinberg Equilibrium? • There can be no situations in which the current distribution of alleles in the population will change over time. • This would be violated under any of 5 assumptions …..

  7. There are 5 assumptions which must be met in order to have a population in equilibrium • There is no selection. In other words there is no advantage for the survival of one genotype over another. • There is no mutation. None of the alleles in a population will change over time. No alleles get converted into other forms already existing and no new alleles are formed • There is no migration (gene flow) New individuals do not bring new alleles into nor take existing alleles out of a population.

  8. Exceptions to Hardy Weinberg cont. • There are no chance events(genetic drift) Populations are sufficiently large to ensure that chance alone does not dictate which alleles are passed on and which are lost. • There is no sexual selection or mate choice The determination of who mates with whom is totally random.

  9. Checking for Hardy Weinberg Equilibrium • Working with the person sitting beside you, read and work out problem #6 on page 219 on your text.

  10. How to attack this problem • First figureactualgenotypefrequenciesfrom the actual numbers given in the problem. • Then count the actualA alleles in the population and the actuala alleles in the population to get the allele frequencies. • Then calculate the genotypefrequenciespredictedby the Hardy Weinberg equilibrium equations and compare to the actual numbers.

  11. Directly from the numbers given The genotypefrequencies are: SS =1194/1778 = .67 Ss=526/1778 = .30 ss= 58/1778 = .03 There are two ways that you can calculate the allele frequencies from this data: • Directly from the numbers given the allele frequencies of p (S) and q (s) are: 2914/ 3556 S alleles = 0.82S frequency (2) We can use genotype frequencies to get allele frequencies such that p = p2 + 2pq : so S= p = .67+1/2(.30) = 0.82 then also for q………….. (1) 642 / 3556 s alleles = 0.18 s frequency OR (2) .03 + ½(.30) =0.18 OR (3)q= 1-p = 1-0.82 = 0.18

  12. If the population were in Hardy-Weinberg equilibrium we can apply conclusion #2 • SS genotypefrequency would be predicted by p2 = (.82)2= 0.67 and • Ss frequency would be 2pq or 2 (.82)(.18) = 0.30; and • ss frequency would be q2 = (.18)2= 0.03. • These numbers almost exactly match the measured genotype frequencies - so this population appear to be in Hardy-Weinberg equilibrium.

  13. What else must we also know to be sure? • We would need to check in future generations to make sure that the allele frequencies are not changing. • So in our calculations we confirmed conclusion #2 (that we can predict genotype frequencies from allele frequencies) • but have not yet verified conclusion#1 (that allele frequencies will not change from generation to generation.

  14. Using Chi Square to determine H-W • Use numerical values NOT frequencies or percentages • df = For Mendelian Dominant recessive = # phenotypes - 1 • df for incomplete dominance, codominnace etc. # phenotypes - #alleles • X2 =  (O-E)2 E • Null hypothesis stated as The difference between observed and expected values is simply due to chance. • We want a .05 level for acceptance of Null hypothesis. This means There is a 95% chance that a LARGER Chi Square value indicates that the difference is REAL and we REJECT our null hypothesis

  15. One more problem adding Chi square test The ability of certain people to taste the chemical phenylthiocarbamide (PTC) is governed by the dominant allele T, and the inability to taste PTC by its recessive allele t. The frequency of the T allele in a human population = 0.8, and a sample of 200 yields 90% tasters (T_) of a chemical called PTC and 10% nontasters (tt). • Does the sample conform to the equilibrium expectations? • What is the chi-square value? • How many degrees of freedom exist? • What is the probability that the observed deviation is due chance?

  16. End of H-W review

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