Biology I

Principles of Biology I

Genetics

4/11/2020 Ch 23 HW https://session.masteringbiology.com/myct/assignmentPrintView?assignmentID=9080278 1/16 Ch 23 HW Due: 11:59pm on Monday, February 10, 2020 You will receive no credit for items you complete after the assignment is due. Grading Policy Activity: Genetic Variation from Sexual Recombination Launch the Genetic Variation in Sexual Recombination Activity and answer the questions. Part A Homologous pairs of chromosomes are lined up independently of other such pairs during _____. ANSWER: Correct This describes metaphase I. Part B Crossing over, resulting in an increase in genetic variation, occurs between _____. ANSWER: Correct This process produces chromosomes containing genes inherited from both parents. Part C In human gamete production there is an average of _____ crossover events per chromosome pair. ANSWER: prophase II telophase II metaphase II anaphase I metaphase I sister chromatids of homologous chromosomes nonsister chromatids of nonhomologous chromosomes sex cells and somatic cells nonsister chromatids of homologous chromosomes sister chromatids of nonhomologous chromosomes 4/11/2020 Ch 23 HW https://session.masteringbiology.com/myct/assignmentPrintView?assignmentID=9080278 2/16 Correct These crossover events increase the genetic variation among gametes. Part D Which of these gametes contain one or more recombinant chromosomes? ANSWER: Correct These gametes carry chromosomes produced as a result of crossing over. Scientific Skills Exercise: Using the Hardy-Weinberg Equation to Interpret Data and Make Predictions Number of Seedlings Time (days) Green ( C G C G ) Green-yellow ( C G C Y ) Yellow ( C Y C Y ) Total 7 21 49 47 111 106 56 20 216 173 Is evolution occurring in a soybean population? One way to test whether evolution is occurring in a population is to compare the observed genotype frequencies at a locus with those expected for a nonevolving population based on the Hardy-Weinberg equation. In this exercise, you’ll test whether a soybean population is evolving at a locus with two alleles, C G and C Y , that affect chlorophyll production and hence leaf color. Students planted soybean seeds and then counted the number of seedlings of each genotype at day 7 and again at day 21. Seedlings of each genotype could be distinguished visually because the C G and C Y alleles show incomplete dominance: C G C G seedlings have green leaves, C G C Y seedlings have green-yellow leaves, and C Y C Y seedlings have yellow leaves. Part A - Calculating allele frequencies in a population 9–10 0–1 2–3 10+ 5–6 B, C, D, and G C, D, E, F, and G A, B, and C B, C, and G B, C, F, and G

BIO | 201 | Principles of Biology I | homework | 4/11/2020 Ch 23 HW https://ses

Chapter 11 – Mendel and Genetic Patterns of Inheritance Pre-Lecture Worksheet Name Daromi Please read Chapter 11 and answer the following questions. Please email your answers before class time tomorrow for credit. 1. Which of the following crosses will NOT produce offspring with the recessive phenotype? Select all that apply. A) TT × Tt B) Tt × Tt C) TT × TT D) Tt × tt E) tt × tt 2. In a classic Mendelian monohybrid cross between a homozygous dominant parent and a homozygous recessive parent, which generation is always completely heterozygous? A) F 1 generation B) F 2 generation C) F 3 generation D) P generation 3. Women with an X-linked disorder could pass the genes for the disorder to ________, and men with an X-linked disorder could pass the genes for the disorder to ________. A) only their daughters; only their daughters B) both their daughters and sons; only their sons C) both their daughters and sons; only their daughters D) both their daughters and sons; their daughters and sons 4. The F 2 offspring of a classic Mendelian monohybrid cross between homozygous dominant and homozygous recessive parents would produce the genotype(s) A) AA and AA . B) AA and AA . C) AA , Aa , and aa . D) AA only. E) AA only. 5. If an individual with a dominant phenotype is crossed with an individual with a recessive phenotype, 4 of their 9 offspring show the recessive phenotype. What is the genotype of the first parent? A) AA B) Aa C) aa D) The answer cannot be determined from this information. 6. The pedigree chart depicts the inheritance pattern of ________. Circles depict females and squares depict males. Colored shapes represent affected individuals (expressing a trait) and uncolored shapes are unaffected (do not express a trait). A) an autosomal recessive characteristic with both parents being heterozygous B) an autosomal dominant characteristic with both parents being homozygous dominant C) an autosomal recessive characteristic with both parents being homozygous recessive D) None. 7. If two parents have blood genotypes I A i and I B i , which of the following blood genotypes are possible in their children? A) I A i and I B i only B) I A i , I B i , and I A I B only C) I A I A , I B I B , and I A I B only D) I A i , I B i , and ii only E) I A i , I B i , I A I B and ii 8. Since each child of two heterozygous parents has a 50% chance of receiving a recessive trait from each parent, A) if the first child is phenotypically recessive, then the next child must be phenotypically dominant. B) if the first child is phenotypically recessive, then the next child has a 3/4 chance of being phenotypically recessive. C) if the first child is phenotypically recessive, then the next child has a 1/2 chance of being phenotypically recessive. D) no matter what the first child's phenotype, the next child will have a 1/4 chance of being phenotypically recessive. 9. When an individual of African descent marries and has children with an individual of European descent, their children often have a mid-shade of skin color, but they could also have a light shade or a dark shade. This can best be described as A) X-linked inheritance. B) polygenic inheritance. C) pleiotropic inheritance. D) the dominance of the gene for dark skin tone. 10. A testcross involves an individual with the dominant phenotype but unknown genotype being crossed with an individual that has a(n) ________ genotype. A) homozygous dominant B) heterozygous dominant C) homozygous recessive D) Any of the choices.

