- Home
- Biological Science Homework Help
- Molecular Biology Homework Help
Molecular Biology Homework Help | Molecular Biology Assignment Help
Excel in Exams with Expert Molecular Biology Homework Help Tutors.
Trusted by 1.1 M+ Happy Students
Excel in Molecular Biology with Expert Online Tutoring
Dive Deep into the World of Molecular Biology
Molecular biology, a fascinating branch of biology, delves into the intricate world of biological macromolecules like nucleic acids and proteins, unraveling their roles in the replication of cells and the transmission of genetic information. This field is not just about understanding the complex biological processes; it also involves hands-on manipulation of DNA to explore the vast genetic landscapes and their implications in various biological phenomena.
Comprehensive Molecular Biology Course Topics
Embarking on a molecular biology course, students will navigate through an extensive array of topics, each shedding light on different facets of cellular and molecular functions. These topics include, but are not limited to:
- Cells and genomes
- Cell chemistry and biosynthesis
- The structure and function of proteins
- DNA, chromosomes, and their intricate dynamics
- Mechanisms of DNA replication, repair, and recombination
- Gene expression and its regulation
- Techniques for manipulating proteins, DNA, and RNA
- Advanced visualization methods for studying cells
- Membrane structure and function, including transport mechanisms
- Intracellular compartments, vesicular traffic, and protein sorting
- The fundamentals of energy conversion in mitochondria and chloroplasts
- Cell communication, the cytoskeleton, cell cycle, and apoptosis
- The mechanics of cell division and the molecular basis of cancer
- The immune system and its responses to pathogens
Why Choose Our Molecular Biology Tutoring Services?
Our mission is to provide unparalleled online tutoring and homework help to college students navigating the complexities of molecular biology. Here’s why our services stand out:
Expert Guidance at Your Fingertips
Our team of highly qualified molecular biology tutors brings a wealth of knowledge and experience, ready to offer personalized assistance on a 24/7 basis. Whether you're tackling a challenging homework assignment or preparing for an exam, our experts are here to support your academic journey.
Tailored Learning Experiences
We understand that every student has a unique way of learning. That’s why our tutoring sessions are customized to match your individual learning style, ensuring that you grasp complex molecular biology concepts with ease.
Boost Your Confidence
Working with our expert tutors not only deepens your understanding of molecular biology but also boosts your confidence in the classroom. This newfound confidence empowers you to excel in your current coursework and lays a solid foundation for your future studies.
Convenience and Accessibility
Our fully remote tutoring services make it effortless to seek help whenever you need it, eliminating the hassle of finding a local tutor. With our support, you can focus on mastering molecular biology without any scheduling conflicts.
Leverage Our Comprehensive Support for Molecular Biology
Whether you need help with specific homework assignments, wish to review broad conceptual ideas, or have questions about molecular biology, our tutors are equipped to provide the assistance you need. By choosing our tutoring services, you gain access to:
- In-depth insights for homework assignments
- Simplified explanations of complex topics
- Tailored instruction that fits your learning style
Embark on Your Molecular Biology Academic Adventure
Don't let the intricacies of molecular biology hinder your academic progress. With our expert online tutoring and homework help, achieving a comprehensive understanding of molecular biology is within reach. Reach out to us today and take the first step towards excelling in your molecular biology coursework and beyond.
Recently Asked Molecular Biology Questions
Expert help when you need it- Q1:Write a one paragraph response explaining the four different perspectives below on human cloning. Introduce the authors’ full names (titles aren’t necessary for this paragraph, but they would be for an essay) to indicate each of their different opinions. Consider the most strategic way of how to organize/structure your paragraph to show connections and contrasts between each of the source’s arguments Use transitions carefully, as they will show how you relate the articles to each other. Perspective 1: Patricia Baird, “Should Human Cloning be Permitted?” “...In conclusion, using nuclear-transfer cloning to allow people to have a child introduces a different way of reproduction for our species. Once we breach this barrier, it leaves us with no place to stop. Given all the problems outlined, the reasons for permitting cloning to produce a person are insufficiently compelling. Even in the few circumstances where the case for human cloning seems justified, there are alternative solutions. We are at an appropriate stopping place on a slippery slope. Not all reasons why a person might wish to copy his or her cells are unethical, but given there are other options open to people wishing to form a family, concerns about individual and social harms from cloning are strong enough that it is not justified to permit it.” Perspective 2: Chris MacDonald, “Yes, Human Cloning Should be Permitted.” “The fact that a portion of society—even a majority—finds an activity distasteful is insufficient grounds for passing a law forbidding it....human cloning for reproductive purposes has legitimate, morally acceptable applications—for example for infertile couples, and for gay couples.” Perspective 3: Jacob M. Appel, “Should We Really Fear Reproductive Human Cloning?” “In an ideal world, human reproductive cloning would be safe, legal and rare. I say rare because my guess is that the vast majority of people, myself included, would have little desire to raise cloned offspring. After all, it is now possible to clone pet dogs--but few of us would choose to spend a spare $150,000 on such a venture. Yet thirty-eight years after James Watson's seminal essay, "Moving Toward the Clonal Man" called for increased public debate on this promising and perplexing subject, I don't believe that we should be so quick to greet cloning technology with a permanent injunction. Instead, what human reproductive cloning requires at the moment is a yellow light, telling us to proceed with extreme caution, until we know with confidence whether the technology can ever be used to produce healthy babies.” Perspective 4: Leon Kass, “The Wisdom of Repugnance” “We are repelled by the prospect of cloning human beings not because of the strangeness or novelty of the undertaking, but because we intuit and feel, immediately and without argument, the violation of things that we rightfully hold dear. Repugnance, here as elsewhere, revolts against the excesses of human wilfulness, warning us not to transgress what is unspeakably profound. Indeed, in this age in which everything is held to be permissible so long as it is freely done, in which our given human nature no longer commands respect, in which our bodies are regarded as mere instruments of our autonomous rational wills, repugnance may be the only voice left that speaks up to defend the central core of our humanity.” See Answer
- Q2:b. C. d. What was the independent variable in this experiment? What was the dependent variable? What were the standardized variables?See Answer
- Q3:e. f. g. Do the results support your hypothesis? Explain. At which temperature was the rate of osmosis greatest? What can you conclude about the effects of temperature on the rate of osmosis? Answer in terms of molecular energy.See Answer
- Q4:h. Name at least two possible sources of error in this experiment. How could this experiment be improved?See Answer
- Q5:Variables Table 1 - Osmosis: Effect of Temperature Cold Beaker Water temperature Tube mass Tube firmness Warm Beaker Room Temp Beaker Water temperature Tube mass Tube firmness Before After % Change Observations Water temperature Tube mass Tube firmnessSee Answer
