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Svetlanka [38]
3 years ago
14

Which organism would have the most variation in the DNA of its offspring?

Chemistry
1 answer:
Arada [10]3 years ago
5 0

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.

Learn more in:

brainly.com/question/7002092

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A 20.0-mL sample of 0.115 M sulfurous acid (H2SO3) solution is titrated with 0.1014 M KOH. At what added volume of base solution
Vitek1552 [10]

Answer:

Explanation:

20mL = 0.020L

0.115M = mol/0.02L

mol=0.0023

equivalence point → mols of weak acid = mols of strong base

0.1014 = 0.0023/volume

volume = 0.02268 L → 22.68 mL

8 0
2 years ago
Use your periodic table of elements. Take any element from the second column, except for beryllium and calcium, and answer the f
Alja [10]

Answer:

Radium (Ra)

Metal

Alkaline Earth Metal

Protons & Electrons: 88

Neutrons: 138

Explanation:

6 0
2 years ago
Read 2 more answers
It is desired to make 1.00 liter of 6.00 M nitric acid from concentrated 16.00 M HNO3.A) How many moles of nitric acid are in 1.
natima [27]

Answer:

A) 6.00 mol.

B) 0.375 L or 375 mL

C) 6.00 M

Explanation:

Hello,

A) In this case, from the definition of molarity, we compute the moles for the given volume and concentration:

n=M*V=1.00L*6.00mol/L=6.00mol

B) In this case, from the stock solution, the required volume is:

V=\frac{6.00mol}{16.00mol/L}=0.375L

C) In this case, we apply the following formula for dilution process:

M_1V_1=M_2V_2

Thus, solving for the final molarity, we obtain:

M_2=\frac{M_1V_1}{V_2}=\frac{16.00M*0.375L}{1.00L}\\  \\M_2=6.00M

Regards.

5 0
3 years ago
Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 8.00 g of octane is m
DerKrebs [107]

Answer:

11.3 g of H₂O will be produced.

Explanation:

The combustion is:

2C₈H₁₈ +  25O₂→  16CO₂  +  18H₂O

First of all, we determine the moles of the reactants in order to find out the limiting reactant.

8 g / 114g/mol = 0.0701 moles of octane

37g / 32 g/mol = 1.15 moles of oxygen

The limiting reagent is the octane. Let's see it by this rule of three:

25 moles of oxygen react to 2 moles of octane so  

1.15 moles of oxygen will react to ( 1.15 . 2)/ 25 = 0.092 moles of octane.

We do not have enough octane, we need 0.092 moles and we have 0.0701 moles. Now we work with the stoichiometry of the reaction so we make this rule of three:

2 moles of octane produce 18 moles of water

Then 0.0701 moles of octane may produce (0.0701 . 18)/2= 0.631 moles of water.

We convert the moles to mass → 0.631 mol . 18 g/1mol = 11.3 g of H₂O will be produced.

4 0
3 years ago
When the following equation is balanced with the lowest whole number coefficients possible, what is the coefficient in front of
jasenka [17]
2Al + 3ZnCl₂ ----> 3Zn + 2AlCl₃

What is the coefficient in front of AlCl₃? ==>> 2
4 0
3 years ago
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