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Chapter 1: Heritable Material
Integrating Questions and Answers
Section 1.1: What is biological information?
No Integrating Questions
---------------------------------------------------------------------------------------------------Section 1.2: What is the heritable material?
1. Identify the positive and negative controls in Figure 1.3. Is it always clear when to call a
control either positive or negative?
It is not always clear which control is the positive or the negative one. The first column
looks like a positive control because S cells were given to the mouse and S cells were
isolated from the dead mouse. The next column could be considered a negative control
because heat-killed cells failed to grow in the host. However, it is not unreasonable to call the
first column a negative control (mouse died) and the second column a positive control
(mouse lived). The important point is that both extremes were controlled for – live and dead
cells behaved as expected. The third column of R cells would be the same as the first column
of S cells; we can call that a different positive control. The fourth column is the experimental
treatment that has an uncertain outcome.
2. Which of the four experiments demonstrates the existence of an “S factor?” Can you use
Griffith’s data to determine whether DNA or protein is the heritable material?
This fourth column is the experiment that reveals the existence of S factor that converts
benign R cells into pathogenic S cells. However, we cannot tell if the S factor is DNA or
protein from this experiment.
3. Summarize how biologists used mathematics to draw the reasonable conclusion that proteins
would be a better heritable material than DNA.
As explained in BME 1.1, a 20 member code (20 amino acids) is more complex than a 4
member code (4 DNA nucleotides). Given the diversity of life forms in the world,
mathematically it made sense to predict proteins (made of amino acids) would be the
heritable material.
4. Since the “S factor” was loosely attached to the bacterial cell wall/membrane, Griffith and
his contemporaries concluded that proteins were the heritable material. Evaluate the strength
of their evidence.
Their conclusion is based on data showing that cell wall and membrane contain proteins,
sugars and lipids. Lipids and sugars had already been ruled out as heritable material, so they
logical conclusion was that proteins were the heritable material. However, this conclusion is
based on incomplete information about how chromosomal DNA is attached to the inner
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surface of cell membranes. Their conclusion is logical, but incorrect, and based on the
available but incorrect understanding of bacterial structure.
5. After washing, the cell wall/membrane no longer transforms R cells into S cells. If nucleic
acids are normally located in the cytoplasm, are they hydrophilic or hydrophobic? What is
the connection to the solubility of nucleic acids and Griffith’s removal of the S-factor by
washing with buffer?
Nucleic acids are hydrophilic as indicated by their being dissolved in the watery
cytoplasm. By washing the membranes with a water-based solvent (buffer), the DNA was
physically removed from the membrane as a solute. Once the DNA was washed off, the
membrane no longer contained the S-factor.
6. Where is the S-factor in Avery’s protocol? Hypothesize what the white stringy material is in
Figure 1.4A. (You may have performed a similar procedure during a lab at some point in
your education.)
The S-factor is the white stringy material (which we now know is DNA). When mixed
with ethanol, the DNA is no longer soluble and it precipitates out of solution and becomes
visible. It is stringy because the long individual strands adhere to each other the same way
short cotton fibers can be twisted into long threads. Many students have performed a DNA
extraction from strawberries and dishwashing soap. (see
https://www.youtube.com/watch?v=vPGKv53zSRQ).
7. If the heritable material was isolated from mostly cytoplasm in the absence of cell
wall/membrane, what is the likely source of heritable material? Apply Avery’s protocol to
hypothesize what Griffith was washing off from his cell wall/membrane material.
It is difficult to put yourself back into the mindset of Avery (1944) and Griffith (1928).
They were very clever biologists but they did not know what we know now. It is clear today
that Avery was washing DNA off the membranes.
8. Table 1.1 highlights the nitrogen to phosphorous (N/P) ratio of the transforming material.
Determine the N/P ratios for the five amino acids in Figure 1.5. Why was the N/P ratio so
important to Avery’s interpretation of his data?
Proteins normally lack phosphorous, so you would expect the N/P ratio to be a very large
number if the heritable material was protein. For proteins, the denominator (P) would be very
small compared to N which is present in amino acids (proteins) and nucleic acids (DNA).