- Q6:Variable Water Temperature Tube Mass Variable Water Temperature Tube Mass Before Tube Mass Table 1 B Room Temperature Table 1B Before weight in grams of pseudocell containing 40% sucrose, and weight after 40 minutes incubation n water at room temperature in degrees Celsius. After % Change in weight Before 102, POLAR W 10.7 After ACCULAB 109 11.6 % Change in weight 9 Table 1 C Warm Beaker Table 1C Before weight in grams of pseudocell containing 40% sucrose, and weight after 40 minutes incubation in warm water. Variable Before After % Change in weight Observations Water Temperature 16.6 Observations ObservationsSee Answer
- Q7:HANDS-ON ACTIVITY#1: 1. Get 2 transparent glasses (same size) 2. Put the same volume of water in both of them. In one of the glasses put cold water and in the other glass put hot water. 3. Place the 2 glasses side by side and let the water sit for a few seconds. 4. Put a drop of food dye in each of the glasses. Use your cell phone to take a picture of this hands-on activity. You will need to submit this picture and your photo ID to DB 2. Answer the following questions: Why does 1 drop of the dye stain the entire volume of water? Does the staining of the entire volume of water happen faster/slower/or at the same speed in the hot water compared to the cold water? Why? HANDS-ON ACTIVITY#2: This experiment will take you 3 days to complete. Perform the experiment and answer questions 1 to 7 in this hand-out. Use your cell phone to take pictures of this hands-on activity. You will need to submit these pictures and your photo ID to DB 2. An unfertilized chicken egg contains a large cell surrounded by egg white, a shell membrane, and an egg shell. You will investigate how the size of an egg changes when the eggshell is removed and the egg is placed in different types of liquid. ➤ Get 2 eggs. To begin, record the weight or circumference of each egg in the day 1 row in the table. (Measure the circumference around the widest part, not lengthwise.) Day 1 2 Egg 1 Egg 2 Weight (grams) (or circumference (cm)) Weight (grams) (or circumference (cm)) (with shell) (with shell) (after a day in vinegar most of shell removed) (after a day in vinegar: most of shell removed)See Answer
- Q8:3 (after a day in water) (after a day in corn syrup) Put each egg in a container labeled Egg 1 or Egg 2. Pour in enough vinegar to cover the egg. Cover the container. Do you see bubbles forming around the egg? These are bubbles of CO2 which result from the chemical reaction between the acetic acid in the vinegar and the calcium carbonate in the eggshell. This reaction will dissolve most of the eggshell by day 2. Day 2 ➤ Observe your eggs. Notice that most of the shell has been dissolved by the acetic acid in the vinegar. The shell membrane around the egg is fairly strong. However, the egg without its shell is fragile, so you will need to handle your eggs very gently and carefully! Rinse each egg and measure the weight or circumference of each egg. Record your results for day 2 in the above table. 1a. Did the eggs become heavier/larger or lighter/smaller 1b. What do you think happened to cause the change in the eggs' weight/size? ➤ Empty the vinegar from the container for egg 1 and rinse the container. Put egg 1 back in the container and add water to cover the egg. ➤ Empty the vinegar from the container for egg 2 and rinse the container. Put egg 2 back in the container and add corn syrup to cover the egg. Day 3 2. Compare and contrast the appearance of the egg that has been in water vs. the egg that has been in corn syrup. 3. You may be able to see a layer of water on top of the corn syrup. Where do you think this water came from? Rinse the corn syrup off of egg 2. Measure and record the weight and/or circumference of the egg for day 3 in the table on page 1. 4. Why did the egg placed in water get heavier and bigger? Where do you think the additional weight/volume came from?See Answer
- Q9:5. What do you think happened to cause the change in weight/size of the egg placed in corn syrup? 6a. Recall that each egg is surrounded by a shell membrane. Based on your observations, which of the following do you think can cross this membrane? a. both water and the proteins in the egg white b. water, but not the proteins in the egg white c. the proteins in the egg white, but not water d. neither water nor the proteins in the egg white 6b. What evidence supports your conclusion? 7. The shell membrane that surrounds the egg is a selectively permeable membrane. Explain why "selectively permeable" is a good way to describe this membrane.See Answer
- Q10: Imagine you have been recently hired by a biotech company. On your first day of work, the manager wants to see your ability to engineer proteins, so they ask you to design a protein expression system for making any protein you wish, but it should have some way after expression in E. coli BL21 to be purified by affinity chromatography. The manager gives you some advice and recommends you produce the protein with a HisTag so you can purify your protein with nickel affinity column chromatography which the company uses regularly. To make things easier since its your first day, the manager gives you a tube containing a proprietary expression vector that only your company has called pET 14p and it is this vector that you will need to use to insert the gene for the protein that you choose to make and purify. Before getting started on the actual experiments, the manager wants you to design the experimental setup and provide a report. The following 6 items (marked w/ **) should be included in your report along with an explanation for each step. Steps to follow in this design project for your report 1) Choose a protein that you are interested in purifying. -This can be any protein (there are millions possible so no two students would by chance choose the same protein) - You can use resources like NCBI (https://www.ncbi.nlm.nih.gov/protein/ ) or KEGG (https://www.genome.jp/kegg/pathway.html), among others to find a protein you are interested in making. **(List the name of the protein you plan to make and the organism it comes from) 2) Find the amino acid sequence of that protein and then find the DNA sequence needed to make that amino acid sequence. -The amino acid sequence may be found at the links above. DNA sequences may be found at NCBI or KEGG as well or deduce it here: https://en.vectorbuilder.com/tool/codon-optimization.html **(List the amino acid sequence and its corresponding DNA sequence for this protein you plan to make) 3) Choose a set of restriction enzyme that you can use to cut your pET14p expression vector but that will not cut the important protein coding regions of the gene you are inserting into pET14p -NOTE: You should make sure that the coding sequence of your selected protein you choose does not already contain the sequences recognized by those restriction enzymes you selected to cut your pET14p vector or else it will cut your gene at a place you don't want. You can use this tool http://nc2.neb.com/NEBcutter2/ to check if your gene (DNA sequence that codes for your protein of interest when translated) contains the sequences recognized by that restriction enzyme that you have chosen. If the gene does contain a sequence that will be cut by your enzymes selected to cut the pET14p vector then you'll either have to choose a different enzyme or select a different codon for those positions on that gene so that it can make the same protein but not be cut improperly by the restriction enzymes you plan to use. Because you will be expressing your protein for purification using Nickel affinity chromatography, make sure that your choice of cutting location does not remove the HisTag sequence from your vector. **(List the names of the restriction enzymes you will use to cut the pET14p vector and their recognition sites. Also show where in the pET14p vector that it cuts) 4) Imagine you have access to the DNA template of this gene. Choose the appropriate forward primer and reverse primer that you will need to PCR this gene from your template. Make sure that your