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Alex787 [66]
3 years ago
12

Give correct procedure of activities carried out in an experiment to investigate soil drainage​

Biology
1 answer:
tatiyna3 years ago
5 0

Explanation:

These are 5 inverted cut bottles with different soils:

(Gravel, Sand, Loam, Clay, and Peat)

- The piece of cloth is tied at the mouth of the bottle to let out water at the same rate as the piece of cotton wool in the previous diagram.

- Soil that is very porous holds very little water and therefore allows too much water to pass through it.

- From the above, gravel has very large particles,it allows too much water to pass through it meaning it holds little water.

- Peat has more fine particles than Clay Soil, hence holds more water than Clay. It therefore allows very little water to pass through it.

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Energy is stored in the chemical bonds. 
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What occurs after cytokinesis is completed at the end of meiosis 1
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The two cells go into interphase, preparing to split apart again.
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If the mum is heterozygous and the dad is heterozygous what are the genotype possibilities​
Fiesta28 [93]

Answer:

dominant

Explanation:

overall

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3 years ago
अट नही रही है कविता में किस ऋतु का वर्णन है-<br>(क) ग्रीष्म ऋतु<br>(ख) वर्षा ऋतु<br>(ग) वसंत ऋतु​
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Your answer is,

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(ग) वसंत ऋतु

Hope it helps you!

7 0
4 years ago
In a population of cats, the phenotypic frequency of black cats is 91%, and the phenotypic frequency of white cats is 9%. Assumi
Talja [164]

Answer:

1. Allele frequency of b = 0.09 (or 9%)

2. Allele frequency of B = 0.91 (0.91%)

3. Genotype frequency of BB = 0.8281 (or 82.81%)

4. Genotype frequency of Bb = 0.1638 (or 16.38%)

Explanation:

Given that:

p = the frequency of the dominant allele (represented here by B)  = 0.91

q = the frequency of the recessive allele (represented here by b)  = 0.09

For a population in genetic equilibrium:

p + q = 1.0 (The sum of the frequencies of both alleles is 100%.)

(p + q)^2 = 1

Therefore:

p^2 + 2pq + q^2 = 1

in which:  

p^2 = frequency of BB (homozygous dominant)

2pq = frequency of Bb (heterozygous)

q^2 = frequency of bb (homozygous recessive)

p^2 = 0.91^2 = 0.8281

2pq = 2(0.91)(0.9) = 0.1638

4 0
3 years ago
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