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- A danger in computing heritability values from studiesinvolving genetically related individuals is the possibility that theseindividuals share more similar environments than do unrelatedindividuals. In the experiment shown in Figure 24.8, which dataare the most compelling evidence that ridge count is not causedby genetically related individuals sharing common environments?ExplainEstimates of broad sense heritability for the commercially important trait "first lactation milk" in dairy cows have nearly doubled from ~0.25 in the 1970's to ~0.4 today. Which of the following can explain this increase? All of these can explain the increase O The number of genes impacting the first lactation milk phenotype is higher now than it was in the 1970s Dairy cows are reared in more uniform environments now than they were in the 1970s O Dairy cows are more genetically uniform now than they were in the 1970sThe following data were obtained from a Ghanaian population of 152 people. Blood Type Number of Individuals M 61 MN 64 N 27 Calculate the frequencies of the M and N alleles in this population. What are the expected numbers of individuals of each genotypic class in this population? Assume Hardy-Weinberg conditions. Use the chi-squared test to determine if these data fit the Hardy-Weinberg equilibrium model. The degrees of freedom for this test should be 1. Why is this appropriate?
- This Punnett square shows allele combinations for all possible genetic crosses in the wildflower population described in Part A. CR E p= 0.7 CW q= 0.3 p= 0.7 ? CW CR CR CR CW q= 0.3 ? ? CR CW CW CW ? What are the expected genotype frequencies in the offspring generation? O 0.49 CRCR (red flowers), 0.42 CRCW (pink flowers), 0.09 CWCW (white flowers) 0.49 CRCR (red flowers), 0.21 CRCW (pink flowers), 0.30 CWCW (white flowers) O 0.09 CRCR (red flowers), 0.21 CRCW (pink flowers), 0.42 CWCW (white flowers) 0.09 CRCR (red flowers), 0.42 CRCW (pink flowers), 0.09 CWCW (white flowers)The ratio for offspring in a specific population is .45 Homozygous AA, .10 Heterozygous Aa, .45 Homozygous for aa. If the A gene is dominant orange and the a gene is recessive blue what is the phenotype ratio that will be observed from this population? T T T T Paragraph v 3 (12pt) 只i8公 Arial O S Mashups - 1 国曲田 田田 NTML Css Path: p Words:0 28 O stv MacBook Air 80 DI esc F1 F2 F3 F4 F5 F7 FB F9 F10 ! @ #3 $ & 一 2 3 4 5 6 7 8. Q W R Y U P tab A S F G J L aps lock > Z C V M ft fn control option command command opt V - * 00 BThe common garden pea (Pisum sp.) can have green or yellow pea pods. Green peas are either homozygous dominant or heterozygous, while yellow peas are homozygous recessive. In a population of 500, there are 150 yellow individuals. Calculate q, p, p^2, 2pq, q^2.Then determine the number of mottled and orange individuals in the population. Select all the answer below that are correct. Note p? and p^2 are equivalent notation.
- Calculate the allelic, genotypic, and phenotypic frequencies for the following population: Genotype Number of individuals Number of Y alleles Number of y alleles YY 150 Yy 50 25 УУ Totals 225 Frequency of the Y allele? Frequency of the y allele? Frequency of the YY genotype? Frequency of the Yy genotype? Frequency of the yy genotype? Frequency of the yellow phenotype? Frequency of the green phenotype?In a population that meets Hardy-Weinberg equilibrium, the dominant allele frequency (A) is 0.7. What is the frequency of homozygous recessive individuals? Write how you reach answers in steps using the number provided. List all equations used. T T T F Paragraph := - E - Arial 3 (12pt) % D O Q e ES O f. * Mashups - 1 E E E - - H HTML CS5No Service x HOUBLON A study on reproductive success showed a positive relationship between hatching failure and F, the coefficient of inbreeding, for a population of chickadees in northern Maine. This relationship is best explained by which consequence of inbreeding? Number o that fail to hetoh Inbreeding coefficient (F) O Inbreeding depression O Increased phenotypic variation O Decreased paternal care 12:46 AM O Increased heterozygosity ✓ @ 51%
- The following variances were calculated for two traits in a herdof hogs.Trait VP VG VABack fat 30.6 12.2 8.44Body length 52.4 26.4 11.70 Calculate broad-sense (H2) and narrow-sense (h2) heritabilitiesfor each trait in this herd.- Gramm X P Pearson X G translat x 7 Traduct x Textboc x b Bongo x b Virt X Unit 58 3144/viewContent/112835/View pruce Harbor, Mai.. Aprende a hablar u. speaking, how many genes are involved in influencing such characters? b. How do the measurements of quantitative characters distribute in a population, i.e., according to what kind of curve? c. Give an example of a 2 quantitative characters other than those mentionedin the class materials. 1) 2) Q4. A person has symptoms of fatigue and lethargy. The doctor suspects that, for some reason, blood sugar levels are abnormally low. But testing indicates that blood sugar levels are normal. Further testing reveals that sugar levels inside cells are indeed abnormally low. a. What could be going on here? Why might the sugar levels inside the cells be low even though in the bloodstream they are normal? State in terms of genes. b. One way to explain the situation in part a is that a gene at one locus is affecting the expression of a gene at a…Webbed fingers, or syndactyly, are caused by a dominant trait found in 1 in 3000 births. What is theallele frequency of webbed fingers in the population