primers have 5' ends that possess the corresponding restriction sites needed to later insert the PCR product into your pET14p vector. - NOTE: You will need your primers to have approximately the same melting temperature (within 2 degrees of each other) and they should be between 55°C to 65°C. You may find this tool useful http://www.biophp.org/minitools/melting_temperature/demo.php (you can select basic Tm). Remember that your reverse primer sequence would be the reverse complement of the gene sequence you had chosen (if you are looking at the coding sequence in the proper way) and as such you may find the following tool useful for getting the reverse complement: http://www.bioinformatics.org/sms/rev_comp.html **(Provide the forward and reverse primer DNA sequences from 5' to 3' and their expected melting temperature...underline the restriction sites on each primer) 5) Show your plasmid containing the recombinant DNA for your gene. Imagine that you have now conducted the PCR successfully with your primers on your gene of interest to get a PCR product and that you have cut the PCR product using the same restriction enzymes from step 3 that you used to cut the vector. After ligating the cut PCR product into the cut pET14p vector you will have a complete plasmid. **(Show what the DNA sequence will look like for the ligated DNA of your final complete plasmid but just show the region between the T7 promoter and T7 terminator after you have ligated your gene into the vector making sure to underline the restriction sites and placing in bold (or highlight in yellow) the new sequence of your gene that you inserted into the vector) 6) Assume you have now transformed E. coli with this complete plasmid DNA including your gene and it worked to make the protein that you wanted. Show the protein sequence of your final product that would result from expression of this plasmid. **(Show the translated DNA sequence from step 5 so please show the amino acid sequence of what your expressed protein would be. Remember that it will need to included the His-tag in the same frame so you must check if your open reading frame makes sense. You can use an online translation tool like: https://web.expasy.org/translate/. Show what the amino acid sequence will look like if you were to express this protein by translating the region between the first Methionine after the Ribosome Binding Site until you reach the first stop codon). If you do see a stop codon prior to that then your reading frame is likely mis-aligned and you will nee to re-adjust your forward PCR primer to either include an additional nucleotide or two nucleotides to get your insert it into the correct reading frame with the rest of your pET 14p vector. This additional nucleotide (or two) can be placed between the restriction site and the start of your gene when designing your forward primer. PET-14p sequence landmarks T7 promoter T7 transcription start His Tag coding sequence Multiple cloning sites (KpnI - Spe I) T7 terminator pBR322 origin bla coding sequence 646-662 645 554-571 510-526 404-450 Sca (4156) 2845 Pvu I(4046) 3606-4463 Pst I(3921) Eam1105 I(3676). HgiE II(3369) AlwN I(3199) Aat II(4598) Ssp I(4480) EcoR I(4669) Apo I(4669) Cla I(24) Hind III(29) <j「3606-4463) ori (2845) Bpu1102 I(458) Nhe I(229) Kpn I (510) Ball (515) Spel (522) Nco I(580) Xba I(619) Bgl II(677) SgrA I(718) Sph 1(874) EcoN I(934) -Sal I(959) PshA I(1024) PET-14P (4671bp) Eag I(1247) Nru I(1282) ApaB I(1360) BspLU11 I(2783) Afl III(2783) Sap I(2667) Bst1107 I(2554) BsaA I(2535) Tth111 I(2528) BsmB I(2424) Pvu Il(2374) BspM I(1362) Bsm I(1667) Msc I(1754) Bpu10 I(1889) Bsg I(1943) PET 14p Nhe I(229) BglII T7 promoter primer #69348-3 T7 promoter Bpu1102 I(458) Kpn I (510) Ball (515) Spel (522) Nco I(580) Xba I(619) Bgl II(677) SgrA I(718) XbaI You can check the sequence recognized by the restriction sites by searching the names of the enzymes on http://www.neb.com/ rbs means the ribosome binding site ▶rbs AGATCTCGATCCCGCGAAATTAATACGACTCACTATAGGGAGACCACAAC GGTTTCCCTCTAGAAATAATTTTGTTTAACTTTAAGAAGGAGA NcoI His Tag Spe I Ball KpnI TATACCATGGGCAGCAGCCATCATCATCATCATCACAGCAGCGGCC TGGTGCCGCGCGGCAGCACTAGTGGCCAGGTACCGGCTGC TAAC AAAGCCCGA MetGlySerSerHisHisHisHisHisHisSerSerGlyLeuVal ProArgGlySerThrSerGlyGInVal ProA laAlaAsnLysAlaArg Bpu1102 I thrombin T7 terminator AAGGAAGCTGAGTTGGCTGCTGCCACCGCTGAGCAATAACTAGCATAACCCCTTGGGGCCTC TAAACGGGTCTTGAGGGGTTTTTTG LysGluAlaGluLeuAlaAlaAlaThгAlaGluGInEnd T7 terminator primer #69337-3 Same as above but text can be copied for your convenience: AGATCTCGATCCCGCGAAATTAATACGACTCACTATAGGGAGACCACAACGGTTTCCCTCTAGAAATAATTTTGTTTAACTTTAAGAAGGAGA TATACCATGGGCAGCAGCCATCATCATCATCATCACAGCAGCGGCCTGGTGCCGCGCGGCAGCACTAGTGGCCAGGTACCGGCTGCTAACAAAGCCCGA AAGGAAGCTGAGTTGGCTGCTGCCACCGCTGAGCAATAACTAGCATAACCCCCTTGGGGCCTCTAAACGGGTCTTGAGGGGTTTTTTGSee Answer
- Q11:3. The following metabolic pathway exists in a strain of bacteria. The bacteria require product D to survive. The three enzymes required to produce D are controlled by three separate genes (genes 1 to 3). A enzyme 1 B enzyme 2° C enzyme 3*D Assume that this pathway is under the control of an operon. a) In the space provided below, draw the operon for this biochemical pathway. [2 A] b)Illustrate a scenario when the levels of D are HIGH (i.e. the bacteria would not require the enzymes to produce D) [2 A] 4. The gel below, was produced by DNA sequencing. What was the original DNA template strand sequence? [4T] ddCTP ddGTP ddTTP ddATP | | |/n5. During an investigation of the promoter region of a gene you apply heat to two sections of DNA that you believe could be a promoter region. One region unwinds at 80 degrees Celsius and the other unwinds around 60 degrees Celsius. Which of the two strands contains the promoter? Explain. [4 T] 6. Use labelled diagrams incorporating the concept of the "Central Dogma of Molecular Genetics" to illustrate how a mutation can lead to disease. [4 C]See Answer
- Q12:Use the article "Fundamental mechanisms of telomerase action in yeasts and mammals: understanding telomeres and telomerase in cancer cells" (below) to answer the following questions https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5376709/ When reading the research article, read the abstract, section 1.1, 1.2, Figure 1, Figure 2, 3.1, 3.2, 3.4 and 4.1. Questions: 1. Based on what you read, what is telomerase and why is it important? (3 points) 2. Based on what you read, is telomerase evolutionarily conserved? Provide evidence from the article to support your answer (3 points)See Answer
- Q13:Inflammation is one of the crucial processes of innate immunity that alerts the immune system and recruits the needed cells, clears the damaged area, and then sets the stage for subsequent tissue repair or regeneration. Despite the importance of inflammation to immunity and the role of chronic inflammation in many common pathologies, we have only begun to gain a basic understanding of the major events that initiate, regulate, and inhibit the process. Although we have known that neutrophils are among the first cells to arrive on the scene of an inflammatory response, we have only recently begun to understand the important role that these cells play in the process. It was discovered in the mid-2000s that neutrophils form structures dubbed neutrophil extracellular traps, or NETS, when activated in an inflammatory response. Briefly, what are NETS, what is in them, and what purpose do they serve in inflammation and immunity? What are a few of the major pathologies that NETs have been implicated in?See Answer