BIO | 150 | Biology I | lecture notes | Chapter 11 – Mendel and Geneti

Part B: Case Study Questions 1) Answer questions 1-8 in Part III of the case study (8). Questions 1. What is a gene family? What mechanism generates them? - A gene family is sets of many related genes formed by a duplicated single original gene. Genetic variation is the mechanism that generates it. 2. What is a SNP? What is a haplotype? How can SNP haplotypes be determined? - Single nucleotide polymorphism (SNP) is a single base-pair difference in the DNA sequence of individual members of a species and a haplotype is a group of genes within an organism that was inherited together from a single parent. SNP halotypes can be determined by using tag-SNPs that are specific to identifying halotypes. 3. Assuming the PAV allele is the original one, what types of mutations (silent, missense, nonsense, etc.) have changed the PAV allele into the AVI allele? Explain. Based on the information given, how do these changes fit with what you have learned about the “functional enzyme hypothesis”? - PAV allele has proline, alanine and valine while AVI has alanine, valine and isoleucine. There is no mutation that have changed PAV allele into AVI, since they are expressed equally at the RNA level and AVI is not terminated, so there is no mutation that happened. 4. You repeat Snyder’s experiment, but instead of determining the ability of each family member to taste PTC, you determine what alleles they carry: 2 PAV (PAV/PAV), 2 AVI (AVI/AVI), or 1 PAV and 1 AVI (PAV/AVI). Using the table provided below, what are the possible outcomes for off spring if the parents have the genotypes described in the fi rst column? Below the genotype, write the phenotype (taster or non-taster) of each kind of individual. Parent Genotype Combinations Possible Genotypes/Phenotypes of Children Both parents PAV/PAV PAV/PAV – taster or can taste Both parents PAV/AVI PAV/PAV - taster or can taste PAV/AVI - taster or can taste AVI/AVI – non-taster or cannot taste One PAV/PAV, the other PAV/AVI PAV/PAV - taster or can taste PAV/AVI - taster or can taste One PAV/PAV, the other AVI/AVI PAV/AVI - taster or can taste One PAV/AVI, the other AVI/AVI PAV/AVI - taster or can taste AVI/AVI – non-taster or cannot taste Both parents AVI/AVI AVI/AVI – non-taster or cannot taste 5. Go back to Snyder’s results and give all the possible genotypes for the phenotypes listed in that table. - Both parents can taste: Genotype of parents: PAV/AVI × PAV/AVI Genotype of children (taster): PAV/PAV, PAV/AVI, PAV/AVI (Non-taster): AVI/AVI - One parent can taste and other cannot: Genotype of parents: PAV/AVI × PAV/AVI Genotype of children (taster): PAV/AVI, PAV/AVI (Non-taster): AVI/AVI - Neither parent can taste: Genotype of parents: AVI/AVI × AVI/AVI Genotype of children (Non-taster): AVI/AVI 6. For each step of gene expression, explain how DNA differences lead to phenotypic differences in the ability to taste PTC. - DNA differences lead to phenotypic differences and the ability to taste PTC is based on the TAS2R38 gene. Some people can’t taste PTC because of their genetic makeup and others can. 7. Generalizing from this example, what can you say about the connection between Mendelian and molecular genetics? What type of reasoning are you using here? - In molecular genetics, PAV is considered dominant while AVI is recessive. Those with PAV/PAV can taste bitter and not bitter, those heterozygous like PAV/AVI could also distinguish between bitter and not bitter substance or food. And just like in Mendelian, a dominant allele even if it’s a heterozygous, still shows its phenotypic expression as dominant, as a PTC taster. 8. Hypothesize about some consequences of PTC tasting ability in the daily lives of tasters and non-tasters. - The ability to taste PTC is inherited from parents. People who can taste PTC can distinguish foods that are bitter and foods that are not bitter. While those who can’t taste PTC cannot distinguish bitter foods from not. Most people who can’t taste PTC are more likely to smoke than tasters.

