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USPshnik [31]
2 years ago
9

Explain how gravitational contraction, radioactivity, and asteroid and meteorite bombardment heated early earth

Biology
1 answer:
Afina-wow [57]2 years ago
4 0
The surface of the early Earth is often described as hellish by scientists. The high temperature is primarily attributed to the high levels of carbon dioxide in the atmosphere and the molten rock on the surface of the planet. However, these two aren't the only factors in play.  
Another factor that contributed to the extremely high temperature of the early earth is radioactivity. Radioactive decay of the isotopes found underground plus the left over heat from the formation of our planet produced high levels of radiogenic heat.
There's also the kinetic energy from falling asteroids and meteorites. Massive amounts of energy in the form of heat and sound is produced when asteroids and meteorites hit the Earth. Also, the pressure waves that travel radially outwards the center of the impact is similar to the impact of an atom bomb. As such, dust particles envelop the atmosphere, trapping the heat from the impact. 
The last factor is the gravitational contraction, it contributed to the hellish temperature of primordial Earth because the energy generated from such event is enough to trigger hydrogen fusion. 
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Two autosomal genes, J and K, are 60 map units apart. You perform the following testcross: J K / j k x j k / j k. At what freque
Dmitry [639]

Answer:

  • J K / j k = 20%
  • j k / j k = 20%
  • J k / j k = 30%
  • j K / j k = 30%                

Explanation:

To calculate the recombination frequency, we have to know that 1% of recombinations = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.

The map unit is the distance between the pair of genes for which every 100 meiotic products, one of them results in a recombinant one.

So, en the exposed example:

  • J and K are autosomal genes
  • J and K are separated by 60 M.U.
  • 60 M.U. means that there is 60% of recombination.

Cross)             J K / j k                    x                  j k / j k

Gametes) JK  Parental                                     jk, jk, jk, jk

                jk   Parental                                  

                Jk   Recombinant                          

                 jK   Recombinant

One map unit equals 1% of recombination frequency. This means that every 100 meiotic products, one of them is a recombinant one.

1 M.U. -------------- 1% recombination

60 M.U. ------------ 60% recombination

                              30% Jk  +  30% jK

100 M.U. - 60 M.U. = 40 M.U.

40M.U.--------------40 % Parental (Not recombinant)

                            20% JK   +   20% jk

Punnet Square)           JK       jk      Jk      jK

                          jk     JK/jk   jk/jk   Jk/jk   jK/jk

J K / j k = 20%

j k / j k = 20%

J k / j k = 30%

j K / j k = 30%                                

3 0
1 year ago
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