Genetics Dideoxy Sequencing Notes Part 3

7 cm. 1 8 cm. 10 9 cm. 45 10 cm. 120 11 cm. 210 12 cm. 252 13 cm 210 14 cm. 120 15 cm. 45 16 cm. 10 17 cm. 1 Calculate the total number of genes involved in fruit diameter in tomatoes. Answer: -(n) number of genes: 5 Explanation: -The ratio of F2 individuals showing extreme phenotypes is equal to (1/4)^� -n=(ln(1)-ln(1024)) / (ln(1)-ln(4)) Assume that the length of wheat leaves is controlled by three loci, each with two alleles: L and l, W and w, H and h. Determine the differences in leaf length between two homozygous strains of wheat. Assume each allele contributes equally to the length of the plant leaves. One homozygous strain, ll ww hh, has leaves that are 100 mm in length, and the other strain, LL WW HH, has 220 mm leaves. The two homozygous strains are crossed, and the resulting F1 are selfed to produce F2 progeny. What length will the leaves of the ll ww HH genotype plant be? What proportion of the F2 progeny will have the same phenotype as the ll ww HH genotype? Input answer as a decimal. Answer: -140 -0.23 Explanation: -Calculate the length that each dominant allele adds to the leaf length phenotype by determining the difference in length between the homozygous genotypes. Divide the difference by the total number of dominant alleles in the homozygous dominant genotype. -To determine the length of the leaves for the llwwHH genotype, first, determine the total number of dominant alleles. Multiply the number of dominant alleles by the length that each dominant allele adds to the baseline leaf length. Add the result to the baseline leaf length/ -Length: 20mm x 2 + 100mm = 140mm -Proportion: (there are a total of 15 total genotypes with two dominant alleles) 15/64 = 0.23 Suppose there are five genes in pumpkins that work additively to affect pumpkin size. A homozygous lowercase allele pumpkin, aabbccddee, weighs 20 lb. Each capitalized allele additively affects the final phenotype by either adding or subtracting weight from the base pumpkin size. The effect of a single capitalized allele is listed in the table. Allele. Effect A. +5 B. +3 C. −2 D. −4 E. +2 What would the potential size range be for pumpkin offspring produced from the cross between a parent with genotype AaBBCcddEE and a parent with genotype AABbccDDEe? Largest pumpkin? Smallest pumpkin? Answer: -Largest: 36 lb -Smallest: 24 lb Explanation: -Use the information about additive effects to identify the genotypes for each gene that will produce the largest and the smallest offspring. Considering that the question is asking about a range, you only need to focus on the two extreme genotypes and their resulting phenotypes. -The genotype for the largest offspring will be AABBccDdEE. Sum the net change in weight for the AABBccDdEE genotype with the base pumpkin weight. -Large: 20+10+6+0+-4+4=36 -The genotype for the smallest offspring will be aABbCcDdEe. Sum the net change in weight for the aABbCcDdEe genotype with the base pumpkin weight. -Small: 20+5+3+-2+-4+2=24 The kernel color in wheat is a continuous trait determined by two additive genes, each with two alleles, that equally contribute to kernel color determination. The red kernels are determined by two genes and two dominant alleles (R1R1R2R2), and white kernels are determined by two recessive alleles at the same two genes (r1r1r2r2). The duplicate dominant alleles R1 and R2 contribute equally to kernel color and cumulatively control the intensity of the red kernel phenotype. Both these alleles are dominant over the recessive white alleles r1 and r2. A true breeding red plant and true breeding white plant are crossed, and the resulting F1 progeny are selfed. What is the expected phenotypic ratio of the kernel progeny of the F2 offspring? Determine which statement best describes the relationship between the phenotypes of the r1r1R2r2 and r1R1r2r2 genotypes. Answer: -1:4:6:4:1 (1 red: 4 dark pink:6 pink: 4 light pink: 1 white) -The phenotypes of the r1r1R2r2 and r1R1r2r2 genotypes will be the same because the same number of dominant and recessive alleles are contributing in each genotype Explanation: -The color intensity of a kernel depends on the number of dominant alleles in the genotype. Suppose that three loci, each with two alleles (A and a, B and b, C and c), determine the differences in height between two homozygous strains of a plant. These genes are additive and equal in their effects on plant height. One strain (aa bb cc) is 10 cm in height. The other strain (AA BB CC) is 22 cm in height. The two strains are crossed, and the resulting F1 are