ICB Chapters 1-3 Questions
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.
Determine the probability that a couple will give birth to a son with Huntington disease if the father is homozygous recessive and the mother has the disease. Huntington is a dominant disease.
25% boy with disease
The most common form of muscular dystrophy is caused by a mutation in the dystrophin gene located on the X chromosome. Calculate the probability that a couple will have a girl with muscular dystrophy if the mother is a carrier and the father is not. Calculate the probability that this same couple will have a boy with the disease. It will be helpful if you track both X and Y chromosomes in your Punnett square.
25% girl with disease and 25% boy with disease.
In order to understand how proteins are made, we need to examine a cell's RNA content. Look at the RNA gel in Figure 2.3 and find the very bright rRNA bands. How many different sized rRNA molecules are in a ribosome?
3 different sizes
Are active genes methylated more or less than DNA isolated from inactive genes? Speculate why active genes would be differently methylated than inactive genes.
Active genes are methylated less than inactive genes. They are methylated differently because they had different functions?
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.)
Adenine and guanine have two rings and thymine and cytosine have one ring
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 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.
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's goal was to prove the RNase was not protein by adding the protease which destroys protein. Avery did not necessarily prove DNA was the heritable material, but gave evidence to support that protein was not.
Use the laws of segregation and independent assortment to explain the results in Figure 3.11. Be sure to analyze each of the four different plant genotypes that produced the four types of peas.
Because homologous chromosomes are split apart, it is likely to get a mix of results (law of segregation) as the chromatids are picked separately (law of independent assortment)
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?
Because the size of the strands needs to be consistent and if other bases were paired, it would cause an indentation or bump in the strand.
Given that rRNA is not radioactive, when was the rRNA transcribed for this experiment? How can intact ribosomes associate with radioactive RNA if the rRNA is not radioactive?
Before the viral infection. the tRNA and mRNA is radioactive and both interact with rRNA
Compare the four steps in mitosis (see Section 3.4) with Figure 3.22 and the movie depicting meiosis. Write down the major differences that you observe in meiosis. Using what you learned about mitosis, predict what structures are responsible for the movement of the chromosomes during meiosis.
Differences: splits twice, homologous chromosomes line up and separate and then chromatids separate again. The lysosomes are responsible for the movement of the chromosomes.
Locate the two ends of one microtubule in Figure 3.21. To what is each end attached?
Each end is attached to the centromere
Which components from the experiment in Figure 2.20B are necessary for translation to occur? Which parts, if any, were not required for translation?
Energy (ATP and GTP), ribosomes, mRNA and amino acids are necessary for translation to occur. The chromosomes are not necessary.
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.
For every base paring, there are three major rings between two strands of DNA
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?
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.
Use Table 3.5 and the images in Figure 3.19 through Figure 3.21 to summarize what happens inside a parental cell in order to produce two equivalent new cells. Start your summary immediately after the previous cell division. You should be able to draw this process as well as outline it in writing.
G_1 phase: During G_1 phase, also called the first gap phase, the cell grows physically larger, copies organelles, and makes the molecular building blocks it will need in later steps. [Do cells always grow before they divide?] S phase: In S phase, the cell synthesizes a complete copy of the DNA in its nucleus. It also duplicates a microtubule-organizing structure called the centrosome. The centrosomes help separate DNA during M phase. G_2 phase. During the second gap phase, or G_2 phase, the cell grows more, makes proteins and organelles, and begins to reorganize its contents in preparation for mitosis. G_2 phase ends when mitosis begins. In early prophase, the cell starts to break down some structures and build others up, setting the stage for division of the chromosomes. In late prophase (sometimes also called prometaphase), the mitotic spindle begins to capture and organize the chromosomes. In metaphase, the spindle has captured all the chromosomes and lined them up at the middle of the cell, ready to divide. In anaphase, the sister chromatids separate from each other and are pulled towards opposite ends of the cell. In telophase, the cell is nearly done dividing, and it starts to re-establish its normal structures as cytokinesis (division of the cell contents) takes place. Cytokinesis, the division of the cytoplasm to form two new cells, overlaps with the final stages of mitosis.
Which amino acids are encoded by four codons? Which base of these four codons can be altered without changing the encoded amino acids? Are there instances when the third base is changed and it does alter which amino acid is encoded?
Glycine, valine, proline, threonine, and alanine are encoded by four codons. The last base of the codon can be changed and will still code the same amino acid. There are a few cases where the base is changed and it changes the amino acid in the third column of the genetic code.