Ratios about 1.6 indicate the heritable material was unlikely to be protein and more likely to
be DNA. The ratio for the five amino acids would be 7N ÷ 0P, which is an undefined number
and definitely not close to 1.6.
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9. Given that the heritable material was either nucleic acid or protein, what was Avery’s goal
when he conducted experiments with protease and RNase? Did Avery prove DNA was the
heritable material?
Avery proposed that if the S-factor persisted after he destroyed proteins with protease, and
RNA with RNase, then the logical conclusion would be that DNA is the heritable material.
His logic was sound, but insufficient to convince skeptics that some proteins survived the
protease and thus was still the heritable material. The skeptics were correct, some proteins
could have survived so Avery’s experiment was not convincing enough. However, do not fall
into the common misuse of the word “prove”. You cannot prove anything outside of
mathematics and logic. In biology, we can support hypotheses and we can disprove them, but
we cannot prove a hypothesis. (See ELSI 2.2 for deeper exploration of the words prove vs.
disprove.)
Section 1.2 Review Questions
1. Summarize the key experiments demonstrating that DNA is the heritable material and protein
is not.
Figure 1.3 showed that the heritable material could be transferred from one strain to
another. Figure 1.4 and Table 1.1 showed that the heritable material was probably DNA, but
there was room for protein to be the heritable material as well.
2. Use Figure 1.5 to identify the 5' and 3' ends of DNA.
The green phosphate.
3. Explain why it is impossible to prove a negative. Define the word “proof” as used in popular
culture. Explain why this term is inappropriate for science.
You cannot show that something did not happen because their no data that could possible
support the lack of an event. Furthermore, you cannot "prove" anything in biology since
proof means there are no possible alternative explanations. Only in mathematics can you
have the certainty of a proof. Non-scientists think that experiments prove a conclusion which
can lead to a confusion about the degree of certainty vs. confidence in a conclusion.
4. Which should you trust more when reading a scientific paper, the authors’ interpretation as
presented in the written text, or the data presented in figures and tables? Explain your answer.
In a typical science class, students reach a logical conclusion that the words in the book
are fact and the figure are decoration for these words. In ICB and all science publications, the
figures and tables are the facts, and the words are interpretations or opinions based on those
fact. Therefore, as a new science student using this book, you should learn to trust the data
over the words, including the text in this book!
---------------------------------------------------------------------------------------------------Bio-Math Exploration 1.1: Why do amino acids make a better code than nucleotides?
1. Make a two-column table listing the number of possible DNA and protein sequences of
length 5, 10, 20, 50 and 100. Use a calculator and the multiplication principle with exponents
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(e.g., 45 and 205) to save you time. Put your answers in scientific notation so you can easily
see how many digits are in each answer.
The number of possible DNA and protein sequences of each length is shown in the table.
Numbers in the DNA column are computed as 4n and in the protein column are computed as
20n, where n is the length of the sequence.
seq length DNA protein 5 1.02E+03 3.20E+06 10 1.05E+06 1.02E+13 20 1.10E+12 1.05E+26 50 1.27E+30 1.13E+65 100 1.61E+60 1.27E+130 2. Looking at the numbers in your table, can you see why Griffith and his contemporaries
predicted that protein would be the heritable material?
There are more possible protein sequences of any length than there are DNA sequences of
the same length. The difference between the two numbers grows dramatically as the length
increases. In fact, the number of protein sequences is always at least the number of DNA
sequences squared. Note that this is a much greater difference than simply multiplying by 5
(the ratio between the alphabet sizes of 20 and 4). In fact, you have to multiply the number of
DNA sequences by 5n to get the number of protein sequences (because 20n = 4n x 5n).
---------------------------------------------------------------------------------------------------Section 1.3: Can you prove protein is NOT the heritable material?
10. What was the experimental benefit to using a bacterial virus when trying to determine
whether DNA or protein was the heritable material?
Bacteriophage such as lambda are composed of only protein and DNA. The simplicity of
the system meant the investigators had to only distinguish between two possible sources.
11. Based on the data in Figure 1.8, is only one form of radioactivity released to the media after
blending? Which radioactive material was more abundant inside the infected E. coli cells?
