Integrating Concepts in Biology

Integrating Concepts in Biology
PowerPoint Slides for Chapter 1: Heritable Material 1.4 How does DNA’s shape affect its function? by A. Malcolm Campbell, Laurie J. Heyer, & Christopher Paradise Title Page Copyright © 2015 by AM Campbell, LJ Heyer, CJ Paradise. All rights reserved.

Biology Learning Objectives
Draw the structure of DNA showing the double helix and base pairings. Demonstrate how DNA replication is semiconservative. Evaluate experimental design and analyze data from research on DNA as molecular information. Bio-Math Exploration Learning Objectives Estimate the amount of DNA in a sample by finding the area under a curve. Determine the number of generations that have passed in a population of cells. Copyright © 2015 by AM Campbell, LJ Heyer, CJ Paradise. All rights reserved.

Nucleotide Structures
Figure 1.9 Nucleotide structures. Atoms are located at the bent corners, and covalent bonds between them are colored rods of the same colors as the atoms. Carbon is gray, oxygen is red, phosphorous is orange, nitrogen is blue, and hydrogen is white. A, Used for energy and RNA synthesis, adenine triphosphate (ATP) looking at the face of the pentagonal ribose in the middle, three phosphates extending to the left, and the base (adenine) seen on edge. B, Used for DNA synthesis, deoxyribose adenine triphosphate (dATP) lacks one oxygen on the bottom right corner of the pentagon ribose. Talking Point: Students need to recognize the difference between DNA and RNA monomers. You should be able to distinguish deoxyribonucleic acid (DNA) from ribonucleic acid (RNA), Fig.1.9 Copyright © 2015 by AM Campbell, LJ Heyer, CJ Paradise. All rights reserved.

Figure 1.9 You should be able to tell which nucleotide is deoxyribonucleic acid and which one is ribonucleic acid. The arrows direct your attention to the only difference between ATP and dATP. Talking Point: Look at the bottom right corner of ribose to see the presence of oxygen in RNA and absence of oxygen in DNA. Arrow directs your attention to the right position. You should be able to distinguish deoxyribonucleic acid (DNA) from ribonucleic acid (RNA), Fig. 1.9 Copyright © 2015 by AM Campbell, LJ Heyer, CJ Paradise. All rights reserved.

Figure 1.9 RNA ribose has oxygen attached to 2’ carbon but DNA deoxyribose lacks this oxygen. Talking Point: Look at the bottom right corner of ribose to see the presence of oxygen in RNA and absence of oxygen in DNA. RNA nucleotide ATP with ribose on the left and deoxyribose of dATP on the right. RNA ATP DNA dATP Fig. 1.9 Copyright © 2015 by AM Campbell, LJ Heyer, CJ Paradise. All rights reserved.

rotate 90° Figure 1.9 Nucleotide structures. Atoms are located at the bent corners, and covalent bonds between them are colored rods of the same colors as the atoms. Carbon is gray, oxygen is red, phosphorous is orange, nitrogen is blue, and hydrogen is white. A, Used for energy and RNA synthesis, adenine triphosphate (ATP) looking at the face of the pentagonal ribose in the middle, three phosphates extending to the left, and the base (adenine) seen on edge. B, Used for DNA synthesis, deoxyribose adenine triphosphate (dATP) lacks one oxygen on the bottom right corner of the pentagon ribose. C, Rotated view of dATP with the adenine base displayed fully on the right side. D, Line diagram of dATP. Talking Point: Rotate top left panel to see the structure of adenine. Line drawing shows all the atoms in this flattened depiction. dATP 2’ carbon Fig. 1.9 Copyright © 2015 by AM Campbell, LJ Heyer, CJ Paradise. All rights reserved.

ATP or dATP? Figure 1.9 Rotation allows you to see the structure of the adenine base that contains 5 nitrogen atoms (blue). 2’ carbon is the site of the only difference between ATP and dATP. Talking Point: Students should recognize this as ATP with the 2’ OH group. Fig. 1.9 Copyright © 2015 by AM Campbell, LJ Heyer, CJ Paradise. All rights reserved.

Watson and Crick DNA Model
Figure 1.10 Structure of DNA. A, The only figure in the Watson and Crick paper published April 25, 1953, shows the double helix and antiparallel strands (arrows pointing in opposite directions) and the bases as solid bars connecting the ribbon-representation of the sugars and phosphates. B, In their May 30, 1953 paper, they diagrammed a little more clearly how the three parts of the nucleotides would be organized. Talking Point: Original figures published by Watson and Crick are helpful but the one on the right is not accurately portrayed. Fig. 1.10 from Watson and Crick, 1953

Figure 1.10 Structure of DNA. A, The only figure in the Watson and Crick paper published April 25, 1953, shows the double helix and antiparallel strands (arrows pointing in opposite directions) and the bases as solid bars connecting the ribbon-representation of the sugars and phosphates. B, In their May 30, 1953 paper, they diagrammed a little more clearly how the three parts of the nucleotides would be organized. Talking Point: Redrawn double helix showing the two strands of deoxyribose sugars in red and blue and base pairs in green. Fig. 1.10 from Watson and Crick, 1953

Figure 1.10 Structure of DNA. A, The only figure in the Watson and Crick paper published April 25, 1953, shows the double helix and antiparallel strands (arrows pointing in opposite directions) and the bases as solid bars connecting the ribbon-representation of the sugars and phosphates. B, In their May 30, 1953 paper, they diagrammed a little more clearly how the three parts of the nucleotides would be organized. Talking Point: The flattened figure in panel B is supposed to be equivalent to panel A, but panel B contains a small error. Fig. 1.10 from Watson and Crick, 1953

