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Marina CMI [18]
3 years ago
11

In a classic experiment using pea shape, Mendel conducted two separate genetic crosses. In the first cross the parent plants wer

e “true breeding” for pea shape; one had round peas ( R ) and the other had wrinkled (r). The first cross produced a filial 1 generation of all round peas. In the second cross, Mendel bred plants from the filial 1 generation. This cross produced different results. Out of approximately 1000 plants, about 75% were round and 25% were wrinkled. From these experiments, Mendel developed four hypotheses. They include all but ___________. A. One heritable factor may be dominant and mask the other factor. B. Any organism that "shows" a heritable factor must be homozygous.
Eliminate C. An organism has two "heritable factors", now called genes, one from each parent.
D. sperm or egg carries only one heritable factor for each trait in the offspring.
Chemistry
2 answers:
Ivanshal [37]3 years ago
7 0
B. Any organism that "shows" a heritable factor must be homozygous. 

Let's go through all choices:
A. TRUE! One heritable factor may be dominant and mask the other factor called recessive.
B. FALSE! Heterozygous organism can "show" a trait as well. If there is one dominant and one recessive form, a dominant form will mask it and dominant trait will be expressed. 
C. TRUE! An organism gets one form of the gene from the mother, and one from the father.
D. TRUE! That's why sperm or egg are called haploid.
lyudmila [28]3 years ago
4 0

Answer:

A) one heritable factor may be dominant and mask the other factor

Explanation:

any organism that shows a heritable factor must be homozygous.  Mendel did not know  or use terms such as genes, alleles, or homozygous versus heterozygous.

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Dumbledore decides to gives a surprise demonstration. He starts with a hydrate of Na2CO3 which has a mass of 4.31 g before heati
vovikov84 [41]

Answer:

Na₂CO₃.2H₂O

Explanation:

For the hydrated compound, let us denote is by Na₂CO₃.xH₂O

The unknown is the value of x which is the amount of water of crystallisation.

Given values:

Starting mass of hydrate i.e Na₂CO₃.xH₂O = 4.31g

Mass after heating (Na₂CO₃) = 3.22g

Mass of the water of crystallisation = (4.31-3.22)g = 1.09g

To determine the integer x, we find the number of moles of the anhydrous Na₂CO₃ and that of the water of crystallisation:

        Number of moles  = \frac{mass }{molar mass }

Molar mass of Na₂CO₃ =[(23x2) + 12 + (16x3)]  = 106gmol⁻¹

Molar mass of H₂O = [(1x2) + (16)] = 18gmol⁻¹

Number of moles of Na₂CO₃ = \frac{3.22}{106} = 0.03mole

Number of moles of H₂O =  \frac{1.09}{18} = 0.06mole

From the obtained number of moles:

                          Na₂CO₃                               H₂O

                           0.03                                    0.06

Simplest

Ratio                  0.03/0.03                         0.03/0.06

                                 1                                      2

Therefore, x = 2    

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Answer: What ball

Explanation:

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Answer: They all bend light.

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4 years ago
For the following galvanic cell, represented in line notation, determine what balanced half-reactions occur at each electrode. (
Alex787 [66]

Answer:

Anode (oxidation): Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻

Cathode (reduction): Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)

Explanation:

Let's consider the notation of a galvanic cell.

Cr(s) | Cr³⁺(aq) || Ag⁺(aq) | Ag(s)

On the left, it is represented the anode (oxidation) and on the right, it is represented the cathode (reduction).

The half-reactions are:

Anode (oxidation): Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻

Cathode (reduction): Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)

To have the global reaction, we have to multiply the reduction by 3 (so the number of electrons gained and lost are the same) and add both half-reactions.

Global reaction: Cr(s) + 3 Ag⁺(aq) ⇒ Cr³⁺(aq) + 3 Ag(s)

6 0
3 years ago
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