No, both 32P and 35S were released to the extracellular media. However, the radioactive
lambda DNA (32P) was more abundant inside the bacteria that outside the bacteria which
pointed to DNA being the heritable material.
12. Are these data more compelling than Avery’s data, or could protein still be the heritable
material?
The data in Figure 1.8 are more compelling than Avery’s S-factor that was not destroyed
by RNase or protease. However, ~20% of the protein (35S) was inside the cells so protein
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could have been the heritable material.
13. Based on the data in Table 1.2, which type of molecule is largely excluded from infected E.
coli? Is this molecule 100% excluded from the infected host cells?
Protein is 99% extracellular which is more compelling evidence that the 70%
intracellular DNA is the heritable material. To hold on to protein as the heritable material,
you would have to ignore the70% intracellular DNA and focus on 1% not-extracellular the
protein. You don’t know for sure that this 1% is intracellular or stuck on the outside of
infected cells. Using Occam’s Razor, DNA is the best choice as the heritable material.
Supporting this conclusion is the fact that some second generation lambda phage were also
radioactive with 32P which was inherited from the parental phage DNA.
14. Hershey and Chase stated, “We infer that sulfur-containing protein has no function in phage
multiplication and that DNA has some function.” Do you think they have sufficient data to
support this claim? Why did they specifically limit their conclusion to sulfur-containing
proteins?
Yes, their evidence supports their claim of S-containing protein having no function and
DNA having some function in reproduction. They can only say this about the S-containing
protein because that is the only protein they could follow given the radioactive tag was 35S.
Any protein lacking methionine or cysteine would also lack sulfur and thus could not be
labeled with 35S.
Section 1.3 Review Questions
1. Summarize the key experiments demonstrating that DNA is the heritable material and protein
is not.
In addition to Figures 1.3, 1.4 and Table 1.1 from Section 1.2, Figure 1.8 and Table 1.2
demonstrated that 99% of the viral protein remained outside infected bacteria but 70% of the
viral DNA was inside the infected E. coli host. Furthermore, the investigators reported that
some phage progeny contained radioactive DNA which clearly demonstrated DNA was the
heritable material and not protein.
2. Evaluate the experimental design by Hershey and Chase. Explain why it is considered to be a
classic experiment.
Hershey and Chase designed a elegant experiment due to its simplicity and certainty to
find the answer. There were two possible sources of heritable material and their experiment
was going to find the answer whether it turned out to be protein or DNA. Rather than just
trying to demonstrate it was DNA or just trying to prove protein was not the heritable
material, their experiment did both simultaneously. The brilliance of their design was best
revealed by choosing a bacteriophage as the model system since it contains only protein and
DNA.
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---------------------------------------------------------------------------------------------------Ethical, Legal, Social Implications 1.1: Who owns your DNA?
10. List some common misconceptions about DNA evidence. Is DNA evidence best suited for
convicting people or proving their innocence?
TV shows and movies have made DNA evidence seem infallible and instantaneous. DNA
samples used in real criminal cases considers only a small portion of the human genome, and
therefore two people could match by chance. If your DNA matched that from a crime scene,
you could be the perpetrator, or not. However, if your DNA does NOT match the evidence,
then you must be innocent.
11. Have you left DNA evidence behind after leaving a room? How could your DNA be used to
frame you for a crime you did not commit?
Yes, we all leave DNA behind everywhere we go. Hair and skin cells are sloughed off
constantly. Also, if you drank or ate, then you left cells behind from your lips, gums and
tongue. Criminals have been identified using DNA isolated from the back of a licked stamp
or from chewing gum.
---------------------------------------------------------------------------------------------------Section 1.4: How does DNA’s shape affect its function?
15. In order to understand the structure of DNA, you need to understand the structure of
nucleotides. Which number carbon in dATP (see Figure 1.9B) is missing the oxygen found in
ATP (see Figure 1.9A)? Which deoxyribose carbon is attached to the string of three
phosphates? Which deoxyribose carbon is bound to the base adenine? You can also use an
interactive Jsmol Figure to help you answer these questions.
2’ OH in RNA of ATP is 2’ H of dATP.
5’ C is attached to the phosphate groups.
