Home
Class 12
BIOLOGY
With four bases the number of possible t...

With four bases the number of possible triplet codons is

A

24

B

32

C

48

D

64

Text Solution

Verified by Experts

The correct Answer is:
D
Promotional Banner

Topper's Solved these Questions

  • GENE, ITS EXPRESSION AND REGULATION

    ARIHANT NEET|Exercise Chapter Exercise (B) Medical entrance s special format questions (Statement based Questions )|13 Videos
  • GENE, ITS EXPRESSION AND REGULATION

    ARIHANT NEET|Exercise Chapter Exercise (B) Medical entrance s special format questions (Assertion reason)|8 Videos
  • GENE, ITS EXPRESSION AND REGULATION

    ARIHANT NEET|Exercise Check point 8.4|20 Videos
  • FUNGI

    ARIHANT NEET|Exercise Chapter exercise C ( Medical entrances gallery (Collection of question asked in NEET & various Medical Entrance Exams )|29 Videos
  • GROWTH, REGENERATION AND AGEING

    ARIHANT NEET|Exercise Chapter exercises (Medical entrances gallery)|11 Videos

Similar Questions

Explore conceptually related problems

A triplet codon means :

Four dice are rolled then the number of possible out comes is :

The number of triplet codons having all the three bases same in 64 triplet codons is :

If a,b, and c are positive integers such that a+b+c le8 , the number of possible values of the ordered triplet (a,b,c) is

If there are 34 amino acids and DNA contains only two nitrogenous bases, what would be the minimum number of bases per codon that code for amino acids?

The set of four quantum number not possible from the following .

If a, b, c are three natural numbres in A.P. such that a+b+c =21, then possible number of odered triplet (a, b, c), is

n-similar balls each of weight w when weighed in pairs the sum of the weights of all the possible pairs is 120 when they are weighed in triplets the sum of the weights comes out to be 480 for all possible triplets,then n is:

If A={14,-11,19,11,7,41,16,23}, find the avernge of the averages of all he possible triplets of numbers takes from the given set A.

Given L.C.M.of three natural numbers a,b,c is 1000 and if the possible number of ordered triplets (a,b,c) is p^(2), where p is a prime number,then find the value of p