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?
Go to the National Center for Biotechnology Information (NCBI) to see how many chromosomes are present in various species Directions: Go to the genome website. Click the "Filters" button (top right) and choose a range of organisms (e.g., eukaryotes, animals, mammals). For each group of organisms, sort from largest to smallest number of chromosomes by clicking on the "Chr" column heading. Click on the name if you want to see what species you have identified. Which organism has the most chromosomes? Do humans have the most chromosomes of all mammals (change the default of 50 organisms to 200)? Why are humans (Homo sapiens) listed with 24 instead of 23 or 46?
Humans have the most chromosomes because we are the most developed animal. We are listed with 24 because of the sex chromosomes.
In science, credit for a discovery is given to those who publish first. To whom would you give credit for discovering the structure of DNA? Explain your answer using the data provided in this chapter.
I would like to give credit to F&G because they actually found the clear picture of the structure of DNAT
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?
Image B is more informative and is a higher quality and Watson and Crick used that data to solve the structure of DNA
RNA can only enter the gel if it was loaded into a well. Near "lane 2", you can see a circular spot of white that is outside the vertical lane. Do you think this is really RNA, or an artifact of the entire process that was captured on film? Explain your reasoning.
It is an artifact because RNA has to be in the wells and lanes. Also the equipment is old so it could be an error from that.
Why does the band of DNA get narrower as the amount of time increases in Figure 1.15A?
It starts out spread evenly and then eventually starts to collect at its density gradient naturally
The laws of independent assortment and segregation are not the only mechanisms for generating diverse gametes. If you were following two loci on the same chromosome, what might happen to a pair of alleles on one chromosome as a result of recombination in prophase I (covered in detail in Section 3.5)? Would recombination increase or decrease gamete diversity? You may want to revisit this question after reading Section 3.5.
It would increase DNA diversity because new traits can be included and the combination of genes is varied.
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 phosphorus and sulfur were released. The 35S was more abundant in the infected cells.
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, meaning they dissolve in water. If the Nucleic Acids were attached to the cell wall, but water was run on them, they would be gone and not be able to convert the cells.
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.
Describe the structure of the pea flower that made it an ideal experimental organism for genetics. Since peas typically self-fertilize within a single flower, what manipulations would Mendel have to perform to mate two different plants?
Pea plants have traits that are easy to identify, are quick to grow, and either self pollinate or cross pollinate. When cross pollinating, Mendel could easily control the results of the crosses. To cross pollinate, Mendel has to move the pollen from one flower to the stigma of another.
At what point during meiosis do the pairs of chromosomes recombine? What is the outcome of chromatid recombination? What advantage does chromatid recombination provide to a species?
Prophase 1. It allows for each chromosome to be completely unique and greater genetic diversity
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.
What category of molecules do you think are responsible for the constriction seen between paired cells in Figure 3.18 and the movie illustrating cell division? Do you think carbohydrates, lipids, proteins, or nucleic acids constrict the cells to divide in two new cells?
Proteins are responsible for the constriction between paired cells.
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.
Is RNA polymerase bigger than a DNA nucleotide, or is it smaller? Use the data from Figure 2.12 to support your answer. Look at the aligned sequences inside the red box of Figure 2.13. Can you visually detect a sequence pattern for promoters? Is there a rule that describes which base will be the first one to be transcribed?
RNA polymerase is bigger than one nucleotide. Promoters all have Ts for the first letter and As for the second to last letter.
Organize a list of properties that mitosis and meiosis have in common. What distinguishes meiosis from mitosis?
Similarities: diploid to haploid, each a type of reproduction Differences: Meiosis: two phases, reconstruction Mitosis: one phase, identical cells
Summarize how biologists used mathematics to draw the reasonable conclusion that proteins would be a better heritable material than DNA.
Since bacteria is made up of proteins, lipids, sugars, and nucleic acids, the S Factor had to be one of those. Lipids and sugars had already been ruled out, so protein was concluded to be the heritable material because it is made up of twenty amino acids that could potentially hold more information where DNA and RNA are only made up of four different nucleotides and in theory would hold less information.
Look at the human chromosomes in Figure 3.19 and notice how they are numbered. What criteria were used to determine which chromosome was number 1, number 2, and so on, all the way to number 22?
Size and bending patterns were used to number the chromosomes with the biggest chromosomes being 1 and the smallest being 22.
Use the data in Figure 2.18 to support the conclusion that the steroids bind to protein receptors.