1’ C is attached to the base.
16. How many rings are in adenine and each of the other three DNA bases shown in Figure 1.5B
and online? (Don’t count the ring in the deoxyribose pentagon sugar.)
C and T have one ring. (tip: C U T bases are short with one ring.)
A and G have two rings.
17. In DNA, the number of A’s equals the number of T’s and likewise for G’s and C’s. How do
these equal numbers of A’s and T’s, or G’s and C’s, compare with the number of rings in
each base that you counted in the previous question? Your answer to this question will help
you understand how the DNA structure was determined.
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For every base pairing, there are three rings between the two strands of DNA.
18. Speculate why a triple-stranded DNA molecule with the negatively charged phosphates in the
center would be unstable and thus not a good model. Think of forcing the negative ends of
two magnets together. Answering this question puts you on par with Watson and Crick when
they were trying to discern DNA’s structure.
Linus Pauling’s incorrect triple-stranded model would have placed three repulsive charges
next to each other which would have destabilized and destroyed any triple strands in this
hypothetical configuration.
19. It is important to distinguish diagrams from genuine biological structures. Look at Figure
1.10B, and find the misleading aspect of the diagram because the two strands are supposed to
be antiparallel. What should be done to correct this historic figure?
This model has phosphates on all four ends rather than only on the two opposite ends.
Replace phosphates on the bottom left and top right with OH groups instead.
20. If the DNA had a uniform diameter of 20 Å, why did the bases have to pair one A for each T
and one G for each C base? Why would it be impossible for an A to pair with a G?
The uniform distance between the strands is consistent because the base pairings always
have a total of 3 rings in the bases. A + G would be four rings wide and would produce a
bump in the two strands. T + C would only have 2 rings and would produce an indentation in
the two strands.
21. What type of bond is depicted by the dashed lines in Figure 1.10B? Consult Figure 1.11 to
verify your answer.
The weakest of the bonds we have studied so far, hydrogen bonds.
22. Look at the two ribbons representing the two antiparallel strands of DNA. Are the two
strands equally spaced apart in the horizontal axis? What about the vertical axis?
Horizontally, they are equally spaced. But in the vertical direction, there are alternating
small and big spaces (called minor and major grooves, respectively).
23. Although you are not an expert in this field, you can tell which image in Figure 1.12 is the
more informative and the one with the highest quality. Whose data do you think Watson and
Crick used to solve the structure of DNA?
Figure 1.12B shows more detail and thus more information. Watson and Crick must have
used this image produced by Franklin and Gosling, about which they hinted in their paper.
24. In science, credit for a discovery is given to those who publish first. To whom would you
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give credit for discovering the structure of DNA? Explain your answer using the data
provided in this chapter.
Credit should go equally to all three sets of authors for the three published papers.
However, Franklin died before the other three lead authors were awarded the Nobel Prize
which is awarded only to living recipients.
25. Use the ruler in Figure 1.13 to measure the distance between paired atoms shown to form Hbonds. Compare the number of H-bonds between G:C base pairs and A:T base pairs in the
figure.
All dashed lines are between 1 and 2 Å long. GC has three and AT has two.
26. Use the ruler to determine if any additional H-bonds could form between either of the base
pairs in Figure 1.13. Which base pairing did Watson and Crick get wrong with regards to the
number of H-bonds?
A third H-bond can be formed between the opposing H and O at the bottom of the G+C
diagram. Watson and Crick got this detail wrong in their famous paper.
27. Go back to Figure 1.10A, and locate the major groove which is
larger than the minor groove. These two grooves describe the
spaces between the two outer helices of DNA. Can you see why
Watson and Crick proposed DNA as a carrier for proteins
tucked into the major groove?
There is room for proteins to stick into the major groove and
hold on to the DNA. You will see some examples of this later in
the book.
28. Why does the band of DNA get narrower as the amount of time increases in Figure 1.15A?
Initially, the DNA is evenly distributed in the salt gradient. Over time, it collects at the
position in the density gradient that matches its own density. An analogy would be people
distributed at all levels of the ocean and the sky and after some time, they would all come to
rest at the surface of the water since they are denser than air and lighter than water.
