Biology I

Genetics

Principles of Biology I

5.1: Macromolecules are polymers, built from monomers 1) A _________ is a long molecule consisting of many similar building blocks  Polymer 2) These small building-block molecules are called?  monomers 3) What are three of the four classes of life’s organic molecules are polymers  Carbohydrates  Proteins  Nucleic acids  Lipids do not fall into this category 4) A ___________ occurs when two monomers bond together through the loss of a water molecule  dehydration reaction 5) Polymers are disassembled to monomers by __________ , a reaction that is essentially the reverse of the dehydration reaction  hydrolysis 6) ____________ include sugars and the polymers of sugars  Carbohydrates 7) What is the difference between glucose and fructose? Are they polar  Glucose is a aldoses and fructose is a ketoses  Yes the are due to the fact that -OH is present 8) _________________have molecular formulas that are usually multiples of CH 2 O  monosaccharides 9) ____________ is the most common monosaccharide  Glucose ( C 6 H 12 O 6 ) 10) _____________, the polymers of sugars have storage and structural roles. ____________ , a storage polysaccharide of plants, consists entirely of glucose monomers  polysaccharides  starch 11) The simplest form of starch is?  Amylose 12) __________ is a storage polysaccharide in animals. This is store mainly in the liver and muscle  Glycogen 13) The polysaccharide __________ is a major component of the tough wall of plant cells  Cellulose 14) The difference is based on two ring forms for glucose?  Alpha ( α ) and beta ( β ) 15) Polymers with ___ glucose are helical  α 16) Polymers with ___glucose are straight  β 17) __________ are the one class of large biological molecules that do not form polymers  Lipids 18) ________ are constructed from two types of smaller molecules: glycerol and fatty acids  Fats 19) A __________ consists of a carboxyl group attached to a long carbon skeleton  fatty acid 20) In a fat, three fatty acids are joined to glycerol by an ester linkage, creating a?  triacylglycerol, or triglyceride 21) _______________ fatty acids have the maximum number of hydrogen atoms possible and no double bonds  Saturated fatty acids 22) _____________ fatty acids have one or more double bonds  Unsaturated fatty acids 23) In a ____________ , two fatty acids and a phosphate group are attached to glycerol  Phospholipid 24) When phospholipids are added to water, they self-assemble into a _______, with the hydrophobic tails pointing toward the interior. Phospholipids are the major component of all ____________

BIO | 201 | Principles of Biology I | homework | 5.1: Macromolecules are polyme

Worksheet 7: Polymerase Chain Reaction 25 points 1. (6 pts). Predict the outcome if: a. A control PCR tube contains A,G, C, and T nucleotides (dNTPs) and a second, experimental PCR tube does not. In the experimental tube, you would expect: a. No target DNA copies made b. Fewer target DNA copies made c. Same number of target DNA copies made d. More target DNA copies made Highlight ONE and explain your answer. The experimental tube doesn’t contain any nucleodes, which is the building block for DNA b. A control PCR tube contains taq DNA polymerase and a second, experimental PCR tube contains human DNA polymerase. In the experimental tube, you would expect: a. No ta rget DNA copies made b. Fewer target DNA copies made c. Same number of target DNA copies made d. More target DNA copies made Highlight ONE and explain your answer. The second tube contains human DNA which would not copy properly because human DNA already contains its own nucleodes c. A control PCR tube undergoes 30 cycles of PCR (95 o , 60 o , 72 o ) at 1 minute each. A second, experimental tube undergoes PCR but only has 1 95 o step at the beginning and then 30 cycles (60 o and 72 o ) at 1 minute each. In the experimental tube, you would expect: a. No target DNA copies made b. Fewer target D NA copies made c. Same number of target DNA copies made d. More target DNA copies made Highlight ONE and explain your answer. Fewer targets would be made because a step was skipped, it didn't go through the full heang and cooling process of PCR that its suppose to go through 2. (2 pts). In your own words, explain why PCR can make crime scene investigations more effective and efficient? PCR is more eﬀecve because you can use a samples of blood, saliva, or a strand hair from a person to idenfy them using PCR 3. (2 pts). The following is a typical target sequence of DNA. The underlined regions at the upper left and lower right are where the primers will bind. Note that the underlined sections are different for the upper and lower strand. -ATCG TTCATTCATTCATTCATTCATGGCAT- upper strand -TAGCAAGTAAGTAAGTAAGTAAGTACCGTA - lower strand What is the sequence of the upper strand primer: TAGC What is the sequence of the lower strand primer: GGCAT Perform the lab simulation or watch the video linked on Canvas. 4. (3 pts). The simulation says “primers…are powerful tools for copying very specific DNA sequences, since there is almost no chance they will target the wrong site”. Explain why there is almost no chance they will target the wrong site. The primers are already made to target there speciﬁc site of DNA sequence 5. (2 pts) In the simulation, why did they use 2 primers (Primer 1 and Primer 2)? They used two primers because they were using two strands of DNA for copying 6. (2 pts) Do you think you could perform PCR without a Thermal Cycler? Why or why not? No, without a thermal cycler we can't heat or cool the DNA at the accurate temperatures in order to copy the DNA correctly

BIO | 150 | Biology I | homework | Worksheet 7: Polymerase Chain 

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

Worksheet_5.docx.docx-Worksheet 5 Name: Enzymes Section: 25

Name: Melanie Bennett Section: Friday @ 11am Chi-Square Calculation Assignment based on Corn Data (12 pts) 1) In looking at kernel color (purple or yellow) and endosperm shape (starchy or sweet), what is the Null Hypothesis (in terms of bead sampling) for this data (1)? H o = No difference between data 9:3:3:1 (P/ST, P/SW, Y/ST, Y/SW) ratio 2) Based on this data and your Null hypothesis stated above, what do you think that Mode of Inheritance is for this phenotype (1)? AABB X aabb 3) What are your O (observed) and E (expected) values (1)? Color of corn Observed Expected P/ST 276 (9/16) (563) = 317 P/SW 134 (3/16)(563) = 106 Y/ST 118 (3/16)(563) = 106 Y/SW 35 ( 1/16)(563) = 35 Total= 563 4) How many degrees of freedom are there (1)? DF = N -1 DF = 4 – 1 DF = 3 5) What is your X 2 value (1)? X 2 = ( ¿¿ ) X 2 = 5.30 + 7.40 + 1.36 + 0 X 2 = 14.06 6) What is the probability (p value) that this data resulted from chance (1)? P value = 0.007 (between 0.01 and 0.001) 7) Based on this chi-square analysis can you reject or fail to reject your null hypothesis? Explain your answer and state what this means about your data (2). Based on our chi-square analysis we have to reject the null because our p value is less than 0.05. This means that our data could have been counted incorrectly or there could have been an incorrect calculation. 8) You are investigating the claim that estrogen within the water supply is resulting in more female births. The total number of births in this region was 284, with 162 of them being Name: Melanie Bennett Section: Friday @ 11am female and the rest boys. Using a chi-square analysis, determine of these claims might be true. Explain your answer (4). H o = No difference between data 1:1 (F:M) ratio. Hypothesis: Ff X ff F = female f = male Observed Experimental F 162 (1/2) (284) = 142 M 122 (1/2) (284) = 142 Total = 284 X 2 = ¿¿ X 2 = 5.63 DF = 2-1 DF = 1 P value = 0.03 (between 0.05 and 0.01) Reject the null. This chi-square analysis tells us that the estrogen in the water supply is not resulting in the increase in female births.

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