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mart [117]
3 years ago
7

periodic table which pair of elements would most likely have a similar arrangement of outer electrons and have similar chemical

behaviors?
Biology
2 answers:
alina1380 [7]3 years ago
7 0
I'm not exactly sure, but i would guess the noble gasses. The reason for this is that the noble gasses are stable. There outer shells are completely full. That means they have the same number of electrons.
Over [174]3 years ago
5 0

Answer: Carbon and nitrogen

Explanation:

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Natural selection is an evolutionary mechanism by which _______.
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According to this theory, organisms that are better adapted to their environmental conditions survive and increase their population by reproduction. So they pass the advantageous trait to the next generation. In other words, it reflects the survival of the fittest.

Due to this, the advantageous trait becomes more common in the population.

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Which of the following is a factor that determines the rate of impulse propagation, or conduction velocity, along an axon?a) len
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c) degree of myelination of the axon

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6 0
3 years ago
Genes A and B are 10 map units apart and genes B and C are 20 map units apart, with gene B located between genes A and C. What w
Marizza181 [45]

Answer:

2% of the progeny will be double crossovers for the trihybrid test cross

Explanation:  

By knowing the positions of genes, we can estimate the distances in MU between them per region.

  • Genes A and B are 10 map units apart (Region I)
  • Genes B and C are 20 map units apart (Region II)
  • Genes A and C are 30 map units apart

----A-------10MU--------B-------------20MU-------------C---

             Region I                      Region II  

We can estimate the recombination frequencies by dividing each distance by 100.

• recombination frequency of A-B region = 10MU / 100 = 0.10

• recombination frequency of B-C region = 20MU / 100 = 0.20

Now that we know the recombination frequencies in each region, we can calculate the expected double recombinant frequency, EDRF, like this:

EDRF = recombination frequency in region I x recombination frequency in region II.

EDRF = 0.10 x 0.20 = 0.02        

2% of the progeny will be double crossovers for the trihybrid test cross

   

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3 years ago
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