29. Since the two isotopes of nitrogen varied by only a single neutron, how could they physically
separate heavy and light DNA sufficiently to take a photograph of the separated DNAs?
Keep in mind that DNA is a long polymer and every nucleotide contains multiple N atoms.
They could not have separated single nucleotides, but the polymers had accumulated mass
differences which could be separated. Just as it would be difficult to discern the weight
difference between one pea and one bean but it would be much easier to distinguish the
weight difference in one million of each one. The little bit of difference amplified by one
million is easy to detect.
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30. Did Meselson and Stahl successfully devise a method to separate heavy and light DNA
molecules? Support your answer with data from Figure 1.15. Use Figure 1.15C to determine
if they really had equal amounts of heavy and light DNA, or was one form more abundant
than the other?
Yes, the centrifugation method was able to separate long polymers of light/normal (14N)
DNA from heavy (15N) DNA. You can see the physical separation by density in the two
bands of Figure 1.15B. By estimating the area under the two curves (area of two triangles) in
Figure 1.15C, you can detect more 14N DNA than 15N DNA. (See BME 1.2 for further
explanation of area under the curve.)
31. Look closely at the two growth curves with two Y-axes in Figure 1.16, comparing the density
of cells at a couple of different times. Are these two populations of cells similar, or
substantially different? Support your answer with data.
At any given time (e.g., time zero), both populations have 108 cells. The key to answering
this question is to use the left Y-axis for cells in experiment #1 and the right Y-axis for cells
in experiment #2.
32. Describe the change in DNA density between generation times of 0 and 1.0 in Figure 1.17.
Do the data from these two time points allow you to eliminate any of the three possible
models of DNA replication outlined in Figure 1.14?
The DNA has gone from 100% all heavy (single band under the red lined labeled 2) to a
single band half way between red lines 1 and 2. From these data, you can eliminate the
conservative replication model which predicted that after one round of replication, one copy
of DNA would be all heavy and the other copy would be all light; you should see two bands
instead of one.
33. Compare the bands at generation times 0, 1.0, and 1.9. How many bands of DNA are present
in Figure 1.17 after approximately two rounds of DNA replication? Do the data from the time
point of 1.9 allow you to eliminate any of the three possible models of DNA replication
outlined in Figure 1.14?
At 1.9 generations, there are two bands: one is half way between heavy and light; the
other band is fully light and under red line 1. These data eliminate the possibility of the
mosaic method for DNA replication. The mosaic model predicted there would always be
only one band of replicating DNA and band would gradually shift half way towards the light
DNA line (red 1).
34. Use the only viable model in Figure 1.14, and predict what DNA bands would be present
after four generations. Is your prediction validated by the data in Figure 1.17?
After four generations, there would be 16 times more DNA than at time zero (24 or four
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doublings = 16). Fourteen of the 16 would be all light and the remaining 2 copies would be
half heavy and half light. Therefore, you would predict the presence of a very intense DNA
band under the red 1 line and a much fainter band (one seventh as intense) half way between
the red lines 1 and 2. The data in Figure 1.17 support this mathematical prediction.
35. Look at the bands in Figure 1.17 after four generations. Determine which band is more
intense than the other, and explain why this is consistent with the correct model of DNA
replication.
The left band is more intense as explained in the answer above. This result is consistent
with the semiconservative model of replication and disproves the other two possible models
of replication. See previous answer for further details.
Section 1.4 Review Questions
1. Use Figure 1.9, and describe the difference between DNA and RNA nucleotides.
DNA lacks the 2' OH and has only and H instead. RNA has an OH attached to the 2'
carbon. Don't be fooled into thinking that DNA is always double stranded and RNA is
always single stranded. As often happens in biology, there are exceptions to the number of
strands but not exceptions to the 2' carbon attachments.
2. Use Figure 1.10 to help you explain the term antiparallel used to describe dsDNA structure.
Double-stranded DNA has two parallel strands but their orientation makes them
antiparallel. This means that the 5' carbon (with the attached phosphate) on one strand is
adjacent the 3' carbon (with the attached OH) on the other strand.
3. Which bases are paired in dsDNA? What types of bonds hold paired bases together, and how
many of these bonds are there for each of the two base pairings?