- Q14:/n 2/9/24, 2:52 PM https://nerd.wwnorton.com/nerd/122222/r/goto/cfi/158!/4 Deep in the DNA Deep in the DNA Each day in the United States, on average 22 people die while waiting for an organ transplant. Over 113,000 men, women, and children are currently on the national transplant waiting list, each hoping their name is called before it's too late. Researchers have explored many ways to grow and store organs for transplantation-from freezing them to building them from scratch-but one of the most promising, if you can look past the mud and flies, is pigs. Our porcine friends have long been considered an excellent potential source of organs because their organs-including the heart, liver, and kidneys-are relatively close in size to human organs and because pigs and humans have similar anatomies (Figure 9.2). In addition, pigs are an easier sell to the public: people tend to prefer the idea of transplants from pigs over transplants from mammals more closely related to us, such as baboons. If we were able to transplant organs from nonhuman animals into humans, a process called xenotransplantation, healthy organs could be available in essentially limitless supply. Kidney Pancreas Liver Heart Lung Figure 9.2 Pig organs and human organs are remarkably similar in size For a nonhuman-to-human organ transplant to succeed, the nonhuman organ must fit into the space where the human organ was removed. Organs that are too big will not fit. Organs that are too small will not function at the level necessary to sustain life. 1/4 2/9/24, 2:52 PM https://nerd.wwnorton.com/nerd/122222/r/goto/cfi/158!/4 Deep in the DNA Yet there has been a barrier to harvesting pig organs for humans: the pig genome is dotted with DNA from a family of viruses called porcine endogenous retroviruses, or PERVS. Because of the presence of this viral DNA in the pig genome, pig cells produce and release PERVS-and two of the three subtypes of PERVS can infect human cells, making it risky to transplant pig organs into people for fear of making the recipients sick. After pigs acquired the viruses, PERV DNA slipped easily into the pig genome because it has the same structure as pig DNA. In fact, all living things share the same DNA structure, and species often share and swap DNA with each other. As discussed in Chapters 3-4, DNA is built from two parallel strands of repeating units called nucleotides. Each nucleotide is composed of the sugar deoxyribose, a phosphate group, and one of four bases: adenine, cytosine, guanine, or thymine. We identify nucleotides by their bases, using "adenine nucleotide" as shorthand for "nucleotide with an adenine base." The nucleotides of a single strand are connected by covalent bonds between the phosphate group of one nucleotide and the sugar of the next nucleotide. The two DNA strands are connected by hydrogen bonds linking the bases on one strand to the bases on the other, like the rungs that connect the two sides of a ladder (Figure 9.3). Covalent bonds are strong, which is important in maintaining the specific order of the nucleotides. The weaker hydrogen bonds allow the two DNA strands to be pulled apart for replication. A pairs only with T The nucleotides in one strand are paired with the nucleotides in the complementary strand. C pairs only with G. The two strands of DNA are held together by hydrogen bonds (dotted lines) between the bases. Nucleotides are linked together by covalent bonds to form one strand of DNA. Q1: Name two base pairs. SHOW ANSWER 0000 SHOW ANSWER Phosphate Sugar (deoxyribose) Sugar-phosphate Base Nucleotide Nucleotide bases: Adenine Figure 9.3 A molecule of DNA consists of two complementary strands of nucleotides that are twisted into a spiral around an imaginary axis, rather like the winding of a spiral staircase. Thymine Guanine Cytosine Q2: Why is the DNA structure referred to as a "ladder"? What part of the DNA represents the rungs of the ladder? What part represents the sides? Q3: Is the hydrogen bond that holds the base pairs together a strong or weak chemical bond? Why is that important? 2/4 2/9/24, 2:52 PM https://nerd.wwnorton.com/nerd/122222/r/goto/cfi/158!/4 SHOW ANSWER Deep in the DNA You can also see Appendix A for answers to the figure questions. The term base pair, or nucleotide pair, refers to two nucleotides held together by bonds between their bases; that is, a base pair corresponds to one rung of the DNA ladder. The ladder twists into a spiral called a double helix (Figure 9.4). Within the long, winding double helix of the pig genome, short sections of DNA from PERVS are scattered about. These PERV sections are made up of the same four nucleotides as the rest of the DNA, but they encode information for viral proteins instead of pig proteins. Figure 9.4 What DNA actually looks like In November 2012, Italian researchers used an electron microscope to directly visualize DNA for the first time. This is the single thread of double-stranded DNA that they saw. Nucleotides do not form base pairs willy-nilly. As shown in Figure 9.3, the adenine (A) nucleotide on one strand can pair only with thymine (T) on the other strand; cytosine (C) on one strand can pair only with guanine (G) on the other strand. These base-pairing rules, which provide complementary base-pairing between two nucleic acid strands, have an important consequence: when the sequence of nucleotides on one strand of the DNA molecule is known, the sequence of nucleotides on the other, complementary strand of the molecule is automatically known as well. The fact that A can pair only with T and that C can pair only with G allows the original strands to serve as "template strands" on which new strands can be built through complementary base-pairing. (In Chapter 10, we delve more deeply into building new DNA strands, including how RNA can pair with DNA, which CRISPR takes advantage of.) Still, the four nucleotides can be arranged in any order along a single strand of DNA, and each DNA strand is composed of millions of these nucleotides, so a tremendous amount of information can be stored in a DNA sequence and in a genome. The genome of the domestic pig, for example, has about 3 billion base pairs, the human genome has about 3.2 billion base pairs, a tomato has only about 900 million base pairs, and the bacterium Escherichia coli (better known as E. coli) has a measly 4.6 million. The sequence of 3/4 2/9/24, 2:52 PM https://nerd.wwnorton.com/nerd/122222/r/goto/cfi/158!/4 Deep in the DNA nucleotides in DNA differs among species and among individuals within a species, and these differences in genotype can result in different phenotypes (Figure 9.5). MIN G A C SHOW ANSWER A Human A Figure 9.5 The sequence of bases in DNA differs among species and among individuals within a species The sequence of bases in a hypothetical gene is compared for two humans (A and B) and a pig. Base pairs highlighted in blue are variant; that is, they differ between the genes of persons A and B and between the same genes in humans and pigs. SHOW ANSWER Human B Q1: If all genes are composed of just four nucleotides, how can different genes carry different types of information? Pig SHOW ANSWER Q2: Would you expect to see more variation in the sequence of DNA bases between two members of the same species (such as humans) or between two individuals of different species (for example, humans and pigs)? Explain your reasoning. Q3: Do different alleles of a gene have the same DNA sequence or different DNA sequences? You can also see Appendix A for answers to the figure questions. In the mid-1990s, scientists became very excited about the idea of using pig organs in humans, but testing stalled because of the fear that humans would become infected with PERVS. Just breeding pigs in sterile conditions can't get rid of the virus; it's integrated right there in the double helix. The Harvard Medical School team believed CRISPR might be able to solve that problem by inactivating the PERV DNA in pig cells once and for all. 4/4See Answer
- Q15:2. What is the difference between a nucleoside triphosphate and a trinucleotide?See Answer
- Q16:4. Shown is a representation of a DNA molecule being transcribed. a. Identify every 3' end and every 5' end in the picture. b. Identify the template strand. c. The nontemplate strand is also called the "sense strand." Explain.See Answer
- Q17:6. Using the pK₂ data in Table 4.1 and the Henderson-Hasselbalch equation, calculate the approximate net charge on each of a the four common ribonucleoside 5'-monophosphates (rNMPs) at pH 3.8. If a mixture of these rNMPs was placed in an electrophoresis apparatus, as shown, draw four bands to predict the direction and relative migration rate of each. Application zone +See Answer