Steroids are shown to bind with protein receptors in how 2.16 shows the steroids bonded to mostly RNAse which is involved with making proteins
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 a single band to another single band. This eliminates the conservative model that shows one round of replication.
You can see TATA in the V-fd and V-SV40 viral promoters in Figure 2.13, but do any human genes also have TATA in their promoters? In this screen shot (see below) from the human genome browser, you can see a human gene that is important to your immune system called interferon gamma (IFNG). Does the human IFNG promoter contain TATA? Look at about 30 bases before the beginning of the mRNA. Can you match the sequence logo with IFNG?
The IFNG does have TATA as it is the beginning of the promoter.
View an interactive molecule of RNA polymerase as it is transcribing a gene. Manipulate the structure until you can see the template strand of DNA and the new RNA strand. Does the RNA form base pairs with the DNA? What happens to the non-template strand of DNA?
The RNA does form base pairs from the DNA. The non-template DNA surrounds the RNA and DNA
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. I predict the material is DNA.
Which sequence initiated transcription the best in Figure 2.14B? Rank the three tested promoters from strongest to weakest.
The TATA initiated transcription the best while TAAA initiated transcription the worst
Are promoters sequence-specific, or can all strings of As and Ts work equally well? How did the TAGA promoter contradict the hypothesis that AT-rich promoters are the best?
The TATA worked the best because it is the beginning of almost every promoter.
Calculate the percent of bases that are A or T inside the box for each of the six sequences in Figure 2.13. Compare these values to the percent A or T for the next seven bases downstream, to the right, of the box. If there were no bias in the DNA, you would expect these two percentages to be near 50%. Are they? Review the number of hydrogen bonds between A:T and G:C from IQ #26 in Section 1.4. What functional significance might the percent AT have on the structure of DNA?
The bases of T and As inside the box was higher than the T and As outside the box. This is to ensure the RNA polymerase knows where the promoter is.
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 the cell densities of these two populations similar, or substantially different? Support your answer with data.
The cell densities are similar, based on the counts of each Y axis
What effect does temperature have on the growth rate of E. coli in Figure 3.16? Which flask of cells was growing faster? Should you look at the end point of a line or its slope to determine growth rate? Explain your answer.
The cells at 25 degrees had a slower growth rate than the cells at 30 degrees. The growth rates for both samples are about the same in 37 degrees and are higher in 30 degrees. You should look at slope to determine growth rate.
What happened to cells lacking the added thymidine? Did they reproduce or only grow in size? Integrate the data from both parts of Figure 3.17.
The cells lacking thymidine only grew in size as shown by the steady increase in 3.17A and the large single cell in 3.17B
Speculate how steroids that first bind to proteins in the cytoplasm can become located in nuclei later. Use the data from Figure 2.19 to support your answer.
The cytoplasm moves and can go near the nucleus and the steroid can then enter the nucleus. The cells were also together but then were separated so the steroid could have gotten into the nuclei when they were together
Use the ruler in Figure 1.13 to measure the distance between paired atoms shown to form H-bonds. Compare the number of H-bonds between G:C base pairs and A:T base pairs in the figure.
The distance between paired atoms is about 1.5 A. There are two bonds between both base pairs
Does the graph in Figure 3.15 look like a symmetrical bump, or is the shape more complex? What do you think the shape of the curve might indicate about the distribution of growth rates for cells of different sizes within a population? (Bio-Math Exploration 3.3 helps you discover the answer to this question.)
The distribution of the volumes is slightly skewed right, meaning there are more cells with a bigger volume than a small volume.
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.
The ends would not bond together and therefore would not hold the structure of DNA
Use Figure 2.10 to explain why the β-galactosidase enzyme was produced before permease.
The galactosidase was produced before permase because it gets catalyzed first.
S represents the dominant allele for smooth. Calculate the probability of producing an F1 wrinkled pea when mating SS x ss plants. Design an experiment to produce true-breeding wrinkled peas in as few generations as possible, starting at the P generation of SS x ss. Give the ratio of phenotypes and genotypes that you would expect from each generation in your experimental design.
The genotypes is Ss and the phenotype is all smooth. To produce wrinkled peas as soon as possible, after crossing SS x ss, cross the Ss x Ss and about 25% should be wrinkled. In that generation the genotypes are 25% SS, 50% Ss and 25% ss and the phenotypes are 75% smooth and 25% yellow.
Using E. coli as a model, speculate whether the rate of bacterial reproduction is genetically constrained for a given species, or governed solely by temperature like a series of chemical reactions. Support your answer with data.
