Solution to the DNA pAequor problem

script.js

// Returns a random DNA base
const returnRandBase = () => {
  const dnaBases = ['A', 'T', 'C', 'G']
  return dnaBases[Math.floor(Math.random() * 4)] 
}

// Returns a random single stand of DNA containing 15 bases
const mockUpStrand = () => {
  const newStrand = []
  for (let i = 0; i < 15; i++) {
    newStrand.push(returnRandBase())
  }
  return newStrand
}


let pAequorFactory = (num, array) => {
  return {
    specimenNum: num,
    dna: array,
    mutate() {
      while(true) {
        let randomDna = returnRandBase();
        let randomMutator = Math.floor(Math.random() * 15);
        if(this.dna[randomMutator] != randomDna) {
          this.dna[randomMutator] = randomDna;
          break;
        }
      }
    },
    compareDNA(pae2) {
      console.log(`The DNA of ${this.specimenNum} and ${pae2.specimenNum} is being compared.`);
      let count = 0;
      for ( let i=0; i<pae1.dna.length; i++) {
        if(this.dna[i] == pae2.dna[i]) {
          count++;
        }
      }
      console.log(this.dna);
      console.log(pae2.dna);
      console.log((count/this.dna.length) * 100);
    },
    willLikelySurvive() {
      let count = 0;
      for(let i=0; i<this.dna.length; i++) {
        if(this.dna[i] == 'C' || this.dna[i] == 'G'){
          count++;
        }
      }
      if (count/this.dna.length >= 0.6) {
        return true;
      } else {
        return false;
      }
    }
  }
}
let survivingPAequor = [];
let count = 1;
while (true) {
  let pae = pAequorFactory(count, mockUpStrand());
  if(pae.willLikelySurvive()) {
    survivingPAequor.push(pae);
    count++;
  }
  if (count == 31) {
    break;
  }
}
console.log(survivingPAequor);