- Q18:14. The gel electrophoresis pattern in Figure 4.23 was determined by soaking the gel in a solution of ethidium bromide (EtBr). This is a fluorescent molecule with a planar structure: H₂N- -NH₂ NBr- CH₂ CH3 Ethidium bromide (EtBr) The flat molecule intercalates, or fits directly between two adjacent base pairs in a double helix. In doing so, it unwinds the double helix by 26° for each ethidium molecule bound. a. If EtBr was added to relaxed, closed circular DNA, would you expect positive or negative supercoiling to occur? Explain. b. If the circular DNA were nicked (had a single-stranded break) on one strand, what would be the effect on supercoiling? c. If negatively supercoiled DNA is titrated with EtBr, the electrophoretic mobility decreases at first but then increases at higher EtBr concentrations. Explain.See Answer
- Q19:18. a. What two enthalpic factors stabilize DNA in double-helical form at low temperature? b. What entropic factor destabilizes helical DNA at high temperature? c. Why is the double-helical structure of DNA stabilized at moderate to high ionic strength?See Answer
- Q20: Writing Lab Reports General Comments The objective for scientific report writing for this course is to teach you how to prepare Figures and Tables that are notionally suitable for submission to a scientific journal for publication. The style that will be used is partially based upon the American Chemical Society (ACS) journal Biochemistry and is described in detail below. You are NOT required to write complete journal type scientific reports for the experiments that you will perform this term. Instead, you will analyze your data and prepare any graphs or tables that are required. You should be aware that preparing titles and captions for Figures and Tables is somewhat subjective and that different levels of detail can be found between different articles in the Biochemistry journal. However, for purposes of standardization and marking fairness you are required to adhere to the format provided in the relevant sections below even if you can find contrary examples in Biochemistry. No sample calculations are required for your reports Collecting and Reporting Data Recording Data Students are expected to record their data as they collect it. There are two places for data to be recorded: 1. Within the body of the protocol, and 2. In the data record sheet that can be found at the end of each experimental protocol. The reason for organizing data recording in this manner is so that you can record your data in your manual while working through the experimental protocols without having to flip pages to do so. The recording of data in the data record sheet is done as you must submit a picture of this data sheet with your submitted lab reports so that the marker can verify that information entered into the excel data sheet that you submit to your submission folder is correct. You are required to complete your data record sheet prior to leaving the lab as you are required to have it signed by your TA before departing. Significant figures Any measured value has error, and significant figures are used to provide information about the precision and uncertainty associated with any measurement and calculations based upon them. The number of significant figures in a number can be unambiguously indicated through the use of scientific notation, and the number of significant figures that are to be used when reporting a value is determined by the equipment used to make the measurement. (See the discussion above regarding recording data-remember the last digit on the right indicates the level of certainty about a measurement). Determining the Number of Significant Figures (All numbers in a number are significant unless they are zeros. Zeroes may or may not be significant according to the following rules: 1. Zeros are always significant when they are found in the middle of a number. e.g. 407.02 has five significant figures. 157 2. Zeros are significant when they are on the end of a number on the right hand side of a decimal place. e.g. 39.50 has four significant figures; 44.00 also has four significant figures. 3. Zeros between a decimal and a number on the right hand side of a decimal are not significant. e.g. 0.004 has only one significant figure; 0.0040 has two significant figures 4. Zeros at the end of a number but to the left of the decimal are not significant but they generate ambiguity. e.g. 50 has only one significant figure. If you wanted to indicate the number 50 to two significant figures you would have to use scientific notation and write it as 5.0x 10. However, it will be unclear to the reader whether you meant to indicate one or two significant figures, so to eliminate uncertainty always use scientific notation in these situations. Calculations with Significant Figures When you are performing calculations with your data, execute the calculation using all available digits from the equipment you are using (even if you are using numbers with different numbers of significant figures). However when you report the final answer for the calculation you must use the number of significant figures that are appropriate for the type of calculation or marks will be deducted. The rules for performing addition and subtraction calculations with significant figures are different than the rules used for multiplication and division. You must report your data according to the correct set of rules. Addition and Subtraction The number of significant figures is determined by the number of decimals places in the least precise number. This may result in your answer having more or fewer significant figures than the original numbers that were used for the calculation. e.g. 14.421 has five significant figures and is precise to the third decimal point. 0.21 has two decimal points and is accurate only to the second decimal point. If 14.421 is added to 0.21 your answer before taking into account significant figures is 14.631. To report this value using significant figures you report 14.63 to indicate that the uncertainty lies in the second decimal point. If you subtracted 3.22 from 3.68 your answer would be 0.46. The answer only has two significant figures even though your original numbers each had three significant figures. Multiplication and Division The number of significant figures is determined by the number of significant figures in the number with the fewest significant figures. e.g. 6.332 multiplied by 2.40 equals 15.1968, but to report this value using significant figures you would report 15.2 since there are only three significant figures in the number with the least number of significant figures. 158 Presenting Experimental Data as Figures Note: This section will focus upon preparing graphs since most of the figures that you will prepare this term will be graphs. Background Presenting data in a graphical form is intended to give a clear presentation of the results that were obtained for a particular experiment or experiments. Graphs should be self- explanatory and this means that a person with relevant scientific knowledge should be able to look at the graph and understand what it means without reading the entire report. You must use computer graphing software to prepare graphs. You must show your data points on the graph and one of the following types of lines must be present: (i) a straight line, (ii) a smooth curve, or (iii) each point joined together. The particular situation will dictate which type of line is to be used. A best-fit line and its equation of the line is required for calibration curves or when a linearization of graphed data is required. Place the equation of the line on the graph next to the best-fit line. When you use a best-fit line DO NOT also join the points with a second line. A calibration curve is not forced through zero. Column graphs will be used only when you are directed to do so. Axes The ordinate (y-axis) should be used for the dependent variable and the abscissa (x-axis) for the independent variable. (NOTE: dependent is measured, independent is chosen.) The axes should be clearly labeled and the correct units provided in brackets immediately thereafter. The