A binds to T with two hydrogen bonds. G binds to C with three hydrogen bonds. Watson
and Crick made a mistake when they published their figure that has been recreated in Figure
1.13. They missed the third hydrogen bond that spans the two atoms at the bottom of the
figure between guanine and cytosine.
4. Draw the structure of dsDNA showing one nucleotide in detail. You do not need to draw the
bases, but you should be able to represent them with the correct number of rings. You do not
need to draw the spiraling double helix, you may draw it flattened the way Watson and Crick
did.
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5’
3’
3’
5’
5. Rank the three types of chemical bonds in order from strongest to weakest. Explain why each
bond has its relative bond strength.
Covalent bonds are the strongest because they share electrons. Ionic bonds are weaker
because they are only held together by the opposite attraction of charged atoms. Hydrogen
bonds are the weakest because they are held together by partial charges between atoms.
6. Describe which of the three possible models in Figure 1.14 for DNA replication is correct.
DNA is replicated in a semiconservative mechanism (panel A). Meselson and Stahl
disproved conservative replication (panel B) because DNA that has replicated once is a
mixture of half heavy DNA and half light DNA. The mosaic model was disproven because
DNA that had replicated twice was a 50/50 mixture of all light DNA and half light + half
heavy DNA. Mosaic model predicted that all DNA would be 75% light DNA and 25% heavy
DNA. These data are presented in Figure 1.17.
7. Summarize the evidence in Figure 1.17 that disproved the two incorrect models.
After one round of replication, the dsDNA was a mixture of one heavy strand and one
light strand. This disproved the conservative model of replication. After two rounds of
replication, the DNA was an equal mixture of two types of DNA. One type was 100% light
DNA. The other type was the same as DNA after only one round of replication with one
strand being all heavy DNA and one strand being all light DNA. The second round disproved
the mosaic model of replication.
Meselson and Stahl did extra work to show DNA after 4 rounds of replication and even
mixed DNA produced after different amounts of replication. Their work clearly supports
semiconservative replication and disproved the other two possible mechanisms for DNA
replication as predicted by Watson and Crick.
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---------------------------------------------------------------------------------------------------Bio-Math Exploration 1.2: How much DNA is in each band?
1. Use a ruler to estimate the area (in mm2) under each peak in Figure 1.15C.
Although the exact area measured may depend on the size of the image on a student's
screen, the main point is to see that the two triangles have approximately the same area. The
second peak is a little wider than the first one, which compensates for the fact that the first
one is taller. An even more important point is to recognize that area, not height, of the peak is
the desired measure of the amount of DNA.
2. Use the areas under each peak in Figure 1.15C to compute the percentage of DNA in this
sample that was labeled with 14N.
Because each peak has about the same area under it, the percentage of DNA labeled with
14
N is approximately 50%.
---------------------------------------------------------------------------------------------------Bio-Math Exploration 1.3: How are generations measured on a log scale?
1. Approximately how many generations have passed if there are ten times as many cells as
there were at time zero? Explain how you estimated this generation number.
You want to know how many doublings there have been to have 10 times as many cells as
present initially. The number of cells after x doublings is 2x times the number of cells present
initially. Therefore, you want to know the value of x such that 2x = 10. To solve this equation,
you need to take the log base 2 of both sides.
Using the fact that log2 2x = x, you get x = log210 = log1010 / log102, or approximately 3.32.
2. Figure 1.16 shows that there were approximately 108 cells in Experiment #1 at time zero, and
5 x 108 cells at the time of the first cell count after time zero. Use this information to compute
the number of generations that had passed at the time of the first cell count after time zero.
The number of cells at the time of the first cell count is 5 times the number at time 0.
Therefore, this problem is the same as the one in BME IQ #1, with 5 in place of 10. To find
the value of x such that 2x = 5, take the log base 2 of both sides.
Using the fact that log2 2x = x, you get x = log25 = log105 / log102, or approximately 2.32.
3. Figure 1.16 shows that the first cell count after time zero was made after about 1 hour. Use
this fact with your answer to Bio-Math Exploration Integrating Question 2 to determine the
generation time for these cells.