Figure 1.10 Structure of DNA. A, The only figure in the Watson and Crick paper published April 25, 1953, shows the double helix and antiparallel strands (arrows pointing in opposite directions) and the bases as solid bars connecting the ribbon-representation of the sugars and phosphates. B, In their May 30, 1953 paper, they diagrammed a little more clearly how the three parts of the nucleotides would be organized. Talking Point: The flattened figure in panel B is supposed to be equivalent to panel A, but panel B contains a small error. Color coding of panel B helps you relate the double helix in panel A to the flattened version in panel B. Fig. 1.10 from Watson and Crick, 1953

Figure 1.10 Structure of DNA. A, The only figure in the Watson and Crick paper published April 25, 1953, shows the double helix and antiparallel strands (arrows pointing in opposite directions) and the bases as solid bars connecting the ribbon-representation of the sugars and phosphates. B, In their May 30, 1953 paper, they diagrammed a little more clearly how the three parts of the nucleotides would be organized. Talking Point: Slide prompts students to search for error in panel B. Can you spot their error? Fig. 1.10 from Watson and Crick, 1953

Figure 1.10 Structure of DNA. A, The only figure in the Watson and Crick paper published April 25, 1953, shows the double helix and antiparallel strands (arrows pointing in opposite directions) and the bases as solid bars connecting the ribbon-representation of the sugars and phosphates. B, In their May 30, 1953 paper, they diagrammed a little more clearly how the three parts of the nucleotides would be organized. Talking Point: Look for error using original versions of figures. Can you spot their error? Fig. 1.10 from Watson and Crick, 1953

phosphate Figure 1.10 Structure of DNA. A, The only figure in the Watson and Crick paper published April 25, 1953, shows the double helix and antiparallel strands (arrows pointing in opposite directions) and the bases as solid bars connecting the ribbon-representation of the sugars and phosphates. B, In their May 30, 1953 paper, they diagrammed a little more clearly how the three parts of the nucleotides would be organized. Talking Point: Phosphates should be on the 5’ end and the two strands should be anti-parallel. Therefore move the bottom right phosphate to top right. should be anti-parallel strands Fig. 1.10 from Watson and Crick, 1953

Chemical Bonds Figure 1.11 Three types of chemical bonds. Hydrogen bonds are the weakest and form between two partially charged, δ, atoms (frequently nitrogen or oxygen) with hydrogen. Ionic bonds are intermediate in strength and form between two fully charged atoms of opposite polarity, frequently between an acid and a base. Covalent bonds are the strongest, can be single or double bonds, and are formed by sharing electrons. Talking Point: Encourages students to connect three types of chemical bonds with their analogies, the chemical representations of the three bonds, and atoms commonly involved in such bonds within biological systems. Fig. 1.11 Copyright © 2015 by AM Campbell, LJ Heyer, CJ Paradise. All rights reserved.

Chemical Bonds You should be able to match these representations with
the appropriate chemical bonds. Figure 1.11 Three types of chemical bonds. Hydrogen bonds are the weakest and form between two partially charged, δ, atoms (frequently nitrogen or oxygen) with hydrogen. Ionic bonds are intermediate in strength and form between two fully charged atoms of opposite polarity, frequently between an acid and a base. Covalent bonds are the strongest, can be single or double bonds, and are formed by sharing electrons. Talking Point: Encourages students to connect three types of chemical bonds with their analogies, the chemical representations of the three bonds, and atoms commonly involved in such bonds within biological systems. Fig. 1.11 Copyright © 2015 by AM Campbell, LJ Heyer, CJ Paradise. All rights reserved.

Chemical Bonds Figure 1.11 Three types of chemical bonds. Hydrogen bonds are the weakest and form between two partially charged, δ, atoms (frequently nitrogen or oxygen) with hydrogen. Ionic bonds are intermediate in strength and form between two fully charged atoms of opposite polarity, frequently between an acid and a base. Covalent bonds are the strongest, can be single or double bonds, and are formed by sharing electrons. Talking Point: Which bond is similar to a PostIt note, a magnet and a nail hammered into wood? Fig. 1.11 Copyright © 2015 by AM Campbell, LJ Heyer, CJ Paradise. All rights reserved.

Chemical Bonds Figure 1.11 Three types of chemical bonds. Hydrogen bonds are the weakest and form between two partially charged, δ, atoms (frequently nitrogen or oxygen) with hydrogen. Ionic bonds are intermediate in strength and form between two fully charged atoms of opposite polarity, frequently between an acid and a base. Covalent bonds are the strongest, can be single or double bonds, and are formed by sharing electrons. Talking Point: Correct matching for analogies and their bonds. Fig. 1.11 Copyright © 2015 by AM Campbell, LJ Heyer, CJ Paradise. All rights reserved.

Chemical Bonds Match the chemical representations with
the appropriate bonds. Figure 1.11 Three types of chemical bonds. Hydrogen bonds are the weakest and form between two partially charged, δ, atoms (frequently nitrogen or oxygen) with hydrogen. Ionic bonds are intermediate in strength and form between two fully charged atoms of opposite polarity, frequently between an acid and a base. Covalent bonds are the strongest, can be single or double bonds, and are formed by sharing electrons. Talking Point: Prompt for the next set of matching. Fig. 1.11 Copyright © 2015 by AM Campbell, LJ Heyer, CJ Paradise. All rights reserved.

Chemical Bonds Figure 1.11 Three types of chemical bonds. Hydrogen bonds are the weakest and form between two partially charged, δ, atoms (frequently nitrogen or oxygen) with hydrogen. Ionic bonds are intermediate in strength and form between two fully charged atoms of opposite polarity, frequently between an acid and a base. Covalent bonds are the strongest, can be single or double bonds, and are formed by sharing electrons. Talking Point: Which of these chemical representations match each of the chemical bonds? Fig. 1.11 Copyright © 2015 by AM Campbell, LJ Heyer, CJ Paradise. All rights reserved.

