5. The Flemish breed of rabbits have an average weight of 3600 grams. The Himalayan breed has a mean of 1875 grams. Matings between these two breeds produce an intermediate F1 with a standard deviation of + 162 grams. The standard deviation of the F2 is ± 230 grams. Estimate the number of pairs of factors contributing to mature body weight of rabbits. Estimate the average contribution of each active allele.
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- 5. When two alleles are observed independently in a phenotype is called multiple alleles. Human blood group ABO is a good example of multiple alleles. The three alleles are: I^, IB, i. Mother's blood type is AB (IAIP) and father's blood type is B (1³i). Provide the genotype and phenotype ratios using the Punnett square given below. Genotypes 6. When the allele of a trait is found on X or Y chromosome, it becomes sex-linked. The results differ from males to females. One such example is of red green colorblindness. The mother is a carrier (XCX), and the father is colorblind (XY). Provide the genotype and phenotype ratios using the Punnett square given below. Genotypes Phenotypes **CLUSTROS Phenotypes 75-10246. You are studying the physiology of hypertension (high blood pressure) using the rat as a model system and wish to breed a strain of rat which has naturally high blood pressure. Systolic blood pressure is a quantitative trait with a continuous phenotypic distribution. It is measured as millimeters of mercury displaced in a blood pressure meter (mm Hg). You start with a base population of normal domestic rats, which has a mean systolic blood pressure of 120 mm Hg. You select the 10% of the rats which have the highest systolic blood pressure. The average systolic blood pressure value of these selected breeders is 140 mm Hg. You then interbreed the selected rats and measure the systolic blood pressure in their progeny. What would be the average systolic blood pressure in the progeny of the selected breeders if A. heritability = 0.5 ? B. heritability = 1.0 ? C. heritability = 0 ? D. In a second experiment, you go back to that same base population of domestic rats and select the 10% of…26. Snow geese (Chen caerulescens) come in two color types, white “snows” and “blues” with dark bodies. A single gene controls coloration, where the dark (“blue”) allele (D) is dominant. A population of 30,012 geese includes 9236 dark individuals. Genetic testing reveals that 7636 of the 9236 dark individuals are heterozygous (Dd). What is the actual frequency of allele D? 0.819 0.181 0.435 0.347
- 1. A selection experiment was carried out on a population of field mice in the lab had a mean hematocrit of 65. The mice used to create the next generation had a mean hematocrit of 50. Once the offspring from that experiment reached adulthood, the experimenters measure a mean hematocrit of 58. Answer the following questions using this information a). What is the narrow sense heritability for hematocrit in these mice? Show all work b).Using those results, what would be the predicted hematocrit in the offspring population after another generation of selection where only mice with a mean hematocrit 10 points below the parental mean were allowed to breed. Show all work c). Would you expect the narrow sense heritability for a wild population of field mice in Iqaluit to be the same ? Explain.1. In a population of 1000 bison, there are two alleles at the B locus. It acts incompletely dominantly, so that you are able to figure out each animal's genotype simply by observing its phenotype. How convenient. You find 665 BB, 225 Bb, and 110 bb bison. a) What are the allele frequencies of B and b? (round off the number) b) Using the allele frequencies, what numbers (not just fractions, but numbers of actual bison in this population of 1000 and yes you can round off) would you expect to be BB, Bb, and bb? c) Do you think this bison population is in HW equilibrium? d) No matter what your answer to c, if they weren't in HW equilibrium, name 5 possible reasons why.1. In a wild strain of tomato plants, the phenotypic variance for tomato weight is 3.2g². In another strain of highly inbred tomatoes raised under same environmental conditions, the phenotypic variance is 2.2 g². With regard to wild strain a. What is VG b. What is h₂² C. Assuming that all of the genetic variance is additive what is h²
- 5. There is a helpful shortcut for calculating the probability of genotypes resulting from a dihybrid cross (or any cross considering two or more pairs of alleles) that does not require drawing a very large Punnett square. This is done by first determining the probabilities of each trait alone (using a monohybrid cross) and then using the multiplication rule to combine the probabilities of all the genotypes being considered. Using our TtYy plants above, we would first assign each set of alleles to its own Punnett square as follows: I t Y y T TT Tt Y YY Yy t Tt tt y Yy УУ a. Based on the Punnett squares above, what is the probability of being genotype tt in this cross? b. Based on the Punnett squares above, what is the probability of being genotype YY in this cross? c. Using the probabilities that you just calculated for the genotypes tt and YY, what is the probability of one offspring being both genotype tt and YY in this cross (use the multiplication rule)? d. What is the probability…1. For a single locus with two alleles, A₁ and A₂: (a) Draw a graph (using graph paper) showing both the frequency of A₁ A2 heterozy- gotes and A₂ A₂ homozygotes, at Hardy-Weinberg frequencies, as functions of p (the frequency of A₁). Note that both p and the genotype frequencies should have values between 0 and 1. (b) Find the value of p above which A₁ A2 genotypes are more common than A₂42 genotypes. You can solve this algebraically, or estimate it from your graphs. 2. Consider three loci, A, B, and C, each with two alleles, with the frequencies of A₁, B₁, and C₁ all being We look at a population and find that there are four distinct haplotypes, shown here, each with a frequency of: A₁ B1 TT A1 C₁ B1 A₂ B₂ AT C₂ A₂ B₂ C₁ C₂ Of the three pairs of loci (AB, AC, and BC) which pair(s) are in Gametic Equilibrium (D = 0) and which are in Gametic Disequilibrium (D ‡0)? [Hint: Consider each pair separately, ignoring the other locus. For example: for the BC pair, consider the four…1. Consider this graph on running speed in huskies Midosspring running speed (m/s) 16 14 12 10 8 + 2 Husky running speed 4 6 8 Midparent running speed (m/s) 10 12 14 16 a. Approximately what is the heritability of running speed in this kennel of huskies? (You can approximate by eyeballing from the graph, no need to calculate the actual slope) b. If the breeder where to selectively breed the dogs, will the dogs run substantially faster in the next generation? c. What else can the breeder do to increase running speed?
- 3. Set up a Punnett square using the following information: Dominate allele for purple corn kernels = R Recessive allele for yellow corn kernels =r Dominate allele for starchy kernels = T Recessive allele for sweet kernals =t Cross a homozygous dominant parent with a heterozygous parent. Using the Punnett square above: a. What is the probability of producing purple, starchy corn kernels? Possible genotype(s)? b. What is the probability of producing yellow, starchy corn kernels? Possible genotype(s)? c.What is the probability of producing purple, sweet corn kernels? Possible genotype(s)? d. What is the probability of producing yellow, sweet corn kernels? Possible genotype(s)?25. List the three alleles for blood: 26. List the genotypes for blood: 27. List the phenotypes for blood: 28. A type of chicken shows codominance with black (B) being codominant with white (B'). The heterozygous condition is checkered. Cross a checkered rooster with a white hen. Show the following: Genotypes: Phenotypes: Genotypic Ratio: Phenotypic Ratio:eritance / 13 of 15 Black hair color is dominant to white hair color in mice. Interpret the Punnett squrare below to determine the expected phenotypic ratio for the offpring of a homozygous black mouse and a white mouse. В Bb Bb Bb Bb O A. 4 Black: 0 White O B. 3 Black: 1 White O C 2 Black: 2 White O D. O Black: 4 White acer