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3 years ago

in sexual reproduction, of each chromosome in a pair has the same genes, how is genetic variety possible?

Chemistry

2 answers:

Sliva [168]3 years ago

6 0

This answer will depend on if you are speaking about asexual reproduction or not.

Variety is possible here because the the product genes are half from the mother and half from the father. So a child might have blue eyes from his mother, but red hair from his father, etc. Different genetic mutations are the initial cause of all differentiation between genes, which are passed down from each generation to the next through this same process.

In Asexual reproduction, all reproductions are essentially clones of the parent, and no genetic variation, other than mutations will occur.

enyata [817]3 years ago

6 0

Punnet Squares, they are charts to determine genetic variation by genotype and phenotype :)

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Molarity of solution = 1.00 M = 1.00 mol/L

In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

$m = 1 mol\times 111 g/mol = 111 g$

Mass of solution = M

Volume of solution = V = 1L = 1000 mL

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$M=d\times V=1.07 g/mL\times 1000 mL=1,070 g$

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2) The value of %(m/V):

$\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%$

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$Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}$

n = Equivalent mass

n = $\frac{\text{molar mass of ion}}{\text{charge on an ion}}$

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Moles of calcium ion = 1 mol (1 $CaCl_2$ mole has 1 mole of calcium ion)

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$=\frac{1 mol}{20 g/mol\times 1L}=0.050 N$

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Moles of chlorine ion = 2 mol (1 $CaCl_2$ mole has 2 mole of chlorine ion)

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$=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N$

Moles of calcium chloride = 1.00 mol

Mass of solvent = Mass of solution - mass of solute

= 1,070 g - 111 g = 959 g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

$\frac{1 mol}{0.959 kg}=1.043 mol/kg$

Moles of calcium chloride = $n_1=1mol$

Mass of solvent = 959 g

Moles of water = $n_2=\frac{959 g}{18 g/mol}=53.28 mol$

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842$

7) Mole fraction of water =

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816$

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

$m'=d\times v=1.07 g/ml\times 100 g= 107 g$

Mass of 100 mL of this solution 107 grams of solution.

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$11.1\%=\frac{m}{100 mL}\times 100$

m = 11.1 g

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