Chemical Bonds Figure 1.11 Three types of chemical bonds. Hydrogen bonds are the weakest and form between two partially charged, δ, atoms (frequently nitrogen or oxygen) with hydrogen. Ionic bonds are intermediate in strength and form between two fully charged atoms of opposite polarity, frequently between an acid and a base. Covalent bonds are the strongest, can be single or double bonds, and are formed by sharing electrons. Talking Point: Correct answers to matching chemical representations with the appropriate bonds. Fig. 1.11 Copyright © 2015 by AM Campbell, LJ Heyer, CJ Paradise. All rights reserved.

Chemical Bonds Describe the general
rules for each bond type found in biological examples. Figure 1.11 Three types of chemical bonds. Hydrogen bonds are the weakest and form between two partially charged, δ, atoms (frequently nitrogen or oxygen) with hydrogen. Ionic bonds are intermediate in strength and form between two fully charged atoms of opposite polarity, frequently between an acid and a base. Covalent bonds are the strongest, can be single or double bonds, and are formed by sharing electrons. Talking Point: Prompt for the final set of matching atoms with bond types. Fig. 1.11 Copyright © 2015 by AM Campbell, LJ Heyer, CJ Paradise. All rights reserved.

Chemical Bonds Figure 1.11 Three types of chemical bonds. Hydrogen bonds are the weakest and form between two partially charged, δ, atoms (frequently nitrogen or oxygen) with hydrogen. Ionic bonds are intermediate in strength and form between two fully charged atoms of opposite polarity, frequently between an acid and a base. Covalent bonds are the strongest, can be single or double bonds, and are formed by sharing electrons. Talking Point: Which chemical elements typically participate in each type of chemical bonds? Fig. 1.11 Copyright © 2015 by AM Campbell, LJ Heyer, CJ Paradise. All rights reserved.

Chemical Bonds (N or O) and H Fig. 1.11 bases (+) and acids (-)
any element, often Figure 1.11 Three types of chemical bonds. Hydrogen bonds are the weakest and form between two partially charged, δ, atoms (frequently nitrogen or oxygen) with hydrogen. Ionic bonds are intermediate in strength and form between two fully charged atoms of opposite polarity, frequently between an acid and a base. Covalent bonds are the strongest, can be single or double bonds, and are formed by sharing electrons. Talking Point: Final alignment of bonds, analogies, chemical representations and common chemical elements involved in each type of bond. C – C, C – O, C – N, C – H Fig. 1.11 Copyright © 2015 by AM Campbell, LJ Heyer, CJ Paradise. All rights reserved.

X-ray Diffraction of DNA
Figure 1.12 Two x-ray film images of DNA published on the same day. A, On page 738, Wilkins et al. published this diffraction pattern to explain how the structure of DNA was deduced from the data. B, On page 740, Franklin and Gosling published their x-ray diffraction pattern of DNA. Talking Point: Comparison of two sources of data with different levels of quality. Maurice Wilkins’ data Rosalind Franklin’s data Fig. 1.12 from Wilkins, et al., 1953 and from Franklin and Gosling, 1953.

X-ray Diffraction of DNA
Figure 1.12 Two x-ray film images of DNA published on the same day. A, On page 738, Wilkins et al. published this diffraction pattern to explain how the structure of DNA was deduced from the data. B, On page 740, Franklin and Gosling published their x-ray diffraction pattern of DNA. Talking Point: The central X is clearer, including the distinct bands. The arches at the top and bottom are clearer too. Maurice Wilkins’ data Rosalind Franklin’s data Fig. 1.12 from Wilkins, et al., 1953 and from Franklin and Gosling, 1953.

Watson & Crick Base Pairs
Find their mistake. Figure 1.13 Base pairs published by Watson and Crick in their May 30, 1953 paper. Hydrogen bonds (H-bonds) form between the diminutive hydrogens (H) with slight positive charges, and either of two slightly negative atoms—oxygen or nitrogen. Scale bar is 5 angstroms (Å). Talking Point: Colorized version of figure published by Watson and Crick showing the now famous base pairs. Prompt asks you to find the error. Fig. 1.13 modified from Watson and Crick. 1953b

Find their mistake. Figure 1.13 Base pairs published by Watson and Crick in their May 30, 1953 paper. Hydrogen bonds (H-bonds) form between the diminutive hydrogens (H) with slight positive charges, and either of two slightly negative atoms—oxygen or nitrogen. Scale bar is 5 angstroms (Å). Talking Point: Immediately after studying bonds including H-bonds, students are encouraged to see the omission of a H-bond between the GC base pair above. Fig. 1.13 modified from Watson and Crick. 1953b

missed H-bond Figure 1.13 Base pairs published by Watson and Crick in their May 30, 1953 paper. Hydrogen bonds (H-bonds) form between the diminutive hydrogens (H) with slight positive charges, and either of two slightly negative atoms—oxygen or nitrogen. Scale bar is 5 angstroms (Å). Talking Point: Missing H-bond drawn in red. Fig. 1.13 modified from Watson and Crick. 1953b

Figure 1.13 Base pairs published by Watson and Crick in their May 30, 1953 paper. Hydrogen bonds (H-bonds) form between the diminutive hydrogens (H) with slight positive charges, and either of two slightly negative atoms—oxygen or nitrogen. Scale bar is 5 angstroms (Å). Talking Point: H-bonds cannot span more than ~2 Angrstoms so AT cannot have a 3rd H-bond. Fig. 1.13 too far for H-bond modified from Watson and Crick. 1953b

