Inheritance - Dihybrid Crosses

Inheritance Dihybrid Crosses Solve the following questions (HINT: use Punnett Squares) 1. Some dogs bark when trailing, others are silent. The barking trait (B) is dominant over the silent trait (b). Erect ears (E) are dominant over drooping ears (e). An erect-eared barker is crossed with a droopy-eared silent trailer. Litter after litter produces pups with erect ears, but some are barkers and some are silent. What are the probable genotypes of the parents? EEBb x eebb ………………………………………………………………………………………………………………………………………………… (All offspring have erect ears, so trait is probably homozygous in erect-eared parent) ………………………………………………………………………………………………………………………………………………… 2. In Cocker Spaniels, black coat (B) is dominant over red coat (b), and solid colour (S) is dominant over spotted colour (s). A solid black male is mated with a solid red female. They produce a litter of six pups: 2 solid black pups; 2 solid red pups; 1 black spotted pup; 1 red spotted pup. Determine the genotype of the parents. BbSs x Bbss ………………………………………………………………………………………………………………………………………………… (All four phenotype combinations are present, so dominant traits must be heterozygous) ………………………………………………………………………………………………………………………………………………… 3. In fruit flies, long wings (L) are dominant over vestigial wings (l). If a vestigial winged fly is crossed with a homozygous long winged fly, what genotype and phenotype possibilities are expected in the F1 generation? In the F2 generation? F1: L l = 100% long winged ………………………………………………………………………………………………………………………………………………… F2: 1 L L : 2 L l : 1 l l = 75% long winged ; 25% vestigial winged ………………………………………………………………………………………………………………………………………………… 4. In hogs, a white belt around the middle (M) is dominant over being beltless (m), and syndactyly, or fused hooves (F), are dominant over normal split hooves (f). A uniformly coloured hog that is homozygous for fused hooves is mated with a homozygous belted hog with split hooves. Determine the genotypes of the parents and the possible genotypes and phenotypes of offspring. Parents: m m F F x M M f f ………………………………………………………………………………………………………………………………………………… Offspring: m M F f = 100% belted with fused hooves ………………………………………………………………………………………………………………………………………………… 5. In rabbits, short hair (K) is dominant over long hair (k), and brown hair (B) is dominant over black hair (b). A short black-haired rabbit was crossed with a long brown-haired rabbit. In four litters, 38 rabbits were produced, all with short brown hair. Determine the genotypes of the parents and the 38 offspring. Parents: K K b b x k k B B Offspring: K k B b ………………………………………………………………………………………………………………………………………………… (Because all offspring show the dominant traits, parents were likely homozygous for these) ………………………………………………………………………………………………………………………………………………… Gene Linkage Define linkage group A linkage group is a group of genes whose loci are on the same chromosome ………………………………………………………………………………………………………………………………………………….......... Linked genes function as a single inheritable unit and do not follow independent assortment ………………………………………………………………………………………………………………………………………………….......... Identify the factor that determines the likelihood of recombination between two linked genes Distance: The further apart two genes are on a chromosome, the more likely they are to recombine) ………………………………………………………………………………………………………………………………………………….......... ………………………………………………………………………………………………………………………………………………….......... Outline how Morgan discovered non-Mendelian ratios via experimentation with Drosophila He undertook breeding experiments with Drosophila (fruit flies) and found a clear sex bias in the inheritance ………………………………………………………………………………………………………………………………………………….......... of certain traits ………………………………………………………………………………………………………………………………………………….......... Morgan determined that genes located on a shared chromosome would NOT assort independently ………………………………………………………………………………………………………………………………………………….......... (+ have non-Mendelian ratios) ………………………………………………………………………………………………………………………………………………….......... Identify the recombinants in the following cross: !" !" × !" !" Ab ab and aB ab ………………………………………………………………………………………………………………………………………………….......... Polygenic Inheritance Define polygenic inheritance Polygenic traits are characteristics that are controlled by more than two gene loci ………………………………………………………………………………………………………………………………………………….......... ………………………………………………………………………………………………………………………………………………….......... Distinguish between discrete and continuous variation Monogenic traits therefore have a finite pattern of expression (i.e. discrete variation) ………………………………………………………………………………………………………………………………………………….......... Polygenic traits normally exhibit a bell-shaped distribution (i.e. continuous variation) ………………………………………………………………………………………………………………………………………………….......... List two examples of polygenic inheritance Human height 1. …………………………………………………………………………………………………………………………………………………... Skin colour / pigmentation 2. …………………………………………………………………………………………………………………………………………………... Chi-Squared Test Complete the chi-squared test to determine whether two genes are linked Two heterozygous long green plants (LlGg) are crossed, yielding the following F1 frequencies: 296 = long green 19 = long yellow 27 = short green 85 = short yellow 1. Identify Hypotheses There is no difference between observed and expected frequencies (i.e. genes are unlinked) Null Hypothesis: ……………………………………………………………………………………………………………………….......... There is a difference between observed and expected frequencies (genes are linked) Alternative Hypothesis: …………………………………………………………………………………………………………….......... 2. Calculate Frequencies Dihybrid Ratios: Frequencies: 9:3:3:1 LG Lg lG lg LG LLGG LLGg LlGG LlGg Lg LLGg LLgg LlGg Llgg lG LlGG LlGg llGG lg LlGg Llgg llGg Expected Phenotype Observed Long green 296 427 X 9/16 = 240 Long yellow 19 427 X 3/16 = 80 llGg Short green 27 427 X 3/16 = 80 llgg Short yellow 85 427 X 1/16 = 27 (Total × Ratio) 3. Calculate Chi-Squared Value !−! ! ! Long green Long yellow 13.067 46.513 Short green Short yellow 35.113 124.593 219.29 ! ! : …………………...……………………………………………………………………………………………………………………….......... 4. Determine Statistical Significance Probability of Exceeding Critical Value Degree of Freedom 0.90 0.75 0.50 0.25 0.10 0.05 0.01 1 0.016 0.102 0.455 1.32 2.71 3.84 6.63 2 3 0.211 0.584 0.575 1.212 1.386 2.366 2.77 4.11 4.61 6.25 5.99 7.81 9.21 11.34 There IS a significant difference between observed and expected data (genes may be LINKED) Conclusion: ………………………………………………………………………………………………………………………………..........