interbred to produce F2 progeny. Match the F2 phenotypes and expected proportions of progeny. Answer: -Phenotype (Height). Proportion of F2 10 cm. 1/64 12 cm. 6/64 14 cm. 15/64 16 cm. 20/64 18 cm. 15/64 20 cm. 6/64 22 cm. 1/64 Explanation: -Consider using a Punnett square to represent the F2 progeny. Which of the statements describe an aspect of a distribution? Answer: -the variance of the number of vertebrae in the eelpout fish -the mean weight of adult grizzly bears in Canada Explanation: -Keep in mind that distributions describe quantifiable traits in populations. For each statement, consider how you would define the population. Some traits are considered to be quantitative even though phenotypes are classified only as present or absent. Genetic diseases often fit this description. What is the term used to describe this type of trait? What creates the underlying quantitative distribution for this type of trait? Answer: -Threshold trait -Genetic and environmental factors determine an individual's susceptibility to the trait. Susceptibility exhibits continuous variation, and individuals above a certain threshold express the trait. Explanation: -Consider that both genes and the environment influence the magnitude of a quantitative trait. One example of this is the disorder phenylketonuria (PKU). Brain damage results in individuals with high levels of blood phenylalanine, which result from a combination of genetic effects and diet. Consider what kind of mechanisms trigger brain damage in a disease such as PKU. The domestic cat, Felis catus, is a domesticated furry creature often kept as a pet in many countries. Suppose a volunteer at a local animal shelter wanted to find out a little more about the cats in her care. The volunteer carefully measured the height for the 10 fully grown adult cats at her shelter. The height measurements were taken from the base of the foreleg to the shoulder. The table shows the results of the measurements. Sample Height (in) 19 2 12.8 3 7.4 4. 8 5. 8.6 6. 12.3 7 9.2 8 11.1 9. 7.8 10 6. Calculate the mean, variance, and standard deviations for cat height for the sample of shelter cats. Report all your answers to two decimal points. Answer: -Mean: 9.27 in -Variance: 4.50 in^2 Explanation: -Variance is the discrepancy between the observed value and the mean, and the standard deviation measures how spread out the numbers are from the mean. -Mean: (calculate like normal) -Variance: 1) begin by subtracting the mean value from each cat's height measurement. 9-9.27= -.027 12.8-9.27= 3.53 7.4-9.27= -1.87 ..... 2) Next square each difference and divide by (n-1) ((-.027)^2 + (3.53)^2 + (-1.87)^2 + ....) / (n-1) = 4.50 The graph illustrates frequency distributions for a trait influenced by a single quantitative trait locus (QTL). The distribution drawn in orange represents the entire population. (a graph similar to this one) Identify the true statements regarding the graph. Answer: -The total population and the population of heterozygotes have the same mean value of the trait. -The distributions that represent the two homozygous populations have the same variance. -The distribution that represents the BB homozygous individuals has a greater mean value of the trait than the total population. Explanation: -Carefully consider the characteristics of a normal distribution that are represented by the height and width of each curve, as well as the relative measurement of the trait value over which it is centered. Use the Artificial Selection Interactive to increase the average flowering head size of the sunflowers and answer the questions. Why does the variance decrease as you breed towards the selected target? Which cross would you perform in the sunflowers if your goal was to increase the variance of head size? Answer: -because the distribution of the phenotype decreases -cross sunflowers with the most phenotypic variance Explanation: -Consider that variance is a deviation from the mean, which indicates how dispersed the measurements are from the mean. -The variance does not change when the distribution remains the same. The distribution shows the range of measurements in a population. If the distribution changes, the variance also changes. Both genetic and environmental factors contribute to variation in phenotypes of individuals within a population. Match the appropriate definition with each of the terms that relate to phenotypic variance. Answer: -Genetic interaction variance: variance due to the interplay among alleles at multiple loci -Phenotypic variance: Sum of all genetic and nongenetic