axis label should unambiguously identify its variable. In the example graph provided by Figure 45 on the next page, data was obtained by measuring the absorbance of a solution of 4-nitroaniline at different wavelengths. Absorbance was measured so this variable belongs on the y-axis, and the wavelengths were chosen, so they by default belong on the x-axis. Absorbance and wavelength do not require further explanation as their meaning is well-understood by a scientific reader. Absorbance has no units and the units for wavelength are nm. Figure Caption Each graph must be accompanied by a figure caption that is found below the graph. A figure caption has several components, (see sample figure caption provided below): 1. Figure number 2. Figure title 3. Additional Information The totality of the information in the figure caption should answer the following questions: a) What type of data is being depicted? If the figure is a graph you should indicate what type of graph it is, e.g. calibration curve, binding plot etc. b) What compounds/organisms etc. were used to gather the data from which the graph was prepared? c) Under what conditions or in what? For example, if the experiment was performed in a buffer, identify the buffer in full. If a special reagent was used, identify it. 160 Understanding and Using Dilutions Basic calculations Understanding dilutions is an important concept for anyone working in a laboratory. Typically, dilutions are required to ensure that measurements are within the reliable recording range of the instrument/technique that is being used to investigate a problem, or to lower the concentration of stock solutions to a working concentration. Dilutions will be expressed in this lab manual as Fold Dilutions or as a dilution factor which refers to how much the solution of interest has been diluted. Fold dilution can be calculated as follows: fold dilution = final dilution volume original volume of what you are diluting Example 1: if you diluted 15 mL of 0.1 M substance X by adding 85 mL of water the fold dilution would be 100 mL/15 mL which equals a 6.67 fold dilution. If you want to calculate the new concentration of substance X simply divide the original concentration by the fold dilution and you will have your answer. This last sentence can be expressed as: original concentration fold dilution 3.53 new concentrat ion - So in this example the final concentration of substance X after you diluted it is 0.015 M. Example 2: if you were instructed to make 100 mL of a 20 fold dilution of solution Y, just set up the first equation above to solve for the original volume of solution Y that you need. The original volume you need would be 100 mL/20 fold dilution which equals 5 mL. So you would mix 5 mL of Y with 95 mL of water/buffer. Example 3: Let us say you made a calibration curve for some analysis that ranged from 0.1 mg/mL to 1 mg/mL, but when you analyzed a sample from your experiment you obtained a reading that was much higher than the range of your standards. Since you do not know how your calibration curve behaves outside the range you have you do not know for sure the concentration of your sample. So if you took 1 mL of your sample and diluted it 5 fold, (you added 4 mL of water or buffer to your sample and mixed it), and then obtained a new reading of 0.42 mg/mL you would not be done because this is not the concentration of your sample but the diluted concentration. To obtain the original concentration you would multiply the new concentration (0.42 mg/mL) by the fold dilution (5 times) and you would now know the original concentration of your sample is 2.1 mg/mL. Multiple dilutions If you are performing a series of dilutions, the dilution factor for each step is calculated according to the examples provided above. However, the overall dilution for a sequence of dilutions of one analyte is the product of each of the individual dilution factors. Example: If a solution of substance X was diluted by mixing 5 mL of it with 95 of ethanol, and then 10 mL of the substance X and ethanol mixture was diluted with 90 mL of water, the total dilution of the substance X in the water would be 200 fold. This is because the first dilution was 20 fold (100/5) and the second dilution was 10 fold (100/10), and the product of 20 x 10 = 200. If you knew the original concentration of substance X was 10 mM, the final concentration would be 0.05 mM or 50 μM. 159 the number (i.e. don't use bold or italics). Number the figures using Arabic numerals (1, 2, 3 etc.). Follow the number with a period (i.e. don't use a colon). 2. Figure title The figure title should be a brief informative description of what the graph depicts or is about. It is preferable that the figure title is in non-sentence format. The title is where the type of graph (if there is a technical name for it) and the principal subject matter of the graph should be identified-it should not be just a repeat of the labels on the axes. For example, the graph above is obtained by measuring the absorbance of a solution of 4-nitroaniline at different wavelengths. Absorbance versus wavelength is not an appropriate title. An example of a more appropriate title for such a graph might be "Absorption spectrum of 4-nitroaniline." 4- nitroaniline is the subject of the graph and the phrase "absorption spectrum" implies to the scientifically-trained reader a particular relationship in this case the effect of changing the wavelength on the light-absorbing properties of the solution. Information about the various conditions under which this data is obtained is of secondary importance and is not title worthy and therefore is relegated to the additional information portion of the figure caption. For some graphs there may be multiple subjects all of which are title worthy. If there are only two subjects, identify them explicitly in the title, but if there are more than two, use a more generic group name for them and then identify them explicitly in the additional information or legend. For example, you might have a graph that looks at some property of various amino acids; if there were only two amino acids used, name them in the title, but if there were three or more it would be better to just use the word amino acid in the title rather than listing them all out. When you are using generic titles, still try to be specific as possible. For this amino acid example, it would be better to use "amino acids" as the generic group name rather than say "organic molecules". Using "organic molecules" while not technically incorrect as a group name, is too vague and would not really give the reader much insight into what the graph is about. Remember, graphs (and tables too) are intended to convey information in a concise and succinct manner. 3. Additional information You will frequently find that axis labels and the figure title do not provide sufficient information about the graph for an educated reader to understand it, or be aware of important information about it and thus additional information is usually added. The additional information portion of the figure caption is separated from the figure title by a period. The additional information may be a single or multiple sentences depending upon the requirements of the figure. Information that belongs in this part of the figure caption is all available information, which if different, might or would change the results presented in the graph. To elaborate on this point based upon Figure 45, you know that concentration affects absorbance so indicating the concentration of the 4-nitroaniline is crucial. What other compounds are present will often affect measurements, so the final working concentrations of any other substances that are present must be given. In our working example, there is only a buffer, so we give its full final concentration and pH. You will also know that cuvette path length affects absorbance, so specifying the path length is important, but because all the cuvettes used for this course have a path length of 1 cm and we are limiting you to 5 lines for your figure caption, path length will not be a required component of your additional information. 