1 pyrimidines Figure 1.13 Base pairs published by Watson and Crick in their May 30, 1953 paper. Hydrogen bonds (H-bonds) form between the diminutive hydrogens (H) with slight positive charges, and either of two slightly negative atoms—oxygen or nitrogen. Scale bar is 5 angstroms (Å). Talking Point: Pyrimidines have one ring in their structure. 1 Fig. 1.13 modified from Watson and Crick. 1953b

1 2 purines Figure 1.13 Base pairs published by Watson and Crick in their May 30, 1953 paper. Hydrogen bonds (H-bonds) form between the diminutive hydrogens (H) with slight positive charges, and either of two slightly negative atoms—oxygen or nitrogen. Scale bar is 5 angstroms (Å). Talking Point: Purines have two rings in their structure. 1 2 Fig. 1.13 modified from Watson and Crick. 1953b

Always Three Rings Wide
1 3 2 Figure 1.13 Base pairs published by Watson and Crick in their May 30, 1953 paper. Hydrogen bonds (H-bonds) form between the diminutive hydrogens (H) with slight positive charges, and either of two slightly negative atoms—oxygen or nitrogen. Scale bar is 5 angstroms (Å). Talking Point: Purines have two rings in their structure. 1 3 2 Fig. 1.13 modified from Watson and Crick. 1953b

3 Models of DNA Replication
Figure 1.14 Three possible models for DNA replication with one piece of double-stranded DNA (dsDNA) in the center (two dark colors). A, Each original strand is a template for the other strand, and the new dsDNA molecules have one old strand (dark) and one new strand (pale). B, The original molecule remains intact (dark), and the new double helix is composed of two new (pale) strands of DNA. C, Each strand of the old is a template for the new, but all four strands of DNA are a mosaic of old and new DNA. Talking Point: Figure begins a series of annotated slides that explore the three possible modes of DNA replication. Fig. 1.14 modified from Meselson and Stahl

all old DNA What are the implications for each model? Figure 1.14 Three possible models for DNA replication with one piece of double-stranded DNA (dsDNA) in the center (two dark colors). A, Each original strand is a template for the other strand, and the new dsDNA molecules have one old strand (dark) and one new strand (pale). B, The original molecule remains intact (dark), and the new double helix is composed of two new (pale) strands of DNA. C, Each strand of the old is a template for the new, but all four strands of DNA are a mosaic of old and new DNA. Talking Point: Simple prompt to encourage you to think about each model individually. Fig. 1.14 modified from Meselson and Stahl

half old semi-conservative half new Figure 1.14 Three possible models for DNA replication with one piece of double-stranded DNA (dsDNA) in the center (two dark colors). A, Each original strand is a template for the other strand, and the new dsDNA molecules have one old strand (dark) and one new strand (pale). B, The original molecule remains intact (dark), and the new double helix is composed of two new (pale) strands of DNA. C, Each strand of the old is a template for the new, but all four strands of DNA are a mosaic of old and new DNA. Talking Point: Semi-conservative replication produces two dsDNA molecules, each with one old and one new DNA strand. Fig. 1.14 modified from Meselson and Stahl

all old conservative all new Figure 1.14 Three possible models for DNA replication with one piece of double-stranded DNA (dsDNA) in the center (two dark colors). A, Each original strand is a template for the other strand, and the new dsDNA molecules have one old strand (dark) and one new strand (pale). B, The original molecule remains intact (dark), and the new double helix is composed of two new (pale) strands of DNA. C, Each strand of the old is a template for the new, but all four strands of DNA are a mosaic of old and new DNA. Talking Point: Duplication model produces one dsDNA molecule with completely old DNA and the other with completely new DNA. Fig. 1.14 modified from Meselson and Stahl

Figure 1.14 Three possible models for DNA replication with one piece of double-stranded DNA (dsDNA) in the center (two dark colors). A, Each original strand is a template for the other strand, and the new dsDNA molecules have one old strand (dark) and one new strand (pale). B, The original molecule remains intact (dark), and the new double helix is composed of two new (pale) strands of DNA. C, Each strand of the old is a template for the new, but all four strands of DNA are a mosaic of old and new DNA. Talking Point: Mosaic replication produces four strands of DNA with each one an equal mixture of old and new DNA strands. all stands half new mosaic Fig. 1.14 modified from Meselson and Stahl

half old all old semi-conservative conservative half new all new Figure 1.14 Three possible models for DNA replication with one piece of double-stranded DNA (dsDNA) in the center (two dark colors). A, Each original strand is a template for the other strand, and the new dsDNA molecules have one old strand (dark) and one new strand (pale). B, The original molecule remains intact (dark), and the new double helix is composed of two new (pale) strands of DNA. C, Each strand of the old is a template for the new, but all four strands of DNA are a mosaic of old and new DNA. Talking Point: Each model has physical implications that can be experimentally distinguished. all stands half new mosaic Fig. 1.14 modified from Meselson and Stahl

Meselson & Stahl Experiments
low concentration (low density) high concentration of salt (high density) Figure 1.15 Centrifugation of DNA in salt gradient forms a single band. A, Fourteen different samples were centrifuged for different times and then photographed to determine how much of the DNA was collected into the central band. B, A mixture of light (14N) and heavy (15N) DNA was centrifuged and photographed. C, Graphical quantification of the data from panel B. Talking Point: Reminder that salt gradient separates molecules based on their density when centrifuged for many hours. Molecules with higher density will settle further to the right side. Fig. 1.15A from Meselson and Stahl

they wanted the DNA to appear about the same total darkness at each time point backgrounds vary due to different photographic exposure times Figure 1.15 Centrifugation of DNA in salt gradient forms a single band. A, Fourteen different samples were centrifuged for different times and then photographed to determine how much of the DNA was collected into the central band. B, A mixture of light (14N) and heavy (15N) DNA was centrifuged and photographed. C, Graphical quantification of the data from panel B. Talking Point: Each row is a photograph using UV light showing the amount and distribution of DNA from distinct experiments of centrifuging DNA for the durations indicated. Background darkness varies because the investigators wanted the amount of DNA darkness to be consistent so photographic exposure times varied. Fig. 1.15A from Meselson and Stahl