factors that affect expression of a trait within a population -Additive genetic variance: variance resulting from the mean effects of single alleles -genetic-environmental interaction variance: variance due to differential effects of nongenetic factors on specific genotypes -dominance genetic variance: variance due to the interaction between two alleles at a single locus -environmental variance: variance in response to external, nongenetic factors -genetic variance: variance in the alleles among individuals within a population Explanation: -Phenotypic variance is the cumulative effect of additive genetic variance, dominance genetic variance, genic interaction variance, environmental variance, and genetic-environmental interaction variance. The graph indicates the stem lengths observed in a plant population. (HW #16 chapter 17) What is the mean stem length in the plant population indicated in the graph? Answer: -20 cm (it is in the center of the normal distribution) Explanation: -The mean is the average value of a quantifiable trait. Classify each statement about the heritability of a phenotype according to the narrow-sense heritability value it describes. Answer: -0.0 -the phenotypic variance is determined by environmental variance -The phenotype of offspring is not related to the phenotypes of parents -0.5 -half of the observed variation is due to the additive genetic effects of alleles -1.0 -a species of birds passes down the long-beaked trait to their offspring in the absence of environmental factors -This heritability value exists in the absence of environmental changes Explanation: -Narrow-sense heritability takes into account both genetic variance and environmental variance. Suppose that phenotypic variation in the tail length of mice is summarized in the table. Additive genetic variance (��)(VA)= 0.4 Dominance genetic variance (��)(VD)= 0.5 Genetic interaction variance (��)(VI)= 0.4 Environmental variance (��)(VE)= 0.2 Genetic-environmental interaction variance (���)(VGE)= 0.0 What is the narrow-sense heritability (ℎ2) of tail length? What is the broad-sense heritability (�2) of tail length? Answer: -Narrow-sense: 0.3 -Broad-sense: 0.87 Explanation: -Narrow-sense heritability includes only additive genetic effects. However, broad-sense heritability includes all sources of genetic variation, such as dominance and interaction effects. -Narrow: h2= VA / (VA+VD+VI+VE+VGE) h2= 0.4 / (0.4+0.5+0.4+0.2+0) -Broad: A similar strategy is used to calculate broad-sense heritability, but instead the variable VG is used in place of VA. VG is the total variance due to all genetic effects, so it includes VA, VD, VI, and VGE. H2=VG / VP H2= (VA+VD+VI+VGE) / (VA+VD+VI+VE+VGE) H2= (0.4+0.5+0.4+0) / (0.4+0.5+0.4+0.2+0) Suppose Jennifer is a horticulturist that specializes in the cultivation of orchids. She houses the orchids in a temperature-controlled and humidity-controlled greenhouse. A salesperson approaches Jennifer and attempts to sell her a special fertilizer that promises to increase flower production. Jennifer, who dabbles in genetics, insists that the fertilizer will have no effect on flower production. She has calculated the narrow-sense heritability of flower production in her orchids and found it to be 0.98. Jennifer claims that a narrow-sense heritability of 0.98 indicates that 98% of the variance in flower production is determined by genetic differences, so a fertilizer would have little effect on flower production in the orchids. The salesperson disagrees with Jennifer and continues to assert that the fertilizer will increase flower production. Who is correct and why? Answer: -The salesperson is correct. High heritability of a trait within a population does not exclude new environmental factors from affecting the trait. Explanation: -Heritability describes the contribution of genetic differences on the total phenotypic variation in a population. Broad-sense heritability reflects the proportion of phenotypic variance due to any genetic variance, whereas narrow-sense heritability reflects the proportion of phenotypic variance due to additive genetic variance. Suppose that the narrow-sense heritability of ear length in Reno rabbits is 0.4.0.4. The phenotypic variance (��)(VP) is 0.6,0.6, and the environmental variance (��)(VE) is 0.1.0.1. Calculate the additive genetic variance (��)(VA) for ear length in these rabbits. Answer: -0.2 Explanation: -Consider that narrow-sense heritability accounts for only additive genetic variance. narrow-sense heritability: h^2=VA / VP -VA= h^2 x VP =0.4 x 0.6 =0.2 Use the