162/n Experiment 5: Purification of Spinach Genomic DNA and Agarose Gel Electrophoresis Analysis of Lambda DNA Restriction Digests Introduction to molecular biology Since the 1960s the development of the collection of methodologies commonly known as molecular biology has revolutionized the fields of biochemistry, cell biology, evolution and genetics. In particular, techniques such as the polymerase chain reaction (PCR) and DNA cloning have made it possible to study in great detail the molecular basis of the relationship between structure and function in many biological molecules. Molecular biology techniques have also been used to begin to unravel the intricate signalling networks that enable individual cells, tissues and organisms to cope with the dynamic challenges of their environment. During this week you will learn about and perform several of the core methods used in molecular biology extraction of DNA from a tissue; digestion of DNA using restriction enzymes; quantification of purified DNA by absorbance spectroscopy; and analysis of purified DNA by agarose gel electrophoresis. DNA Purification, Restriction Enzymes and Molecular Cloning DNA Purification control. P The first thing that a researcher needs to do if they want to analyze or manipulate DNA is to find a way to get it out of the cells that contain it, and then to purify it to ensure that there are no contaminating molecular species present that may distort experimental results or interfere + with any subsequent reactions (PCR, digestion, ligation etc.) to which the DNA might be subjected. Historically, DNA extraction and purification procedures utilized potentially hazardous organic solvents with complicated, delicate procedures that were difficult for less experienced researchers to replicate. Presently, well-equipped modern molecular biology labs make extensive use of commercially prepared kits that have simplified and standardized the extraction and purification of DNA that make it relatively easy for novice experimenters to obtain good yields of pure, high-quality DNA. In the simplest terms, after the cells of the tissue have been lysed, the kits work by selectively degrading and precipitating non-DNA-biological molecules which are removed by centrifugation or by washing them off a column. The columns for DNA purification are usually silica-based, and DNA molecules will bind tightly to these columns under certain ionic conditions--this allows for the removal of residual contaminants without significant loss of DNA. The DNA can then be eluted out of the column by altering the salt and pH conditions in the column. At this point the DNA can be used immediately for further investigations, or stored at -20°C until required. (column Evaluating the effectiveness of DNA purification When one is isolating or manipulating DNA, (in your case extraction and purification), one needs to determine if the procedure has worked properly, and this means of course that the DNA sample will require some form of analysis. The two most important factors for evaluating DNA quality are purity and structural integrity. The quickest and easiest way to evaluate the DNA purity is by measuring the absorbance of a small volume of the purified sample at 260 and 280 nm. DNA absorbs light well at 260 nm, whereas protein absorbs well at 280 rim (you saw this in your previous experiment), so if the 260/280 absorbance ratio is high, it suggests the DNA is relatively puré. Normally a value for this ratio between 1.8 at 1.9 is desired. However, even if one has successfully obtained pure DNA, it does not mean that it is high quality 85 Loryne 9201 A DNA sample could be pure (ie. free of protein), however, it may not be nigh quality i it is extensively degraded. Fragmentation of eukaryotic genomic DNA during purification is inevitable as the removal of histones and other stabilizing proteins leaves the DNA very fragile, and it invariably degrades into smaller pieces. This itself is not problematic so long as the average pieces are relatively large-say larger than 10 kilobases, because pieces of this size will leave numerous copies of individual genes and regulatory DNA intact, and therefore the DNA sample can be still be used for whatever purposes it was originally purified. If the DNA were smashed into much smaller fragments, it would be much less likely to contain intact pieces of desired sequences, and therefore unlikely to generate good results in the subsequent stages of a longer experimental protocol. Since size is one of the properties of DNA that can tell a researcher something about its quality, having a method that can be used to quickly and cheaply determine the approximate size of DNA is a useful tool to aid the evaluation of DNA quality. One of the simplest and most widespread methods to analyze DNA size is to perform agarose gel electrophoresis upon a few microliters of the sample. Restriction enzymes and molecular cloning One of the most significant technical advances in the history of biochemistry was the development of methods that allow researchers to transfer a small portion of DNA (say a gene of interest) from one DNA molecule into a different DNA molecule and then making numerous /\ copies of the new hybrid molecule. This process of recombining DNA is known as molecular cloning. We will not consider molecular cloning further in this course as it is outside the scope of our objectives, but the ability to do cloning requires the ability to cut DNA in a very precise manner. This ability to cut out a piece of DNA is dependent upon a specialized group of enzymes found in bacteria known as restriction enzymes. Restriction enzymes are proteins that cut the sugar phosphate backbone of DNA only when a specific sequence of nucleotides is present. For example, the restriction enzyme Xhol only cuts between nucleotides C and T of the sequence iausixe oplem 5' CITCGAG 3' 3' GAGCTIC 5' og nitido When a piece of DNA is incubated in the presence of a restriction enzyme, the enzyme will cut the DNA wherever it finds its target sequence, and therefore a series of fragments will be produced depending upon the specific sequence of the DNA and the restriction enzyme used. This incubation and cutting process is known as "digestion", and it is the precise control over the restriction digest fragments that enables experimenters to assemble different pieces of DNA from different DNA molecules into a new DNA molecule with the desired nucleotide sequence. In the lab for this week, you will observe how different fragments are produced when different restriction enzymes are used to cut the same sample of DNA. Electrophoresis Basic Principles DO +1- When charged molecules are placed in solution and exposed to an electrical field, the molecules will have the tendency to migrate to the pole opposite to their charge. This is illustrated in the picture below depicting how negatively charged molecules migrate towards the positively charged anode, whereas the positively charged molecules will migrate towards the negatively charged cathode. Anode doo bo Figure 23. Direction of charged particle movement during electrophoresis. Cathode V= This process can be exploited to separate charged molecules, and not surprisingly, has been extensively utilized in biochemistry and related fields for a wide variety of useful applications which are grouped together under the rather generic name of electrophoretic techniques. 1 Electrophoresis can be described as the examination of the movement of charged particles through solution under the influence of an electric field, and apart from the fact that oppositely charged molecules move in opposite directions, one of the most useful attributes of electrophoresis is the fact that the speed at which a molecule moves towards the oppositely charged pole, the migration velocity, is different for different molecules. The migration velocity of any particular molecule is dependent upon a variety of factors which can be very simply represented by the equation: speed Eq f (1) where v is the velocity of the charged particle; E is the strength of the electric field to which the charged particle is exposed; q is the net charge on the particle, and fis the frictional coefficient of the particle as it moves through the solution. 