The growth rate for bacteria is majorly effected by temperature because as as shown in Figure 3.16, growth rate increased as the temperature increased.
What type of bond is depicted by the dashed lines in Figure 1.10B? Consult Figure 1.11 to verify your answer.
The hydrogen bond is depicted by the dashed lines
Compare the Cp and m5Cp spot intensities for the active genes versus inactive genes in Figure 1.20. Which DNA contained more m5Cp?
The inactive DNA contained more M5Cp while both have about the same amount of Cp.
Figure 2.9 depicts a proposed model of gene regulation with a pair of inhibitors that lactose inhibits. By analogy, construct a sentence that means you like to sleep, but use the word "not" two times in the sentence. Now describe the mechanism of β-galactosidase gene induction in Figure 2.8 using the nouns "inhibitor" and "lactose" and the verb "inhibits."
The inhibitor of galactose inhibits the bacterial protein as an enzyme for the production of galactosidase
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?
The last experiment, where he injected both S and R cells showed the existence of the S Factor. The data can be used to determine that DNA is a heritable material because the DNA of the bacteria was able to be passed down to other mice.
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.
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?
The phosphates are on all ends rather than on opposite ends. Replace the phosphates on the bottom left and top right with OH groups.
Identify the positive and negative controls in Figure 1.3. Is it always clear when to call a control either positive or negative?
The positive control was injecting the S cells and the negative control was injecting the R cells. It is not always clear whether a control is positive or negative so you should always remember the objective of the experiment when identifying controls
Is the promoter in Figure 2.16 a digital on/off switch that only has two states: full on and full off? Use the different promoter lengths to determine which portions of this promoter contained functionally-important DNA.
The promoter is like an on/off switch where it is either full on or full off. Based on the promoter lengths in 2.16, the white part seems to contain the fully functional DNA.
Use the data in Figure 2.11C, Figure 2.15, and Figure 2.16 to speculate what can happen to a promoter when transcription factors bind to it to start transcription. Draw a picture of DNA that is ready for RNA polymerase to start transcription.
The promoter is unwinded and used as the start of where the mRNA chain goes and is changed.
Mendel correctly deduced that variation from a perfect 3:1 ratio of peas from one plant represented chance events. Explain why pea color for crosses #5 and #8 in Table 3.3 had such different ratios and yet the overall ratios were close to 3:1. Use what you already know about meiosis and fertilization to list all the steps that include chance event; meiosis is explored in Section 3.5.
The ratios were very different but overall were the same because the chance of getting the results of the cross was 3:1. At the end of Meiosis, the gametes are haploid and have half the number of chromosomes of a daughter cell after mitosis and then they will fuse with another gamete during fertilization and create an organism. The gametes that are chosen to be potentially fused are by complete chance.
Analyze the data in Figure 3.3. Were the second generation peas a blend of green and yellow? What happened to the green color from one parent in each of these two crosses? Did the color of the second generation peas depend on which parent produced the pollen?
The second generation peas were all yellow. The green color disappeared in the second generation. The results of the second generation did not matter because they produced all yellow with both sets of parents.
Look at Table 3.2, and determine which trait is dominant and which is recessive. Use the data from Tables 3.1 and 3.2, and calculate the F1 and F2 ratio of dominant to recessive phenotypes. Are the ratios the same for both pea color and pea texture? Do these data fit precisely with the expected ratios? Explain your answer using data and your calculations.
The smooth peas are dominant and the wrinkled peas are recessive. F1 ratio is 0:1 and the F2 ratio is about 1:3 for both the pea color and the pea texture. The data is not exactly 1:3 but is very close to that. The ratios are not exact because of the law of independent assortment states how each allele is chosen randomly and therefore the results will be not exact because the plant does not pick alleles to fit a ratio, but by chance.
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.
The source of the heritable material is most likely the cell wall. Avery's team was likely washing off a buffer that protected DNA from the cytoplasm.
View the human receptor for the steroid called glucocorticoid bound to DNA. What must happen before a steroid receptor can bind to DNA?
The steroid must match up with DNA based before binding to the DNA
View this translation animation to watch the how many moving parts are required to translate a protein from an mRNA. Now go to this interactive Jsmol web page. Make sure you can see all the major parts required for translation that are in this visualization. Compared to Figure 2.4, which biologically significant components are missing from this particular view?
The tRNA and the mRNA are missing from 2.4's model.
Calculate the ratio of β-galactosidase to total protein using three points from the diagonal line in Figure 2.8. Use these data to argue that the transcription of the β-galactosidase gene is induced by lactose.