BIO | 240 | Genetics | homework | Part B: Case Study Questions 1

Worksheet 1 Name: Biological Molecules Section: 20 points Date Turned In: 2/12/2019 Introduction: 1. (3 pts) If a lab group combined an Unknown solution with Iodine and observed a color change from yellow to black, which of the following terms would ﬁt this result? a. positive control b. positive result c. negative control d. negative result Circle ONE and explain your reasoning. I’ts positive result because we are dealing with an unkwown that changed colors from yellow to black which indicates a positive test result. 2. (2 pts) If the Biuret test from this lab were quantitative rather than qualitative, explain how the data you recorded would be diﬀerent. If it was quantitative, instead of just revealing the color change, it would have to involve a value with numbers. For instance, it would have a rate of change, or a graph that reveals the amount of concentration that the Biuret would have in the solution. Materials and Methods: 3. (3 pts) Respond to this question from an IMB student in last fall’s class: “Wouldn’t we get out of the lab sooner if we just tested the unknown milk solutions?” No because it would lack validity. We would need to perform the known tests to compare it to the unknown to ensure it is accurate. 4. (3 pts) Respond to this question from an IMB student in last spring’s class: “After we do the controls, wouldn’t it be easier to use one drop of the indicator and one drop of the unknown solution? Wouldn’t we get out of lab sooner?” No because the experiment would be compromised. Each indicator requires diﬀerent speciﬁc measurements and if we were to change it, it would reveal the wrong results because we are lacking the right amount of reagents. Results: 5. (4 pts) For each of the listed unknown solutions, provide the BEST experimental evidence that demonstrates that it is NOT fat free milk. Circle the letter or letters of solutions that could be fat free milk. Evidence against fat free milk Unknown A It was negative for protein. Unknown B It was positive for lipid. Unknown C It was positive for lipid. Unknown D It could be fat-free milk because it was negatie for both lipid and starch and positive for protein. Unknown E It was positive for starch. Discussion: 7. (2 pts) The ﬁgures below shows cells of wheat that have been stained with Iodine. . Decide if you think starch is or is not present. Then answer this hypothetical question: If a cow were to eat this plant and we collected its skin cells and stained them with iodine, do you think starch would appear? Explain why or why not? No, starch wouldn’t be present because although there was starch present on the weath cells, it wouldn’t appear on the cow’s skin cells because they don’t produce starch so it would have been broken down.

BIO | 150 | Biology I | lab | Worksheet 1 Name: Biological M

Chapter 7 - Photosynthesis Pre-Lecture Worksheet Name Daromi Please read Chapter 7 and answer the following questions. Please email your answers before class on July 16 th for credit. 1. Which of the following is a common photosynthetic pigment found in plants? Select all that apply. A) chlorophyll a B) chlorophyll c C) carotenoids D) chlorophyll b 2. Which of the following are characteristics of C 4 plants? Select all that apply. A) C 4 plants store carbon dioxide temporarily as oxaloacetate. B) C 4 plants are found in hot, dry climates. C) C 4 plants have a net photosynthetic rate two or three times greater than C 3 plants. D) C 4 plants experience more photorespiration than C 3 plants. E) C 4 plants separate carbon fixation from the Calvin cycle. 3. What would happen if a drug directly interfered with the functioning of RuBP carboxylase? A) Water would not be split at the start of photosystem II, halting the light reactions. B) Chemiosmosis and ATP production would cease, halting the Calvin cycle. C) Carbon dioxide would not be fixed to   RuBP, stalling the Calvin cycle. D) RuBP would not be regenerated from G3P, halting the Calvin cycle. 4. What would happen if the stomata of a leaf remained closed all day? Select all that apply. A) Carbon dioxide in the air spaces of the leaf would decrease. B) Oxygen in the air spaces would increase. C) The amount of water within the air spaces would decrease. D) Photorespiration would increase in C 3 plants. E) Photorespiration would increase in CAM plants. 5. Protons (H + ) accumulate in the thylakoid space during electron transport between photosystems II and I. Which of the following describes what will happen to these protons next? A) The protons will reenter the electron transport chain and be used to generate water. B) The protons will be used to oxidize NADPH to NADH, generating ATP in the process. C) The protons will diffuse out of the lipid layer surrounding the thylakoid and exit the plant during gas exchange. D) The protons will raise the pH of the thylakoid space until it becomes alkaline at which point the protons are pumped out of the thylakoid, generating ATP. E) The protons will move from the thylakoid space to the stroma through an ATP synthase complex that generates ATP. 6. Characterize these reactions involved in the Calvin cycle. A) These are reactions involved in the regeneration of RuBP. The reactions include an exergonic reaction (ATP breaks down into ADP) and the reduction of the coenzyme NADPH. B) These reactions are involved in the reduction of CO 2 . Energy is required in the form of ATP, which is hydrolyzed to ADP. The substrate BPG is reduced, while the coenzyme NADPH is oxidized. C) These reactions are involved in the fixation of CO 2. Energy is required in the form of ADP, which is hydrolyzed to ATP. The substrate BPG is oxidized, while the coenzyme NADPH is reduced. D) These reactions are part of the light reactions in a noncyclic pathway. ATP and NADPH are used during these reactions.

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