An Sci 361 1st Edition Lecture 10 Outline of Last Lecture N/A (Change of professor/Start of material for Exam 2)Outline of Current Lecture I. Populationa. Calculationsi. Counting Methodii. FormulaII. Hardy-Weinberg Equilibriuma. Calculationsb. 4 forces that change gene frequency Current Lecture- Population: All the individuals of a species that live together in one place at one time.o Can be described by age structure, geography, birth and death rates, and allele frequencieso More diverse than individuals Only a group/population of people can carry all the alleles for traits such as blood types A, B, and O- Population genetics: a branch of genetics that applies Mendelian inheritance to populations and studies the frequency of alleles and genotypes in breeding populations- Calculation of allele (gene) frequencieso p = freq. Of B alleleo q = freq. of b alleleo p + q = 1 How to determine p and q?o 1. Counting method: p = 120/200 = .60, q = 80/200 = .40Number ofIndividualsNumber of allelesB b40 BB 80 0These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.40 Bb 40 4020 bb 0 40100 individuals 120 80o 2. Formula p = freq. BB + 1/2(freq. Bb) = P + 1/2 H = .40 + 1/2 (.40) = .40 + .20 = .60 q = freq. bb + 1/2(freq. Bb) = Q + 1/2 H = .20 + 1/2 (.40) = .20 + .20 = .40- 1st generation of random mating .60 B .40 b .60 B .40 b .36 BB .24 Bb .24 Bb .16 bb Black Blue Blue White o Genetic frequency: 0.36 BB + 0.48 Bb + 0.16 bbo Gene frequency: P = freq. BB + 1/2(freq. Bb) = P + 1/2 H = .36 + 1/2 (.48) = .36 + .24 = .60 q = freq. bb + 1/2(freq. Bb) = Q + 1/2 H = .16 + 1/2 (.48) = .16 + .24 = .40- Hardy-Weinberg Equilibriumo In large, random mating populations, gene and genotypic frequencies remain constant from generation to generation and follow the relationship: (p+q)2 = p2 + 2pq + q2 Where p2 = P, 2pq = H, q2 = Q- Chicken example: o Frequency of B: p = 0.60o Frequency of b: q = 0.40- Therefore: o P = frequency BB = p2 = (0.60)2 = 0.36o H = frequency Bb = 2pq = 2(0.60)(0.40) = 0.48o Q = frequency bb = q2 = (0.40)2 = 0.16o Can not have >50% heterozygoteso When the frequency of one allele is very low that allele is mostly in Heterozygote, and very few in Homozygote, which is very important in Selection - Using Hardy-Weinbergo Pigs: WW, Ww = white belt ww = solid color (no belt)o Population of 1,000 random mating pigs with 910 belted and 90 solid coloro 1. How many of the pigs are expected to be heterozygous?3.009.0q 09.01000/90qQ2 p = 1 – q = 0.742.0)3.0)(7.0(2pq2Ww .freqH - Expected number of Ww animals: 0.42 x 1000 = 420o 2. What proportion of the belted pigs are expected to be homozygous?54.0910/490910/100049.0910/1000p2- More on Hardy-Weinbergo Hard-Weinberg equilibrium states that gene and genotypic frequencies remain constant from generation to generation ---- as long as outside forces do not change frequencies.o Forces that can change gene frequencies: 1. Mutation 2. Migration 3. Genetic Drift (Chance) 4.