I'm not totally sure. But in my view, <span>(4)a salt solution</span> would be used to prove that the gas produced by the yeast in the vacuum bottle could change the pH of the liquid in the flask
The answer is c because they are not identical to parents but have some things in common ( resemble)
Answer:
THE ANSWER ISSSSSSSSS HOST
Explanation:
Oh sweetheart. Leave the person if you're clearly not in love with them as they are with you. You're only hurting yourself, and that other person. They deserve someone who loves them as much as they love. And so do you! Be with whom you love, and set that other person free. Yes its not going to be easy, but life isn't easy. That's just part of life and growing up.
Answer:
The correct answer is: a.
Explanation:
- A Diploid organism possess two copies ( also called alleles) of each autosomal gene, of which one copy of the gene is obtained from one of its parents and the other copy from its other parent.
- During the process of replication, one of the alleles of the GTPase encoding gene undergoes a mis-sense mutation.
- A Mis-sense mutation can be defined as a non-synonymous mutation in which one of the nucleotide in the sequence of the gene gets altered such that it causes a change in the amino acid encoded by the codon (triplet nucleotide message encoding for a single amino acid) formed by the mutated nucleotide. A single amino acid change in the protein, if occurs in the functional site of the protein, can render a protein non-functional.
- Hence, the mis-sense mutation in one of the alleles encoding for the GTPase gene causes the resulting protein produced from the mutated allele to have drastically reduced function (1%).
- A heterozygous cell is one which has one copy of the normal allele and the other copy of the mutated allele of the GTPase encoding gene.
- As one copy of the normal GTPase allele is present in such an individual, it can produce the normal GTPase protein that can assist in the process of cell growth and division.
- Hence, the phenotype(expressed character) of the heterozygous cell will not be affected due to the mutation and it will be able to grow and divide normally.