The transcription of the galactosidase gene is infused by lactase because the data displays a roughly steady incline on the graph
If steroids can diffuse into every cell but activate genes in only a subset of cells, how do organisms regulate which cells will respond to a particular steroid?
Their DNA has on and off switches that tell the gene when to use things that come into the cell.
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?
Their method was successful as they were able to separate 14N and 15N as shown in 1.15B. According to 1.15C there was more of 14N than 15N.
Make a Punnett square with four rows and four columns to represent the possible outcomes of a dihybrid cross of YySs x YySs. Count the number of phenotypes possible, and their ratios.
There are 16 possible outcomes with a 9:3:3:1 ratio. There are 4 phenotypes possible with a 3:1 ratio.
Describe the size distribution of the radioactive RNA in Figure 2.6. Speculate about the function of the radioactive RNA transcribed after viral infection had already begun.
There are many small RNA, some medium RNA, and some large RNA. The ribosomes associate with the mRNA when radioactive.
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 1.9 generations allow you to eliminate any of the three possible models of DNA replication outlined in Figure 1.14?
There are two bands present. This data eliminates the mosaic model which says that there would be one band replicating and one band would gradually shift towards the light
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?
There could be one more pairing in the guanine and cytosine pairing and Watson and Crick got this detail wrong in their paper
Calculate the genotype and phenotype ratios for self-pollination using a Yy heterozygous plant. What is the probability of producing a true-breeding yellow or a true-breeding green pea? Predict the outcome of self-fertilization for plants grown from all the F3 peas in Figure 3.6. Calculate the probability of producing a green pea when mating Yy with yy plants.
There is 25% chance of a true breeding yellow plant, 50% chance of a heterozygous yellow plant, and a 25% chance of a true breeding green plant. Self fertilization of the F3 plants will result in the same ratio. The probability of getting a green plant when crossing a heterozygous yellow plant and a true breeding green plant is 50%.
If you mated YY with yy peas, use Table 3.4 to predict the proportion of homozygous yellow, homozygous green, and heterozygous peas after ten generations ( F10) if all plants were allowed to self-pollinate after the initial mating. Assume that all plants survive throughout your simulated experiment and each mating produces four peas.
There is a 1:2:1 ratio with homozygous yellow, heterozygous yellow, and homozygous green. where 25% are homozygous yellow, 50% are heterozygous yellow and 25% are homozygous green.
The fatal human illness called Huntington disease is caused by a dominant mutant allele. Calculate the probability that a couple will have a child with the disease if the father is heterozygous and the mother is homozygous recessive. Calculate the probability of this same couple having either a boy with the disease or a girl without the disease.
There is a 50% chance of having a child with the disease. There is a 25% chance of the boy with the disease and a 25% chance of a girl without the disease.
Cystic fibrosis is caused by a recessive allele. Calculate the probability that a couple will have a child with cystic fibrosis if the mother is heterozygous and the father is homozygous dominant. Use the example of cystic fibrosis to prepare a policy statement explaining why a brother and a sister conceiving children is a bad idea genetically. You can read about a real sibling case in Germany.
There is a 50% chance of the child having cystic fibrosis. A brother and sister conceiving a chid is a bad idea genetically because their children have a higher chance of getting genetic diseases and disabilities because their parents have similar alleles.
Other than the very prominent rRNA bands, can you see any other size categories of RNA in the two lanes? Describe the size and relative amounts of the different RNA size categories you see. Remember, the gel only appears white where RNA is present.
There is a very slight bit of white at the end of the two bands meaning there are many sizes of RNA
Go back to Figure 1.10A, and locate the major groove which is a larger gap than the minor groove. These two grooves describe the parallel distances 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 protein to stick in the major grooves and hold on the DNA
Look at the two ribbons representing the two antiparallel strands of DNA in Figure 1.10A. Are the two strands equally spaced apart in the horizontal axis? What about the vertical axis?
They are equally spaced horizontally but not equally spaced vertically
When chromosomes are pulled toward the clear zones within a nucleus, which part of the chromosome leads the way? In other words, are they pulled from one end, both ends simultaneously, or somewhere near the middle? Watch the movie to answer this question.
They are pulled apart from somewhere in the middle
When do the future gametes become haploid instead of diploid? Speculate why evolution resulted in a two-phased meiosis rather than a one-phase, four-way separation of all the chromosomes.