artificial selection interactive to answer the questions. Cross two plants with narrow flower diameters in the artificial selection interactive and observe the results. Complete the statement about the experimental data. Next, select the correct conclusion regarding the relationship between parental and offspring flower diameters. Parents with narrow flower diameters produce offspring with ________ flowers on average than the parental generation. What kind of relationship is there between parental and offspring flower diameter? Answer: -Narrower -Direct Use the Artificial Selection Interactive to increase the average flowering head size of sunflowers. Select the statements that describe the change in distribution as the diameter of the flowering heads increases. Answer: -The phenotypic variance of each genotype decreases with each successive generation. -As the diameter of the sunflower head increases, the mean increases. Explanation: -The mean of the data is the average of the flowering head diameters. The variance is the average squared deviation of the flowering head diameters from their mean value. Suppose a strawberry farmer calculates that the average strawberry produced from his garden weighs 2 g.2 g. The farmer's 10 most productive plants grow strawberries with a mean weight of 6 g.6 g. The farmer interbreeds these 10 plants, and the progeny produce strawberries with an average weight of 5 g.5 g. If the farmer interbreeds plants that produce strawberries with an average weight of 4 g,4 g, how much would strawberries produced by the progeny of these selected plants weigh? Answer: -3.5 Explanation: -Use the equation to estimate the response to selection, �,R, in the selected plants. �=ℎ2×� Estimate narrow-sense heritability, ℎ2,h2, using the example trait data from the 10 most productive plants. �S represents the selection differential, or the difference in trait values between the individuals selected as parents and the starting population. R= 5g - 3g = 3g S= 6g - 2g = 4g h^2=R/S =3/4 .75 S = 4g - 2g = 2g �=ℎ2×� =.75 x 2 =1.5 Progeny weight = mean wieght of the orignial population + R Progeny weight = 2g + 1.5g = 3.5g Bipolar disorder is a psychiatric disorder with a strong hereditary basis, but the exact mode of inheritance is not known. Research has shown that siblings of patients with bipolar illness are more likely to develop the disorder than are siblings of unaffected persons. Findings from one study demonstrated that the ratio of bipolar brothers to bipolar sisters is higher when the patient is male than when the patient is female. In other words, relatively more brothers of bipolar patients also have the disease when the patient is male than when the patient is female. Determine what the evidence suggests about the inheritance of bipolar disorder. Answer: -An X-linked locus may contribute to this disorder. Explanation: -Consider which mode of inheritance most often results in a skewed gender ratio of affected progeny. In humans, red blood cells have a number of proteins embedded in the cell membrane. One type of protein, the Rh factor, is controlled by a single gene and is either present or missing from the red blood cells. If present, the individual has the Rh+ phenotype. If missing, the individual has the Rh− phenotype. Rh+ is the dominant to Rh−. Suppose that, in the Welsh population, the frequency of the Rh− phenotype is 0.04 Using the Hardy-Weinberg equations, calculate the frequency of the Rh+ allele to at least two decimal places. Answer: -0.8 Explanation: -For recessive phenotypes, the phenotype frequency is equal to the recessive homozygote genotype frequency. p^2 + 2pq + q^2 = 1 p+q=1 √0.04 = 0.2 1-p=q 1-0.2=0.8 The color distribution for a specific population of geckos is 150 red, 60 orange, and 40 yellow. The allele for the color red is represented by RR, whereas the allele for the color yellow is represented by RY. Both alleles demonstrate incomplete dominance. What are the genotype frequencies in the population? Calculate to at least two decimal places. What is the allele frequency of RR in this population? Calculate to at least two decimal places. Answer: -RRRR: 0.60 -RRRY: 0.24 -RYRY: 0.16 -allele frequency of RR: 0.72 Explanation: -The genotype frequency is the proportion of total individuals in a population that have the same genotype. The allele frequency is the proportion of individuals in a population that have the same allele. (AA)= AA / N Suppose the human trait for dimples is controlled by a simple dominant and recessive relationship at one locus. Dimples, D, is the dominant