evenly distributed DNA Figure 1.15 Centrifugation of DNA in salt gradient forms a single band. A, Fourteen different samples were centrifuged for different times and then photographed to determine how much of the DNA was collected into the central band. B, A mixture of light (14N) and heavy (15N) DNA was centrifuged and photographed. C, Graphical quantification of the data from panel B. Talking Point: When the experiment began, all the DNA was equally mixed throughout the salt density gradient. Fig. 1.15A from Meselson and Stahl

partially equilibrated DNA 15 Figure 1.15 Centrifugation of DNA in salt gradient forms a single band. A, Fourteen different samples were centrifuged for different times and then photographed to determine how much of the DNA was collected into the central band. B, A mixture of light (14N) and heavy (15N) DNA was centrifuged and photographed. C, Graphical quantification of the data from panel B. Talking Point: After 15 hours of centrifugation, the DNA had begun to form a band near the center of the salt gradient. Fig. 1.15A from Meselson and Stahl

Figure 1.15 Centrifugation of DNA in salt gradient forms a single band. A, Fourteen different samples were centrifuged for different times and then photographed to determine how much of the DNA was collected into the central band. B, A mixture of light (14N) and heavy (15N) DNA was centrifuged and photographed. C, Graphical quantification of the data from panel B. Talking Point: After 36 hours of centrifugation, the DNA had stopped moving in the gradient. Centrifuging the DNA 7 more hours did not alter its location in the salt gradient. Now the investigators know how long to centrifuge the DNA (~36 hours is sufficient). 36 fully equilibrated DNA 43 Fig. 1.15A from Meselson and Stahl

density of normal (light 14N) DNA Figure 1.15 Centrifugation of DNA in salt gradient forms a single band. A, Fourteen different samples were centrifuged for different times and then photographed to determine how much of the DNA was collected into the central band. B, A mixture of light (14N) and heavy (15N) DNA was centrifuged and photographed. C, Graphical quantification of the data from panel B. Talking Point: The red line shows the density of normal DNA composed of 14^N found in the air and DNA. Fig. 1.15A from Meselson and Stahl

mix DNA made of 14N (light) and 15N (heavy) centrifuge in salt gradient Figure 1.15 Centrifugation of DNA in salt gradient forms a single band. A, Fourteen different samples were centrifuged for different times and then photographed to determine how much of the DNA was collected into the central band. B, A mixture of light (14N) and heavy (15N) DNA was centrifuged and photographed. C, Graphical quantification of the data from panel B. Talking Point: Explanation of how two bands of DNA were produced by centrifuging light and heavy DNA in a salt gradient. Fig. 1.15B modified from Meselson and Stahl

density of light (14N) DNA mix DNA made of 14N (light) and 15N (heavy) centrifuge in salt gradient Figure 1.15 Centrifugation of DNA in salt gradient forms a single band. A, Fourteen different samples were centrifuged for different times and then photographed to determine how much of the DNA was collected into the central band. B, A mixture of light (14N) and heavy (15N) DNA was centrifuged and photographed. C, Graphical quantification of the data from panel B. Talking Point: The light DNA is the same density as the bottom band from the previous figure. Heavy DNA bands settle further to the right of the light DNA. density of heavy (15N) DNA Fig. 1.15B modified from Meselson and Stahl

DNA (15N) DNA Figure 1.15 Centrifugation of DNA in salt gradient forms a single band. A, Fourteen different samples were centrifuged for different times and then photographed to determine how much of the DNA was collected into the central band. B, A mixture of light (14N) and heavy (15N) DNA was centrifuged and photographed. C, Graphical quantification of the data from panel B. Talking Point: The intensity of the dark DNA is converted to graphs in panel C. The area under the curve, and not the height of the peak, indicates relative amounts of DNA. Fig. 1.15B & C modified from Meselson and Stahl

quantifying DNA in each band (14N) DNA (15N) DNA Figure 1.15 Centrifugation of DNA in salt gradient forms a single band. A, Fourteen different samples were centrifuged for different times and then photographed to determine how much of the DNA was collected into the central band. B, A mixture of light (14N) and heavy (15N) DNA was centrifuged and photographed. C, Graphical quantification of the data from panel B. Talking Point: The BioMath Exploration 1.2 helps your learn how to quantify the DNA using the area of a triangle to approximate the area under the curves in panel C. Fig. 1.15B & C modified from Meselson and Stahl

Figure 1.16 Experimental protocol for Meselson and Stahl. Two flasks, labeled Exp. #1 and Exp. #2, of genetically identical cells were grown in 15N media and sampled at the indicated times to track their growth over 15 hours. At time zero, the cells were transferred to fresh media containing 14N. Sampling continued as indicated in the graph. Note the two different Y-axes, there is one for each experiment. Talking Point: The experiment was performed twice as indicated by the two lines and two Y-axes. experiment performed twice Fig. 1.16 modified from Meselson and Stahl

Figure 1.16 Experimental protocol for Meselson and Stahl. Two flasks, labeled Exp. #1 and Exp. #2, of genetically identical cells were grown in 15N media and sampled at the indicated times to track their growth over 15 hours. At time zero, the cells were transferred to fresh media containing 14N. Sampling continued as indicated in the graph. Note the two different Y-axes, there is one for each experiment. Talking Point: Experiment #1 grew the cells until there were about 10^8 cells (left Y-axis) when the media was change to 14^N. Notice the growth curve levels out at the 8 hour time point. Fig. 1.16 modified from Meselson and Stahl