They become haploid after the second time they separate. They evolved to have two phases to the gametes can have a more diverse population or create unique babies.
How could a population of bacteria be 0.8 µm3 or 0.4 µm3 at the same time, as shown at 2 and 4 hours in Figure 3.17A? Did the researchers really average all of the cells and plot only one size at a time? For the cells with thymidine, is there a pattern that might indicate how often cells divided in this experiment?
They divided the cells into older cells that have not divided yet and cells that are newly divided as the cells were going through mitosis and were actively dividing. With thymidine, there is a pattern with how often the cell divides because it shows different sizes as time goes on to demonstrate the cell getting smaller.
If ribosomes assemble amino acids into proteins, speculate about the function of tRNA molecules that bind to one amino acid at a time.
They read the codons and apply it to an amino acid to do its job
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 in Figure 1.15B? Keep in mind that DNA is a long polymer and every nucleotide contains multiple N atoms.
They separated groups of the nitrogen so they could easily tell which was which as opposed to just one big group.
Why did they use U's instead of T's in their synthetic mRNA? Which temperature was able to translate the poly-U code the fastest in Figure 2.21? How can you tell? Hypothesize why the amount of phenylalanine polymer produced at 37° C began to decline after 5 minutes.
They used uracil over thymine because RNA has uracil not thymine. The slope at 37 degrees was the highest which means the DNA translated the fastest at 37 degrees. Rapidly declines after 5 minutes because it runs out of substrate.
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.
This wasn't a very strong conclusion because while the cell membrane did originally change R cells to S cells, it was easily washed off the wall meaning it wasn't what made up the wall, but something touching it. So just because the wall did convert the cells initially, it was not the wall itself but what was on the wall
Speculate about possible advantages for chromosomes condensing before separating if they have to relax again later for genes to be transcribed.
To make sure all the parts of the genes are copied and separated
Look at the full genetic code in Figure 2.23; three codons do not encode any amino acid. What is their function? In most species, the first amino acid in a protein is methionine (M). What codon encodes for this amino acid and has the nickname of the "start codon?" Be sure you are able to translate an mRNA using the genetic code.
UGA, UAA, and AAG do not code proteins, but are stop codons to end the amino acid chain. The "start codon"that creates methionine is AUG
What was the experimental benefit to using a bacterial virus when trying to determine whether DNA or protein was the heritable material?
Viruses are not cells so they would only have the DNA molecules or the protein molecules , not both. They also reproduce quickly, and if controlled do not effect humans.
Determine the genotypes of all the F3 peas in Figure 3.12. What are the ratios of the peas that result from the crosses in Figure 3.12? Draw the Punnett squares for each half of Figure 3.12. Do Mendel's original data match the expected ratios? Explain why or why not.
YS, Ys, ys, yS are the genotypes produced. Mendel's original data did not match the expected ratio because it did not include yellow smooth.
Calculate the ratios of the four phenotypes in a dihybrid cross, using the real numerical data in Figure 3.9. Use your Punnett square from the previous question to determine the expected ratio of pea phenotypes in Figure 3.9. Mendel's second famous ratio is the overall ratio of phenotypes presented in the Punnett square of Figure 3.10. What is this famous ratio?
Yellow smooth: 57%, Yellow wrinkled: 18%, green smooth: 19%, green wrinkled: 6%. There is about a 9:3:3:1 expected ratio of the cross.
Are these data more compelling than Avery's data, or could protein still be the heritable material?
Yes because DNA allowed the bacteria to stay more intact unlike the proteins.
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 because most of the proteins were excluded. They limited their conclusion to sulfur-containing proteins because it was how they could convince scientists DNA was the heritable material and not proteins.
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.
How can you tell when a promoter has transcription factors bound to it in Figure 2.15? Do the three transcription factors and RNA polymerase each bind independently to the promoter in Figure 2.15?
You can tell a promoter has transcription factors bound to it shows up block in figure 2.15. The three factors bind together to the promoter, as shown in figure 2.15.
UV light measures the presence of RNA regardless of radioactivity. Which type of RNA is the most abundant in Figure 2.6? What differences do you see compared to Figure 2.3?
rRNA is most abundant. It shows the radioactivity is increased more. rRNA is not radioactive, mRNA is radioactive.
Use the data in Figure 2.5A to determine which form of RNA binds to individual amino acids prior to protein production. Did the amino acid leucine directly interact with rRNA (ribosomes) where proteins are produced? Support your answer with data.
tRNA and leucine does not directly interact with rRNA because rRNA is not radioactive.