allele, and a lack of dimples, d, is the recessive allele. In a college genetics class, the professor takes a tally of students who have dimples and of students with a lack of dimples. In this class of 117117 students, 3737 have dimples. Calculate the frequency of the dominant allele, D, and the heterozygous genotype, Dd. Express the frequencies in decimal form rounded to the nearest thousandth. Assume the class is in Hardy-Weinberg equilibrium for this trait. Answer: -frequency of D: 0.173 -frequency of Dd: 0.286 Explanation: -Use the population data and the Hardy-Weinberg equations, �+�=1p+q=1 and �2+2��+�2=1 to calculate the genotype and allele frequencies. -Students with recessive genotype = 117-37 = 80 Frequency of recessive genotype =80 / 117= 0.684 Frequency of recessive allele = √0.684 = 0.827 Frequency of dominant allele= 1-0.827= 0.173 frequency of heterozygotes= 2 x 0.173 x 0.827= 0.286 Suppose the given data was collected on five populations of sheep. Breed DD. Dd. dd Adal 0.36~ 0.48 ~0.16 Chios 0.09~ 0.42~ 0.49 Dala 0.16~ 0.53~ 0.31 Columbia 0.23~ 0.63 ~0.14 Dorset 0.46~ 0.50~ 0.04 Which of the populations are in Hardy-Weinberg equilibrium?Columbia Answer: -Chios -Adal Explanation: -The equation for Hardy-Weinberg equilibrium is �2+2��+�2=1,p2+2pq+q2=1, where �p is the allele frequency of the dominant trait and �q is the allele frequency of the recessive trait. (HW #5 chapter 28) Simulations were run on a population with an effective population (Ne) size of 1000 and15 , respectively, to simulate the change in allele frequencies over time. Ten runs were simulated across 100 generations for each population. The beginning frequencies for each allele were set at 0.50. The graphs show only the wildtype allele. (graphs similar to this one) 1000 Ne population 15 Ne population Compare the graphs and determine which evolutionary force is occurring in the population of 15 Ne individuals and not in the population of 1000 Ne indviduals. Answer: -genetic drift Explanation: -Compare the allele frequencies in each of the graphs to determine how the wildtype allele may fluctuate over time. Each line represents the theoretical change in allele frequency in the population. W genotype = (number of offspring from the genotype) / (number of offspring from the most prolific genotype) ... According to the biological species concept, what is the definition of a species? organisms that can interbreed in nature to produce healthy offspring How is anagenesis different from cladogneesis? Answer: -Anagenesis -occurs within a single lineage -results in evolution of a species -cladogenesis -occurs when a lineage diverges -results in speciation Explanation: -when a lineage diverges, it forms a clade of closely related species Classify the examples as prezygotic or postzygotic barriers. Answer: -prezygotic -one fly species mating in the morning, whereas another fly species mates in the afternoon -one bird prefers open areas, whereas another bird species prefers wooded areas -postzygotic -individuals from two different species mate, but the resulting embryo dies before birth -A horse and donkey mate to produce a sterile mule Explanation: -Prezygotic barriers prevent a zygote from being created, whereas postzygotic barriers prevent two species from generating viable offspring or generating offspring that are able to reproduce. The Northern spotted owl and Mexican spotted owl arose from geographically isolated populations of the same ancestral owl species. The geographic ranges of the Northern spotted owl and the Mexican spotted owl are shown in the image. The evolution of the Northern spotted owl and Mexican spotted owl is an example of which type of speciation? Answer: -allopatric speciation Explanation: -The ancestral owl populations that gave rise to the Northern spotted owl and Mexican spotted owl species were separated geographically into the Northern and Southern populations. How does the concept of molecular clocks contribute to addressing the question of genetic identity between different species? helps estimate time since species had a common ancestor In plants, an increase in ploidy is one common mechanism of evolution. How does an increase in ploidy contribute to evolution? Which statements address why polyploidy is more common in plants than in animals? Answer: -One copy of a gene retains its function, whereas the other gene copy develops a new function. -Self-fertilization in plants allows new polyploids to mate with themselves. -Behavioral isolation mechanisms prevent different animal species from interbreeding.