The generation time is the amount of time per doubling. Here, there were 2.32 doublings
in 1 hour, or 1/2.32 hours per doubling = 0.43 hours per doubling = 0.43 x 60, or
approximately 26 minutes per doubling.
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---------------------------------------------------------------------------------------------------Section 1.5 Is all genetic information encoded linearly in the DNA sequence?
36. It is helpful to understand a new method before analyzing data. Look at Figure 1.19 and
explain how the different colored molecules were separated in two dimensions by TLC.
Panel B shows the colored molecules separating based on their solubility in the solvent
and their interaction with the solid matrix on the TLC sheet. When rotated 90º and dipped
into a different solvent (panel C), the colored molecules migrate according to their solubility
and interaction with the matrix of the TLC sheet. Note that the purple molecules did not
migrate very far in the first solvent but migrated the furthers in the second solvent. The
purpose of this method is to determine how many different molecule types there are in a
mixed sample.
37. Compare the Cp and m5Cp spot intensities for the active genes versus inactive genes in
Figure 1.20. Which DNA contained more m5Cp?
DNA from active genes/nuclei have more Cp (darker spot) than m5Cp (lighter spot).
Genes that are inactive have more m5Cp than Cp.
38. Are active genes methylated more or less than DNA isolated from inactive genes? Speculate
why active genes would be differently methylated than inactive genes.
Methylation correlates with silenced genes and may be causative but this experiment
cannot determine causation. This is a form of epigenetic regulation of genes since the DNA
sequence is not changed.
39. Does reduced methylation of DNA increase fetal hemoglobin production?
Yes, the addition of methylation inhibitor increases the amount of fetal hemoglobin and
larger does, the fetal hemoglobin persists longer.
40. Use Figure 1.21 to determine the effect of higher doses of 5-azaC on hemoglobin levels.
What additional clinical data would you want to see before moving forward with human
trials?
Higher doses prolong the production of fetal hemoglobin and do not increase the amount of
fetal hemoglobin. I would want to see if other genes are similarly affected which could
indicate off-target consequences and undesirable side effects. Do the animals become sick
over time?
41. You can see another relationship between the structure and function of DNA if you view this
Jsmol interactive web page that shows a DNA methyltransferase in action. From which
groove, the major or minor, does the enzyme reach the cytosine base in order to methylate it?
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For this short piece of DNA, it is difficult to distinguish between major and minor grooves.
However, based on the large space between the two strands, it appears the protein has
reached into the major groove and placed some of its protein where the cytosine base used to
be located in the DNA. You can see the orange cytosine protruding into the yellow enzyme
where it will become methylated.
Section 1.5 Review Questions
1. Describe how the method of TLC works and what type of data it produces.
TLC separates a mixture of molecules based on their chemical and physical properties.
The mixture is spotted onto a special matrix and then dipped into a solvent. The molecules
are wicked up the matrix at different velocities depending on how they interact with the
matrix.
Two dimensional TLC performs this twice as shown in the figure, with the second
dimension running perpendicular to the first dimension.
2. Summarize the conclusions drawn from the TLC experiment in Figure 1.20.
The investigators found that active genes tended to have very little methylated cytosine
DNA bases while silenced genes were more heavily methylated. They deduced that
methylation is able to silence genes without mutating the DNA. This was the first example of
epigenetic regulation in eukaryotes.
3. What are the four standard bases of DNA? What base is sometimes called the fifth base of
DNA? Provide an example of how the fifth base can affect gene activity.
GCAT (guanine, cytosine, adenine and thymine) are the four DNA bases. The "fifth base"
of DNA would be methylated cytosine (meC). As shown in Figures 1.20 and 1.21, heavily
methylated (hypermethylated) genes are not active while unmethylated (hypomethylated)
genes can be active.
4. Explain how the experiment in Figure 1.21 was a clinical failure but a basic research success.
The experiment demonstrated that by reducing the methylation of the fetal hemoglobin
gene, the normally silent gene became active in a dose-dependent fashion. Altering gene
expression by epigenetic regulation was a big research success, but the investigators realized
this would not be a clinical success. Too many off-target genes would also be activated and
the side effects would be too widespread.
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