Problem Set #1-BIOS 230 Spring 2023

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Lewis University *

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CHEM-301

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Biology

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Feb 20, 2024

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©Karin Nelson 2023 Print your name + UIN:________________________________ BIOS 230-Spring 2023 Problem Set #1 - 47 pts. Due SATURDAY, February 11 at 11:59 p.m. Instructions: Do your own work (but ask us if you need help). Show all your work for full credit. Write legibly —if we can’t read it, you don’t get credit. Use all digits during calculations, but take all final answers out to 4 decimal points unless otherwise instructed. Circle your final answer. Submit your work through Blackboard Assignments in PDF, JPEG, or Word formats ONLY we cannot read or accept HEIC format. DON’T post this assignment to any website. No late homework will be accepted! https://srelherp.uga.edu/snakes/thasir.htm https://www.boredpanda.com/unusual-animal-colors/?utm_source=google&utm_medium=organic&utm_campaign=organic In Cirtalus’ belt ed snake, a single gene determines skin color. The gene has two alleles, B and b ; simple dominance rules apply the dominant allele causes the common brown coloration; homozygous recessive individuals are multi-colored (rainbow). In a sampled population, it is determined that the b allele has a frequency of 0.031. 1. What is the frequency of the B allele? Show your work and carry to 4 decimal places. (2 pts.) 2. What are the three possible genotypes? (3 pts.)
©Karin Nelson 2023 3. What are the frequencies of each of these three genotypes? (6 pts.) 4. In a population of 8,000 Cirtalus’ belt ed snakes, how many would you expect to be rainbow colored (bb)? Round off to the nearest whole animal. (2 pts.) *************************************************** Tail shape in royal empennaged guppies (a type of freshwater fish) is determined by one gene with two alleles, R (round shape) and r (fan tail); simple dominance rules apply. In an aquarium with a population of 150 empennaged guppies, you observe the following genotypes: RR (Round tail) 67 individuals Rr (Round tail) 45 individuals rr (Fan tail) 38 individuals *** For the following questions, do NOT assume population is in Hardy-Weinberg Equilibrium. Take all your final answers out to 4 decimal places. *** 5. Calculate the allele frequencies of R. (2 pts.) 6. Calculate the allele frequencies of r. (2 pts.)
©Karin Nelson 2023 Congratulations your guppies had babies! In this second generation of guppies, the genotypes are: RR (Round tail) 230 individuals Rr (Round tail) 85 individuals rr (Fan tail) 135 individuals 7. Calculate the allele frequencies of R among the babies. (2 pts.) 8. Calculate the allele frequencies of r among the babies. (2 pts.) 9. Is the population of empennaged guppies above in Hardy-Weinberg Equilibrium? Explain (3 pts.)
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©Karin Nelson 2023 *************************************************** Phenylketonuria (PKU) is a recessive genetic disease that causes the amino acid phenylalanine to build up in the body. PKU is caused by a mutation in the phenylalanine hydroxylase (PAH) gene. A normal PAH gene creates the enzyme needed to break down phenylalanine, which is found naturally in many foods. Without this enzyme, phenylalanine builds up to toxic levels. If left untreated, PKU causes microcephaly, developmental delays, seizures, brain damage, and death. Because of its seriousness, newborn babies in the U.S. are screened for this disorder shortly after birth. In the United States, about 1 in 12,000 babies are born with PKU every year. The disease occurs at different frequencies in different populations. In Japan, 1 in 125,000 babies are born with PKU. In Italy, 1 in 4,500 babies are affected. In Turkey, the rate is 1 in 6,000. Among the ethnic Roma, the rate of PKU is 1 in 40. 10. What is the frequency of the PKU allele in the U.S. population? (2 pts.) 11. What is the frequency of the PKU allele in the Japanese population? (2 pts.) 12. What is the frequency of the PKU allele in the Italy population? (2 pts.) 13. What is the frequency of the PKU allele in the Turkey population? (2 pts.) 14. What is the frequency of the PKU allele in the Roma population? (2 pts.)
©Karin Nelson 2023 15. What percentage of Japanese people will be carriers of the trait, but not have the disease? (2 pts.) 16. What percentage of people of Roma heritage will be carriers of the trait, but not have the disease? (2 pts.) *************************************************** We want to determine whether time to sexual maturity (egg laying) in Orpinton silky chickens is a heritable trait. A population of 100 chickens takes an average time of 153 days to reach maturity. We select ten chickens that reached maturity earliest; their average time to adulthood is 127 days. We breed those 10 early-adult chickens with each other and monitor the resulting chicks. We watch to see when they mature; the average time to maturity is 138 days. 17. Calculate the selection differential. (2 pts.) 18. Calculate the response to selection. (2 pts.) 19. Calculate heritability (h 2 ). (2 pts.) 20. Interpret the heritability of maturity in Orpinton silky chickens from above. (3 pts.)