Figure 1.16 Experimental protocol for Meselson and Stahl. Two flasks, labeled Exp. #1 and Exp. #2, of genetically identical cells were grown in 15N media and sampled at the indicated times to track their growth over 15 hours. At time zero, the cells were transferred to fresh media containing 14N. Sampling continued as indicated in the graph. Note the two different Y-axes, there is one for each experiment. Talking Point: Experiment #2 grew the cells until there were about 10^8 cells (right Y-axis) when the media was change to 14^N. experiment #2 Fig. 1.16 modified from Meselson and Stahl

Figure 1.16 Experimental protocol for Meselson and Stahl. Two flasks, labeled Exp. #1 and Exp. #2, of genetically identical cells were grown in 15N media and sampled at the indicated times to track their growth over 15 hours. At time zero, the cells were transferred to fresh media containing 14N. Sampling continued as indicated in the graph. Note the two different Y-axes, there is one for each experiment. Talking Point: All the cells grew for 14 hours in heavy (15^N) nitrogen. all cells (DNA) grown in heavy 15N Fig. 1.16 modified from Meselson and Stahl

Figure 1.16 Experimental protocol for Meselson and Stahl. Two flasks, labeled Exp. #1 and Exp. #2, of genetically identical cells were grown in 15N media and sampled at the indicated times to track their growth over 15 hours. At time zero, the cells were transferred to fresh media containing 14N. Sampling continued as indicated in the graph. Note the two different Y-axes, there is one for each experiment. Talking Points: Hours of growth are indicated in negative numbers leading up to the time of switching from heavy to light nitrogen at zero hours. harvest cells (DNA) at time zero Fig. 1.16 modified from Meselson and Stahl

Figure 1.16 Experimental protocol for Meselson and Stahl. Two flasks, labeled Exp. #1 and Exp. #2, of genetically identical cells were grown in 15N media and sampled at the indicated times to track their growth over 15 hours. At time zero, the cells were transferred to fresh media containing 14N. Sampling continued as indicated in the graph. Note the two different Y-axes, there is one for each experiment. Talking Points: Both sets of cells grew in light nitrogen (14^N) until the cells had replicated at least 4 times. all cells (DNA) switched to light 14N Fig. 1.16 modified from Meselson and Stahl

harvest cells (DNA) at several time points Figure 1.16 Experimental protocol for Meselson and Stahl. Two flasks, labeled Exp. #1 and Exp. #2, of genetically identical cells were grown in 15N media and sampled at the indicated times to track their growth over 15 hours. At time zero, the cells were transferred to fresh media containing 14N. Sampling continued as indicated in the graph. Note the two different Y-axes, there is one for each experiment. Talking Points: Investigators sampled the cells throughout their growth in light nitrogen so they could isolate the replicating DNA. Fig. 1.16 modified from Meselson and Stahl

separate new DNA by salt gradient Figure 1.16 Experimental protocol for Meselson and Stahl. Two flasks, labeled Exp. #1 and Exp. #2, of genetically identical cells were grown in 15N media and sampled at the indicated times to track their growth over 15 hours. At time zero, the cells were transferred to fresh media containing 14N. Sampling continued as indicated in the graph. Note the two different Y-axes, there is one for each experiment. Talking Points: Investigators sampled the cells throughout their growth in light nitrogen so they could isolate the replicating DNA. Fig. 1.16 modified from Meselson and Stahl

separate mixed DNA by salt gradient Figure 1.17 Combined results from the two flasks. A, DNA isolated after cells were shifted from heavy to light media and spun in the salt gradient to determine DNA density. B, Traces of the bands’ intensity as a measure of DNA quantity, along with the generation number of the cells. Mixtures of DNA (0 and 1.9; 0 and 4.1) help distinguish relative densities of critical bands. Red lines 1 and 2 can be used with a straight edge so that you can distinguish slight shifts in band positions. Talking Points: Introduction to summary figure from Meselson and Stahl experiment. Fig. 1.17 modified from Meselson and Stahl

time = 0 (in generations) DNA centrifuged 43 hours all old/heavy (15N) high salt density on right side Figure 1.17 Combined results from the two flasks. A, DNA isolated after cells were shifted from heavy to light media and spun in the salt gradient to determine DNA density. B, Traces of the bands’ intensity as a measure of DNA quantity, along with the generation number of the cells. Mixtures of DNA (0 and 1.9; 0 and 4.1) help distinguish relative densities of critical bands. Red lines 1 and 2 can be used with a straight edge so that you can distinguish slight shifts in band positions. Talking Points: At time point zero, all the DNA was heavy (15^N) and produced a single band after cetrifugation in salt gradient. Fig. 1.17 modified from Meselson and Stahl

quantify DNA in band high salt density on right side Figure 1.17 Combined results from the two flasks. A, DNA isolated after cells were shifted from heavy to light media and spun in the salt gradient to determine DNA density. B, Traces of the bands’ intensity as a measure of DNA quantity, along with the generation number of the cells. Mixtures of DNA (0 and 1.9; 0 and 4.1) help distinguish relative densities of critical bands. Red lines 1 and 2 can be used with a straight edge so that you can distinguish slight shifts in band positions. Talking Points: Quantifying graph shows a single band. Fig. 1.17 modified from Meselson and Stahl

all 15N DNA Figure 1.17 Combined results from the two flasks. A, DNA isolated after cells were shifted from heavy to light media and spun in the salt gradient to determine DNA density. B, Traces of the bands’ intensity as a measure of DNA quantity, along with the generation number of the cells. Mixtures of DNA (0 and 1.9; 0 and 4.1) help distinguish relative densities of critical bands. Red lines 1 and 2 can be used with a straight edge so that you can distinguish slight shifts in band positions. Talking Points: Cells growing in light nitrogen were sampled four times as the cells replicated all their DNA and doubled the number of cells (1 new generation). The DNA became lighter (moved to the left). All of the DNA is in a single band. Fig. 1.17 half 15N, half 14N DNA modified from Meselson and Stahl

