If there were five different types of bases in mRNA instead of four, what would be the minimum codon size (number of nucleotides) required to specify the following numbers of different amino acid types: (a) 4, (b) 20, (c) 30?
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Identify the relevant information provided that will enable you to answer the question. | rBBC6ww4fdpL2n1SfriINw== |
Use the formula and relevant information to solve the specific question asked. | iQrv5b+I5xMOyJWWX4Ow9A== |
Identify the mathematical relationship between the parts of the question. | ERtoTFIKsuAK6xNnbvJriQ== |
Identify the question being asked. | AtTzzErpwCspTRZjRVC4AA== |
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If there were five different types of bases in mRNA instead of four, what would be the minimum codon size (number of nucleotides) required to specify the following numbers of different amino acid types: (a) 4, (b) 20, (c) 30?
a. There is/are p7luMvsfRXBdHbqVMv01QBAk14HudK9cj7pM4p6DA45U67d9hojR7HbhyH9PNhQ7 base(s) in this question, instead of the normal I7uGW8GrAsdHg9eL17Bnhixn/0fsjXE+RmLOCQ== base(s).
b. Codon length, which is normally three bases, is now /LtHuNB/r6u0GxJ47z1rDMQg9a1guvLcolgtt+/ZH8fTWxZl bases.
Use the notepad to highlight or take note of relevant information as you read the question. This will help you keep it in mind while considering the ultimate questions being asked.
Visualize a card table with a space marker for laying down only one card. The space represents a single codon. If you have 5 different cards (bases), how many different codons will you be able to make?
Visualize a card table with a space marker for laying down two cards. The two spaces together represent a single codon (amino acid). If you have 5 different cards (bases) for each of the two spaces, how many different codons (amino acids) will you be able to make?
What mathematical formula can you use to arrive at these answers?
5 bases with 1 codon length = 5 codons/amino acids, can be viewed as 5 ____ 1 = 5.
5 bases with 2 codon length = 25 codons/amino acids, can be viewed as 5 ____ 2 = 25.
What mathematical symbol could you put into the empty space to make both answers correct? Addition, subtraction, division, multiplication, power, square root, log?
Use the formula from the previous step. Remember, in reality more than one codon can usually specify a particular amino acid. You just need at least four codons/amino acids, but there can be more.
Use the formula from the previous step. Remember, in reality more than one codon can usually specify a particular amino acid. You just need at least 20.
Correct.
This type of question is one where you can use a known scenario to build a model and then modify it to see how the model has to change to accommodate the new hypothetical information. Your background knowledge is that in real life, three nucleotides/bases make a codon and there are four bases (adenine, guanine, cytosine, and uracil). You can draw out or find manipulatives like the cards to generate this scenario. Three nucleotides per codon and four bases like this:
So, for 4 bases and 3 base “spots” per codon, we could have AAA or ACA or AGA or AUA and so on. You can simply list out all the possibilities to get the count and then find the mathematical equation that will give you that answer. So, more generally,
possible codons/amino acids = (number of bases)codon length .
If you look at the actual codon table, there are 64 different codons, but one amino acid can specify more than one codon (eg glycine is encoded by GGA, GGC, GGG, and GGT.) Here, x = 43 = 64 possible codons or amino acids using the formula.
So, now all you would need to do is change your model to accommodate 5 bases and an unknown codon length. Notice this requires that you work your problem backwards. But you are familiar with it now.
Incorrect.
This type of question is one where you can use a known scenario to build a model and then modify it to see how the model has to change to accommodate the new hypothetical information. Your background knowledge is that in real life, three nucleotides/bases make a codon and there are four bases (adenine, guanine, cytosine, and uracil). You can draw out or find manipulatives like the cards to generate this scenario. Three nucleotides per codon and four bases like this:
So, for 4 bases and 3 base “spots” per codon, we could have AAA or ACA or AGA or AUA and so on. You can simply list out all the possibilities to get the count and then find the mathematical equation that will give you that answer. So, more generally,
possible codons/amino acids = (number of bases)codon length .
If you look at the actual codon table, there are 64 different codons, but one amino acid can specify more than one codon (eg glycine is encoded by GGA, GGC, GGG, and GGT.) Here, x = 43 = 64 possible codons or amino acids using the formula.
So, now all you would need to do is change your model to accommodate 5 bases and an unknown codon length. Notice this requires that you work your problem backwards. But you are familiar with it now.
Use the formula from the previous step replacing all variables with the values you know. For example, if there were 5 bases and the codon length was 8, the formula would be 58 codons/amino acids or 390625 codons/amino acids. Remember, in reality more than one codon can usually specify a particular amino acid. You just need at least 30.