Q: 32 An extraterrestrial life form is discovered on Mars. It has a genetic code much like that of organisms on Earth, except that there are 6 different DNA bases and the base sequences are translated as quadruplets. HOW MANY DIFFERENT AMINO ACIDS could be accommodated by this genetic code? Show the CALCULATIONS fully.

Q: 32 An extraterrestrial life form is discovered on Mars. It has a genetic code much like that of organisms on Earth, except that there are 6 different DNA bases and the base sequences are translated as quadruplets. HOW MANY DIFFERENT AMINO ACIDS could be accommodated by this genetic code? Show the CALCULATIONS fully.

Biology 2e

2nd Edition

ISBN:9781947172517

Author:Matthew Douglas, Jung Choi, Mary Ann Clark

Publisher:Matthew Douglas, Jung Choi, Mary Ann Clark

Chapter16: Gene Expression

Section: Chapter Questions

Problem 35CTQ: Protein modification can alter gene expression in many ways. Describe how phosphorylation of...

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43 For the amino acid sequence shown, could more than one sequence of bases in MESSENGER RNA encode this peptide? MET-PRO-LYS-GLY SER Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a YES NO Unanswered Save 44 A mutation that changes the amino acid sequence in a protein sometimes increases the fitness of the individual. Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a TRUE FALSE
1CFD CALCIUM-FREE CALMODULIN How long is the protein’s primary sequence?Does this protein have any secondary, tertiary and/or quaternary structures present?Are there any molecular additions to this protein that are not amino ac
LS1-1 Which of the following can you conclude about the hemoglobin in individuals with sickle cell anemia? Choose two answers that apply. O The amino acids sequenced for the abnormal hemoglobin replaced a Glucine (Glu) for a Leucine (Leu). The amino acids sequenced for the abnormal hemoglobin was the same as it was for the normal hemoglobin. There was an error in the transcription from DNA to RNA. O The amino acids sequenced for the abnormal hemoglobin replaced a Glucine (Glu) for a Valine (Val). There was an error in the nucleotide sequence in the original DNA strand.
Part II: Information Transfer Background Information - Key Points The background information provided for this lab has given you a general overview of some of 24 the key terms and definitions necessary to understand the transfer of information from gene to protein. The information included below will help you work through the specific problems included in your Tutorial 4 Assignment. When working on the problems remember the base mu to pairing rules (Table 3). Table 3: Rules for nucleotide base pairing. cytosine (C) - guanine (G) adenine (A)- thymine (T) DNA RNA For Transcription: ● ● ● ● cytosine (C) - guanine (G) adenine (A)- uracil (U) Initiation is determined by the recognition of the promoter sequence in the DNA by the RNA polymerase. Stef The transcription start site is downstream of the promoter and is designated as the +1 site. Aspartic acid Alanina Valine Arginine Serine Lysine Asparagine Glutamic TEOPO|0C|AGUCAG|UC|AG/DCAG/3G/ CAGUC UGU A C A Threonine G Methionine Isoleucine…
Sequence: CCACCTGTACCCGGACACACCCTGGTGTCC 1. Identify the gene from which the querysequence originates (Name of gene) 2. Provide the FULLprotein sequence encoded by the gene. 3. Are different splice variants known for this gene? 4. What human disease has been connected to this gene? 5. Calculate molecular weight (kiloDalton, kD) and calculated pI (the pH where the protein carries no net electrical charge) of the protein.
Sequence: CCACCTGTACCCGGACACACCCTGGTGTCC What human disease has been connected to this gene? Calculate molecular weight (kiloDalton, kD) and calculated pI (the pH where the protein carries no net electrical charge) of the protein.
Okazaki fragments in bacterial cells are 1000–2000 nucleotides in length. Those in eukaryotic cells are much shorter, usually only 100– 200 nucleotides in length. Propose some possible explanations for why Okazaki fragments in bacteria are longer than those in eukaryotes. What difference between the two cell types could lead to the difference in Okazaki fragment length?
CAN YOU PLEASE ANSWER 1b 1a) The phosphate groups in the sugar-phosphate backbone of each strand of a DNA molecule have a pKa of ~2.14. Normal physiological pH in most cells is 7.0-7.5, which is significantly above the pKa for the phosphate groups, so every phosphate group in the backbone of both strands of a DNA molecule is deprotonated and carries a -1 charge - but DNA in cells is normally double-stranded. Briefly explain, in your own words, why the large negative charges on the backbones of the two strands of a double-stranded DNA molecule don't cause the strands to separate under normal cellular conditions. 1b) If you raise the pH to 12, but keep everything else about the solution in which the double-stranded DNA is dissolved the same, the two strands of DNA will separate/the DNA will denature. Briefly explain, in your own words, why this happens - why does DNA denature at pH 12 but not at pH 7?
Original sequence: Consider the following coding 71 nucleotide DNA template sequence (It does not contain a translational start): 5’-GTTTCCCCTATGCTTCATCACGAGGGCACTGACATGTGTAAACGAAATTCCAACCTGAGCGGCGT GTTGAG-3’ Question: 4) In a mutant you discovered that the underlined nucleotide has been deleted. What would the resulting peptide sequence be? What type of mutation is this? 5’-GTTTCCCCTATGCTTCATCACGAGGGCACTGACATGTGTAAACGAAATTCCAACCTGAGCGGCGT GTTGAG-3
Select TRUE or FALSE for each of the following statements: 1. Only one of the three phosphate groups present in each nucleotide precursor remains present in a DNA polymer. 2. Starch and cellulose are alike in that both contain sugars bonded together in identical ways. 3. The coding strand of DNA is complementary in sequence to the corresponding MRNA. 4. Ribosomal RNA (rRNA) is synthesised by ribosomes in the process of translation. 5. Polyribosomes speed up the rate of transcription.
Q.)The Bacteria Escherichia coli DNA genome has a molecular mass of about 3.1 X 109 In your answers, show how you came up to each result? (a) How many base pairs does this bacterium contain? (b) How many full double-helical turns does this DNA contain? (c) How long is this DNA in mm?
Complete the following problems. 1. Draw a molecule of phosphatidyl serine using palmitic acid and linoleic (Δ9,12) acid 2. Consider the following coding strand sequence of DNA to a corresponding amino acid sequence. This represents the first amino acids to be translated. Underline the codon where translation will be initiated, and give the amino acid sequence of the peptide produced. 5’-CGATGCTCAAGTTTC-3’ 3. Within the matrix of the mitochondria, the pH is approximately 6.8 ([H+] = 1.6 X 10-7), while it is 7.4 ([H+] = 4.0 X 10-8) in the intermembrane space outside the matrix. Calculate the free energy driving transport of hydrogen ions from the intermembrane space to the matrix. Remember to include both the concentration and the membrane potential factors. 4. Draw the adenosine-5’-diphosphate.