加速 When you apply an electric field (E) to charged molecules in solution they will start to accelerate towards the pole of the opposite charge. How fast they accelerate depends upon both the strength of the field (E), and how much charge (g) the molecule carries. However, as the molecule begins to move faster through solution the viscous drag that results from moving through solvent molecules also increases, and this drag is influenced by the size and shape of the molecule. The viscous drag is a force of opposite direction to that of the migrating particle. The size and shape of the molecule is represented by the frictional coefficient (f). Thus, the overall of = V (Steady stute velocity effect is that almost immediately after the voltage is applied and maintained at a constant level, the force accelerating the molecule forward, (Eq), is equalized by a force. (. decelerating it, and a STEADY STATE velocity or migration speed through the solvent is attained. This steady state velocity is denoted by the term from the above equation. It should be obvious upon examination of equation 1 that in any specific electrical field that the E is the same for all molecules in that field, and therefore the migration velocity of any particular molecule will depend upon its q/f ratio. This is sometimes mistakenly referred to its "charge-to-mass ratio", but in the context of solution-based electrophoresis, the magnitude of fis extremely complex and as indicated in the previous paragraph, is determined in part by the size and shape of the charged particle. We will not delve into the details of how the frictional coefficient can be modelled, and instead our attention will turn to the application of the basic principles of electrophoretic theory to practical considerations for the electrophoresis of nucleic acids. Electrophoresis of DNA The picture below is a representation of the structure of DNA at a neutral pH, and the structural element that is most relevant for electrophoresis is the regularity of the negatively charged phosphate groups (yellow circles) along the length of the structure. H₂C 03. 05 H₂C H₂C base H base Enter base 88 Figure 24. Structure of DNA. The phosphate groups impart an overall negative charge to the molecule which means that it will migrate towards the positively charged pole when voltage is applied. However, since the amount of negative charge on a DNA molecule is proportional to its length, DNA molecules of different length have a similar q/f ratio and in theory migrate at a similar steady state velocity. This means of course that all DNA molecules regardless of their size would migrate at the same speed in electrophoresis. This should lead to the question of how electrophoresis can be useful to the researcher if all the DNA molecules behave the same. The answer to this can be found in the utilization of various types of gels. Agarose gel electrophoresis To separate DNA molecules of different lengths we need to perform the electrophoresis in a porous, solvent-filled mechanical support that is called a gel. Gels for molecular biology can be made from any material that forms a sieving matrix that has minimal interaction with the molecules that one is trying to separate. For DNA, the gels are usually made from agarose. Agarose is a linear polysaccharide that forms a mesh network in aqueous solution when it is melted in buffer and allowed to re-solidify. Agarose gels are typically used for the electrophoretic separation of large biomolecules such as nucleic acids and also of proteins with a mass greater than 250 kDa. The mesh network formed by solidification of molten agarose creates a gelatinous like substance that is filled with pores of varying sizes. One can yary the average pore size by changing the concentration of the agarose, with higher agarose concentrations resulting in a lower average pore size. When you apply a DNA sample to one end of a gel and activate the voltage, the negatively-charged DNA will begin to migrate towards the positively charged pole at the opposite end of the gel (see pictures below). While the solvent molecules can still easily pass KI Side view: Sample loaded into well Negative (-) Electrode Gel Electric field and direction of migration 89 Plastic gel box. buffer Positive (+) Electrode Figure 25. Photograph of DNA samples being loaded on an agarose gel (left), and Side view drawing of the electrophoresis chamber for an agarose gel (right). through the gel unimpeded, larger molecules like DNA can only migrate through the larger pores. Since there is a distribution of pore sizes, shorter DNA molecules will have a greater number of pores that they can migrate through compared to the larger DNA molecules. Since the smaller DNA molecules are able to more frequently find a pore that they can fit through, they migrate faster through the gel than the larger DNA molecules because they in effect travel a shorter distance for any linear distance along the axis of the gel in their journey towards the other end of the gel. We of course see this as them moving further along the gel compared to larger DNA molecules. This is what makes electrophoresis such a useful technique-it separates DNA molecules on the basis of their size. Once the electrophoresis is completed, one must have some way of identifying how far the DNA has moved-on-the-gel. Since DNA is not readily visible on the gel, chemicals that are easily detected and interact only with the DNA are used to "see" the DNA. There are many different dyes that can be used to visualize DNA, and the dye that you will be using in this lab is known by the trade name as Red SafeTM. This dye will be incorporated into the molten agarose solution before you come to the lab. To determine the size of the DNA molecules subjected to electrophoresis, one compares the migration distance of the DNA sample to the migration distance of a standard solution of DNA molecules of known sizes that were run on the same gel as the sample. Samples of DNA of known sizes designed to be run on gels are called DNA ladders.See Answer
Popular Subjects for Molecular Biology
You can get the best rated step-by-step problem explanations from 65000+ expert tutors by ordering TutorBin Molecular Biology homework help.Get Instant Molecular Biology Solutions From TutorBin App Now!
Get personalized homework help in your pocket! Enjoy your $20 reward upon registration!
"After using their service, I decided to return back to them whenever I need their assistance. They will never disappoint you and craft the perfect homework for you after carrying out extensive research. It will surely amp up your performance and you will soon outperform your peers."
"Ever since I started using this service, my life became easy. Now I have plenty of time to immerse myself in more important tasks viz., preparing for exams. TutorBin went above and beyond my expectations. They provide excellent quality tasks within deadlines. My grades improved exponentially after seeking their assistance."
"They are amazing. I sought their help with my art assignment and the answers they provided were unique and devoid of plagiarism. They really helped me get into the good books of my professor. I would highly recommend their service."
"The service they provide is great. Their answers are unique and expert professionals with a minimum of 5 years of experience work on the assignments. Expect the answers to be of the highest quality and get ready to see your grades soar."
"They provide excellent assistance. What I loved the most about them is their homework help. They are available around the clock and work until you derive complete satisfaction. If you decide to use their service, expect a positive disconfirmation of expectations."
TutorBin helping students around the globe
TutorBin believes that distance should never be a barrier to learning. Over 500000+ orders and 100000+ happy customers explain TutorBin has become the name that keeps learning fun in the UK, USA, Canada, Australia, Singapore, and UAE.
