Question 3: Complementarity & Genetic Code. 5 points A. 2 point. Imagine you hav

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Question 3: Complementarity & Genetic Code. 5 points A. 2 point. Imagine you have a DNA double helik that is 200 bp in length (meaning both the top and bottom strands are each 200 nucleotides long). If this dsDNA molecule has 60 adenines altogether (counting the total number of As on the top and bottom strands), specify how many of each of the other nucleotides must exist in this molecule. If there is not enough information, specify "X # of thymines: # of cytosines; # of guanines: B. 1 points. As you know the genetic code is redundant' because most amino acids can be specified by more than one codon. For example, the amino acid proline can be specified by these 4 different codons: CCU, CCC, CCA, cCG. This means that there are 4 different tRNA genes in the genome that get transcribed and then conjugated to the amino acid proline. These 4 tRNAs each have a different anti-codon to match the codon sequences for proline List the 4 tRNA anti-codons that match codons for proline. Be sure to write each tRNA with the polarity as specified, 3' to 5 3' 5 5 3' 5 3' 5 C. 2 points. Imagine you decide to build an organism from scratch, using the same genetic code that we currently have. However, to save time and money, you decide to use only one of the possible codons for each amino acid. That means you only have to make 20 kinds of tRNA genes, one for each of the 20 amino acids. You will also have to reserve 1 codon to be the "stop" codon. Once you do this, how many unused codons will you have? Note: this question has real-world applications because you can use the unused codons to make an texpanded genetic code", e.g. see: https:/len wikipedia.org/wiki/Expanded genetic code" To get credit, show the equation you used to derive your answer.

Explanation / Answer

A) Here, DNA double helix has length of 200bp which means that there are total 400 nucleotides (200 nucleotides in top strand and 200nucleotide in bottom stand)

now problem says that number of adenines = 60 which can basepair with 60 thymidines

and rest of remaining nucleotide will be guanines and cytosines = 400-(60+60) = 280

guanines and cytosines will be equal in number = 280/2 = 140 each

hence, your answer is number of Thymidines =60, number of guanines= 140 and number of cytosines=140

B) C base pair with G and vice versa while U base pair with A and vice versa

so list of tRNA anticodon that will match codons for proline in   3' GGA 5'    3' GGG 5'     3' GGU 5'and 3' GGC 5'

c) codon is made of 3 nucleotides which can be the combination of any of the 4 nucleotides ( A U C and G)

so total number of possible codon = 43 =64

out of 64 codons, 3 are stop codons. so 61 codons are designated to 20 amino acid

As you have used only one codon for each amino acid i.e. number of leftover codon for amino acid= 61-20 = 41

and you have reserved only one stop codon i.e. the number of leftover stop codon = 3-1 =2

Hence total number of unused codon = 41+2 = 43

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