Question

In section 7.6 of your textbook, you learned how the codon table was filled in. Through logic and simple experiments, it was eventually determined that the minimal number of nucleotides to encode an amino acid is three. It was also determined that the genetic code had no "punctuation" to separate code "words" - nucleotides are read one after the other from beginning to end. Figure 7.30 showed that short, synthetic mRNAs were used to figure out some of the codons. Nirenberg and Matthaei did another experiment, not recounted in your text, that didn't use repetitive polymers, but instead random copolymers with nucleotides of different ratios. The number of codons that resulted could be figured out with the "product rule" where you multiply the abundances of each nucleotide at each of the three positions (you practiced this at this link). Following this same protocol; • Calculate the probabilities of each codon. Round to the 10th (i.e. one decimal place). • In the second column, indicate the three-letter name for each amino acid (Rank from most to least abundant as you move down) • Write percentage expected of each amino acid (round to the 10th). • Put an "X" (no spaces or other characters) in any AA spaces that don't have an amino acid or spots with a zero value. • Check that your percentages add up to 100% ± 0.5% rounding errors. Ensure stop codons (which don't encode amino acids) aren't throwing off your percentages! Uracil = 55%, Guanine = 45% Codon % Probability UUU UUG UGU GUU UGG GUG GGU GGG Amino Acid Total % AA 1 AA 2 AA 3 AA 4 AA 5 AA 6

In section 7.6 of your textbook, you learned how the codon table was filled in. Through logic and simple experiments, it was eventually determined that the minimal number of nucleotides to encode an amino acid is three. It was also determined that the genetic code had no "punctuation" to separate code "words" - nucleotides are read one after the other from beginning to end.
Figure 7.30 showed that short, synthetic mRNAs were used to figure out some of the codons. Nirenberg and Matthaei did another experiment, not recounted in your text, that didn't use repetitive polymers, but instead random copolymers with nucleotides of different ratios. The number of codons that resulted could be figured out with the "product rule" where you multiply the abundances of each nucleotide at each of the three positions (you practiced this at this link).
Following this same protocol;
• Calculate the probabilities of each codon. Round to the 10th (i.e. one decimal place).
• In the second column, indicate the three-letter name for each amino acid (Rank from most to least abundant as you move down)
• Write percentage expected of each amino acid (round to the 10th).
• Put an "X" (no spaces or other characters) in any AA spaces that don't have an amino acid or spots with a zero value.
• Check that your percentages add up to 100% ± 0.5% rounding errors. Ensure stop codons (which don't encode amino acids) aren't throwing off your percentages!
Uracil = 55%, Guanine = 45%
Codon % Probability
UUU
UUG
UGU
GUU
UGG
GUG
GGU
GGG
Amino Acid Total %
AA 1
AA 2
AA 3
AA 4
AA 5
AA 6

Added by Justin A.

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