all light all heavy Figure 1.17 Combined results from the two flasks. A, DNA isolated after cells were shifted from heavy to light media and spun in the salt gradient to determine DNA density. B, Traces of the bands’ intensity as a measure of DNA quantity, along with the generation number of the cells. Mixtures of DNA (0 and 1.9; 0 and 4.1) help distinguish relative densities of critical bands. Red lines 1 and 2 can be used with a straight edge so that you can distinguish slight shifts in band positions. Talking Points: Quantification of DNA in each time point. Note that the DNA moved from 100% heavy to a single lighter band. At 0.3 and 0.7 generations, some DNA has been replicated but not all of it. Dotted triangles denote the location of the first DNA peak of 100% heavy nitrogen. Fig. 1.17 modified from Meselson and Stahl

all light all heavy Figure 1.17 Combined results from the two flasks. A, DNA isolated after cells were shifted from heavy to light media and spun in the salt gradient to determine DNA density. B, Traces of the bands’ intensity as a measure of DNA quantity, along with the generation number of the cells. Mixtures of DNA (0 and 1.9; 0 and 4.1) help distinguish relative densities of critical bands. Red lines 1 and 2 can be used with a straight edge so that you can distinguish slight shifts in band positions. Talking Points: Quantification of DNA shows the DNA moved from 100% heavy to a single lighter band. DNA replicated once half 15N, half 14N DNA Fig. 1.17 modified from Meselson and Stahl

all light all heavy DNA replicates at different times Figure 1.17 Combined results from the two flasks. A, DNA isolated after cells were shifted from heavy to light media and spun in the salt gradient to determine DNA density. B, Traces of the bands’ intensity as a measure of DNA quantity, along with the generation number of the cells. Mixtures of DNA (0 and 1.9; 0 and 4.1) help distinguish relative densities of critical bands. Red lines 1 and 2 can be used with a straight edge so that you can distinguish slight shifts in band positions. Talking Points: Quantification of DNA at 0.3 and 0.7 generations shows some DNA has been replicated but not all of it. Dotted lines denote the location of the first DNA peak of 100% heavy nitrogen. Fig. 1.17 modified from Meselson and Stahl

all 15N DNA 50% 15N DNA Figure 1.17 Combined results from the two flasks. A, DNA isolated after cells were shifted from heavy to light media and spun in the salt gradient to determine DNA density. B, Traces of the bands’ intensity as a measure of DNA quantity, along with the generation number of the cells. Mixtures of DNA (0 and 1.9; 0 and 4.1) help distinguish relative densities of critical bands. Red lines 1 and 2 can be used with a straight edge so that you can distinguish slight shifts in band positions. Talking Points: After a second round of replication, the DNA has shifted from one band to two bands. One band is the same density but the second band is less dense than produced at one replication. DNA replicated ~ twice what has happened? 1.9 Fig. 1.17 all light all heavy modified from Meselson and Stahl

all light all heavy all 15N DNA 50% 15N DNA 1.0 Figure 1.17 Combined results from the two flasks. A, DNA isolated after cells were shifted from heavy to light media and spun in the salt gradient to determine DNA density. B, Traces of the bands’ intensity as a measure of DNA quantity, along with the generation number of the cells. Mixtures of DNA (0 and 1.9; 0 and 4.1) help distinguish relative densities of critical bands. Red lines 1 and 2 can be used with a straight edge so that you can distinguish slight shifts in band positions. Talking Points: Close up of the critical time points of 1 and ~2 generations. what has happened? 1.9 Fig. 1.17 modified from Meselson and Stahl

all light all heavy all 15N DNA 50% 15N DNA 1.0 Figure 1.17 Combined results from the two flasks. A, DNA isolated after cells were shifted from heavy to light media and spun in the salt gradient to determine DNA density. B, Traces of the bands’ intensity as a measure of DNA quantity, along with the generation number of the cells. Mixtures of DNA (0 and 1.9; 0 and 4.1) help distinguish relative densities of critical bands. Red lines 1 and 2 can be used with a straight edge so that you can distinguish slight shifts in band positions. Talking Points: After 2 generations, the DNA is a 50/50 mix of DNA composed of two different types of DNA. half DNA is 100% 14N half DNA is 50% 14N 50% 15N 1.9 Fig. 1.17 modified from Meselson and Stahl

half old all old semi-conservative conservative half new all new Figure 1.14 Three possible models for DNA replication with one piece of double-stranded DNA (dsDNA) in the center (two dark colors). A, Each original strand is a template for the other strand, and the new dsDNA molecules have one old strand (dark) and one new strand (pale). B, The original molecule remains intact (dark), and the new double helix is composed of two new (pale) strands of DNA. C, Each strand of the old is a template for the new, but all four strands of DNA are a mosaic of old and new DNA. Talking Points: Reminder of the three models of DNA replication and their implications. all stands half new mosaic Fig. 1.14 modified from Meselson and Stahl

all 15N DNA 50% 15N DNA 1.0 Figure 1.17 Combined results from the two flasks. A, DNA isolated after cells were shifted from heavy to light media and spun in the salt gradient to determine DNA density. B, Traces of the bands’ intensity as a measure of DNA quantity, along with the generation number of the cells. Mixtures of DNA (0 and 1.9; 0 and 4.1) help distinguish relative densities of critical bands. Red lines 1 and 2 can be used with a straight edge so that you can distinguish slight shifts in band positions. Talking Points: After 1 generation, the DNA is NOT of two types as required of the conservative model. 1.9 Fig. 1.17 modified from Meselson and Stahl

