If you analyze a double-stranded DNA molecule and find that 15% of all the nucleotide bases are Adenines, you know that there must also be [Select] Thymines, [Select] Guanines, [Select] Cytosines and [Select] Uracils. (Count each of the bases in any double-stranded DNA molecule and calculate their percentages to find the simple key for this if you haven't seen it yet.)

If you analyze a double-stranded DNA molecule and find that 15% of all the nucleotide bases are Adenines, you know that there must also be [Select] Thymines, [Select] Guanines, [Select] Cytosines and [Select] Uracils. (Count each of the bases in any double-stranded DNA molecule and calculate their percentages to find the simple key for this if you haven't seen it yet.)

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

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DNA structure depends on base pairing of its four nucleotides, A, C, T, and G. Nucleotide A pairs with T, and nucleotide C pairs with G. This forms a four-letter DNA “alphabet." Because DNA codes for amino acids in sets of three nucleotides, there are 4 cubed (4'), or 64, possible combinations, coding for 20 different amino acids. What is the best explanation for why there is no selective advantage for DNA to have five nucleotides (e.g., A, C, T, G, and E) with C pairing with either G or functionally equivalent E? It would be impossible to form the DNA molecule, because it must have an equal number of Cs and Gs. Because G and E have the same role, there would still be four functional letters of the alphabet. Replication would be inaccurate because sometimes C would bond with G and sometimes C would bond with E. There would be a five-letter alphabet with 125 combinations, which is too numerous. It is impossible because there are not five known nucleotides in the cell.
If a DNA double helix contains 28% T nucleotides, then what is the percentage of A nucleotides?
The following is diagram of a generalized tetranucleotide. Carbons exist at corners on the shapes and phosphate groups are filled circles. A. Is this a DNA or an RNA Molecule? B. Where is the 3’ end of this tetranucleotide? C. Given that the DNA strand which served as a template for the synthesis of this tetranucleotide was composed of the bases 5’-ACAG-3’, where are the expected bases?
The Tm of a DNA strand can be calculated by hand using the formula: (2 ℃)(?????? ?? ? + ?) + (4 ℃)(?????? ?? ? + ?) = ??℃ Using this formular, calculate the Tm for the following DNA sequence: [CTTTCACAGCCACTATCCAGCGGTAC] Note: This formula has several limitations and is not useful for sequences longer than 14 bp. Use the Internet search to find an online Tm calculator. Use this calculator to find the Tm of the above sequence. Using information from your search, identify three factors that can affect the Tm.
Assume that the molar percentage of thymine in a double stranded DNA is 20. What are the percentages of the four bases (G, C, T, A)? b. The base content of a sample of DNA is as follows: A=31% G=31% T=19% C=19%. What conclusion can be drawn from this information?
The figure shows that the average distance between base pairs measured parallel to the axis of a DNA molecule is 3.4 Å. The average molar mass of a pair of nucleotides is 650 g•mol-1. Estimate the length in cm of a DNA molecule of molar mass 5.1x109 4.0 .26 g•mol-1. cm Roughly how many base pairs are contained in this molecule? [4.0 784615 x base pairs CG G C OH TA Thymine Adenine H. H. OH CH2 N-H.......O CH3 H H N……·H–N HLH TA OH 0-P H2C CG G C H. CH2 Guanine H OH Cytosine ... H-N TA AT H' N.....H-N HLH N-H.......O H2C OH H -o. -P3D0 TA 0- OH H. -CH2 (b) The most common structure of DNA, which is a right-handed double helix. The two strands are held together by hydrogen bonds (a) Base-pair formation between adenine (A) and thymine (T) and between cytosine (C) and guanine (G). and other intermolecular forces.
Using the figure below, which end of the double helix,"A" or "B", would be considered the 5 prime end? B (type "A" or "B" in the blank) Which of the following statements would best explain your reasoning for question 1? 1 (type 1, 2 or 3 in the blank) The end of the DNA strand with the phosphate exposed is the 5 prime end The end of the DNA strand that shows the five carbon sugar pointing down is the 5 prime end The end of the DNA strand with the fifth carbon on the sugar molecule is the 5 prime end
The variable structures of a nucleotide are the sugar and the phosphate group. (a). Say if the statement written above is true or false. (b). Give ONE reason for your answer in part (a).
Cytosine makes up 42% of the nucleotides in a sample of DNA from an organism. The composition of the remaining bases is % adenine, % guanine, and % thymine.
Place an asterisks (*) next to the 3' carbon atoms in the polynucleotide shown. -O CH₂ HOHDH Guanine -O-CH₂ H H H Thymine =P-O-CH₂ H OH Answer Bank H H Cytosine
Moira, a biochemistry major, wanted to explore the shapes a single-stranded DNA molecule can take. She sketched the two shapes below. Her professor was impressed with Moira’s imagination and artistic ability, but she informed Moira that only one of her sketches was feasible. In the sketches, the lines indicate complementary base pairing. (d) Would a new double-stranded molecule assume the shape similar to one in the drawing? (e) Why or why not?
What is the difference between a nucleoside triphosphate and a trinucleotide?