Answer:
anaphase II
Explanation:
Nondisjunction in meiosis II results from the failure of the sister chromatids to separate during anaphase II.
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Part A
Plasmids are extracted from the bacteria
The plasmids are spliced at specific regions using restriction enzymes.
The cut plasmids are incubated with the novel gene and DNA ligase
Plasmids that will have incorporated the new gene are transformed back into the bacteria
Part B
The answer is transgenic organisms.
An example of a 3rd class lever is a pair of tweezers!
In a DNA molecule, three nucleotides make up a codon. A codon is creates a specific amino acid. Therefore, in order to work this out, we must calculate the number of codons that will not be affected by the mutation. This is done by finding the highest number that is divisible by 3 but less than 85. This can be done by:
85 / 3 = 28.333
So 28 codons will not be affected.
The correct answer is A. When free hydrogen ions and free hydroxyl ions are removed from solutions of acids and bases, the acidic and basic properties disappear.
The strength of the acid or base depends upon its degree of ionization. Since different acids and different bases dissociate or ionize to different extent in water so their degree of ionization is also different. Strong acids and strong bases ionize in water completely while the weaker acids or bases ionize partially. Ionization of the weak acids in water are in equilibrium. The degree of ionization is the ratio of the number of ionized molecules to the number of molecules which are dissolved in water. In case of acids the ionized molecules are hydrogen ions and in case of bases hydroxyl ions.
