ANSCI 361 1st EditionExam # 2 Study Guide Lectures: 10 - 19Lecture 1 (February 18): Allele Frequencies- 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 80 040 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. SelectionLecture 2 (February 20): Simply Inherited, Polygenic and Threshold Traits- Simply-Inherited Traitso All simply-inherited traits are influenced by a few alleles at one or a few loci o Most simply-inherited traits are: Defined by discrete categories (qualitative) Not affected by the environment- Phenotype = Genotype Examples:- Polled or horned in cattle and sheep- Black or red coat color in cattle and pigs- Rose comb or single comb in chickens- Long ears, medium length ears, or no ears in goats- Black, chocolate, or yellow in Labrador dogs- Polygenic Traitso All polygenic traits are influenced by a few to several alleles at many loci, and most of the alleles have a small effect on the trait o Most polygenic traits: Are continuous in their measurement (quantitative) – the trait can have an infinite number of values Tend to be normally distributed within a large population – they follow a bell-shaped curve Are affected by the environment- Phenotype = Genotype + Environment (Environment usually has greater effect on phenotype than genotype) Examples:- Milk production in dairy cattle, goat, and sheep- Average daily gain in growing beef cattle, pigs, and sheep- Speed in race horses and dogs- Egg production in poultry- Body weight in all animals- From Single Loci to Quantitative Traitso One locus, two alleles: Freq. A = p = 0.5; Freq. a = q = 0.5- A = 1 ounce of increased wool production- a = 0 ounces of increased wool production- Freq. AA = p2 = P = 0.25- Freq. Aa = 2pq = H = 0.50- Freq. aa = q2 = Q = 0.25 Co-dominance: - AA = 2, Aa = 1, aa = 0 ounces of increased woolo Two loci, each with 2 alleles Freq. A = p1 = 0.5; Freq. a = q1 = 0.5 Freq. B = p2 = 0.5; Freq. b = q2 = 0.5- A & B = 1 ounce of increased wool production- a & b = 0 ounces of increased wool productiono As the number of loci influencing the trait increases, the distribution of genotypes and phenotypes becomes continuous and often tends to fit a normal distribution.- Threshold Traitso A special type of polygenic trait Influenced by a few to several alleles at several loci Most of the alleles have a small effect on the trait Affected by the environment- Phenotype = Genotype + Environment Defined by discrete categorieso Examples: Dystocia (difficult birth): - No, some, moderate, or great assistance necessary Disorders (e.g. hip dysplasia in dogs): - Affected or not affected Conception rate: fertility or no fertilitySimply-Inherited PolygenicQualitative Coat color,Polled/HornedCalving Ease,Conception,Litter SizeQuantitative Fat Thickness,Milk Yield*Calving ease = Threshold traito Examples: Dwarfism (body size) Double muscling (muscle mass) - Mutations on Myostatin gene- Breeding Approaches for Simply-Inherited and Polygenic Traitso Notice: Both simply-inherited and polygenic traits are subject to the same Mendelian ruleso F However: With simply-inherited traits: identification of effects of individual genes and gene combinations; test mating to uncover unknown genotypes With polygenic traits: statistical models to predict genetic value of animals (estimated breeding value, expected progeny difference, transmitting ability, etc.)Lecture 3 (February 23): Genetic Model for Polygenic (generally quantitative) TraitsGenetic Model for Polygenic (generally quantitative) Traits- Polygenic Traitso 1. Influenced by alleles at many loci; most alleles have a small effecto 2. Environmental influence is large o 3. Presence or absence of individual alleles generally cannot be determined from the phenotypeBasic Genetic Model-EGP - Where: o P = Phenotypic value