The organism that would have the most variation in the DNA of its offspring is the cat (Option C). Meiosis is a type of cell division that generates more genetic variability than asexual types of reproduction.
Meiosis is a type of reductional cell division by which a parental cell produces 4 daughter cells (gametes), each containing half of the genetic material.
Animals (e.g., cats) generate gametes by meiosis which fuse during fertilization to produce new offspring.
Both amoeba and bacteria reproduce by a type of asexual reproduction called binary fission. Moreover, yeasts also reproduce asexually by a process called budding and fission.
Both asexual and sexual types of reproduction generate genetic variability by the emergence of new mutations in daughter cells.
Meiosis generates much more genetic variability than asexual types of reproduction due to two different processes:
- Random assortment of chromosomes, which produces new allele combinations.
- Recombination, i.e., by the exchange of genetic material (DNA) between non-sister chromatids during Prophase I.
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Answer:
sorry I don't know
Explanation:
have a great time with your family and friends
The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.
Isotope mass amu Relative abundance
1 77.9 14.4
2 81.9 14.3
3 85.9 71.3
Express your answer to three significant figures and include the appropriate units.
Answer: 84.2 amu
Explanation:
Mass of isotope 1 = 77.9
% abundance of isotope 1 = 14.4% = 
Mass of isotope 2 = 81.9
% abundance of isotope 2 = 14.3% = 
Mass of isotope 3 = 85.9
% abundance of isotope 2 = 71.3% = 
Formula used for average atomic mass of an element :
![A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]](https://tex.z-dn.net/?f=A%3D%5Csum%5B%2877.9%5Ctimes%200.144%29%2B%2881.9%5Ctimes%200.143%29%2B%2885.9%5Ctimes%200.713%29%5D)
Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu
12 % salt is present in 125 g mixture of salt and sand.
Keep in mind that the total percentage is always 100 %
Therefore, if 12 % is the salt, remaining 88 % must be
sand.
a. The amount of mixture is 125 q. Here, 12 % of 125 is
12 * 125 / 100 = 15 g of salt is present in 125 g mixture.
b. The amount of sand can be calculated similarly, 88 %
of 125 g is 88 * 125 / 100 = 110 g of sand is present in
125 g mixture.
Answer:
<em>fractional distillation</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>most common method for air separation</em><em>.</em>