all 15N DNA ✓ data 50% 15N DNA 1.0 Figure 1.17 Combined results from the two flasks. A, DNA isolated after cells were shifted from heavy to light media and spun in the salt gradient to determine DNA density. B, Traces of the bands’ intensity as a measure of DNA quantity, along with the generation number of the cells. Mixtures of DNA (0 and 1.9; 0 and 4.1) help distinguish relative densities of critical bands. Red lines 1 and 2 can be used with a straight edge so that you can distinguish slight shifts in band positions. Talking Points: Therefore, the data contradict conservative model’s prediction so complete duplication cannot be correct. 1.9 Fig. 1.17 modified from Meselson and Stahl

all 15N DNA ✓ 50% 15N DNA 1.0 Figure 1.17 Combined results from the two experiments. (a) DNA isolated after cells were shifted from heavy to light media. (b) Traces of the bands’ intensity as a measure of DNA quantity are shown, along with the generation number of the cells. Talking Points: After ~2 rounds of replication, the DNA is NOT uniformly one intermediate density as predicted by the mosaic replication model. half DNA is 100% 14N half DNA is 50% 14N 50% 15N 1.9 Fig. 1.17 modified from Meselson and Stahl

all 15N DNA ✓ 50% 15N DNA ✓ data 1.0 Figure 1.17 Combined results from the two experiments. (a) DNA isolated after cells were shifted from heavy to light media. (b) Traces of the bands’ intensity as a measure of DNA quantity are shown, along with the generation number of the cells. Talking Points: After ~2 rounds of replication, the DNA is NOT uniformly one intermediate density as predicted by the mosaic replication model. half DNA is 100% 14N half DNA is 50% 14N 50% 15N 1.9 Fig. 1.17 modified from Meselson and Stahl

half old all old semi-conservative conservative half new all new Figure 1.14 Three possible models for DNA replication with one piece of double-stranded DNA (dsDNA) in the center (two dark colors). A, Each original strand is a template for the other strand, and the new dsDNA molecules have one old strand (dark) and one new strand (pale). B, The original molecule remains intact (dark), and the new double helix is composed of two new (pale) strands of DNA. C, Each strand of the old is a template for the new, but all four strands of DNA are a mosaic of old and new DNA. Talking Points: You can eliminate a second possible model for DNA replication leaving only one viable option – semiconservative replication. all stands half new mosaic Fig. 1.14 modified from Meselson and Stahl

# replications 4 0 & 2 Figure 1.17 Combined results from the two experiments. (a) DNA isolated after cells were shifted from heavy to light media. (b) Traces of the bands’ intensity as a measure of DNA quantity are shown, along with the generation number of the cells. Talking Points: After 4 rounds of replication, the DNA is almost completely light DNA, but a small amount of intermediate DNA persists. 0 & 4 Fig. 1.17 modified from Meselson and Stahl

all light all heavy # replications 4 0 & 2 Figure 1.17 Combined results from the two experiments. (a) DNA isolated after cells were shifted from heavy to light media. (b) Traces of the bands’ intensity as a measure of DNA quantity are shown, along with the generation number of the cells. Talking Points: Alignment of DNA bands replicated 0, 2 or 4 times highlights the semiconservative mode of DNA replication because of the banding pattern and quantification of each band. 0 & 4 Fig. 1.17 modified from Meselson and Stahl

# replications 4 0 & 2 Figure 1.17 Combined results from the two experiments. (a) DNA isolated after cells were shifted from heavy to light media. (b) Traces of the bands’ intensity as a measure of DNA quantity are shown, along with the generation number of the cells. Talking Points: After 4 rounds of replication, the DNA is almost completely light DNA, but a small amount of intermediate DNA persists. 0 & 4 Fig. 1.17 modified from Meselson and Stahl

percentage 14N mixed 15N # replications 87.5 12.5 0.0 4 = 16 copies 0 = 1 copy +2 = 4 copies 40.0 40.0 20.0 52.5 7.5 40.0 0 = 1 copy +4 = 16 copies Figure 1.17 Combined results from the two experiments. (a) DNA isolated after cells were shifted from heavy to light media. (b) Traces of the bands’ intensity as a measure of DNA quantity are shown, along with the generation number of the cells. Mixtures of DNA (0 and 1.9; 0 and 4.1) help distinguish relative densities of critical bands. Red lines 1 and 2 can be used with a ruler so you can distinguish slight shifts in band positions. Talking Points: The amount of DNA loaded in each of these three experiments was manipulated to highlight the three different densities of DNA. The table on the right approximates the relative abundance of the DNA present in each type of DNA molecule. After 2 rounds of replication, the number of dsDNA has increased from 1 to 4 copies. After 4 rounds of replication, the number of dsDNA has increasdd from 1 copy to 16 copies since the DNA duplicated 4 times (2^4). Fig. 1.17 modified from Meselson and Stahl

DNA is replicated in a semiconservative process Figure 1.17 Combined results from the two flasks. A, DNA isolated after cells were shifted from heavy to light media and spun in the salt gradient to determine DNA density. B, Traces of the bands’ intensity as a measure of DNA quantity, along with the generation number of the cells. Mixtures of DNA (0 and 1.9; 0 and 4.1) help distinguish relative densities of critical bands. Red lines 1 and 2 can be used with a straight edge so that you can distinguish slight shifts in band positions. Talking Points: The elegance of this experiment is revealed in not only the aesthetics of the data, but in the clarity of the outcome. They ruled out two alternative models of DNA replication and determined conclusively that DNA is replicated in a semiconservative manner. Fig. 1.17 modified from Meselson and Stahl

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