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Gnoma [55]
2 years ago
7

1. Unlike catastrophism, _______ states that mountains formed slowly over time. 2. William Smith was working as a surveyor in th

e late eighteenth century. He noticed that certain fossils were found in certain layers of sedimentary or igneous rock, that the same order of strata could be found in areas that were far apart, and these strata contained similar _______. 3. The Permian Period lasted from 299 Ma to 251 Ma. This means that the Permian Period lasted _______ years.
Biology
1 answer:
Viefleur [7K]2 years ago
7 0

Answer:

uniformitarianism, Fossil groups, 48 million years ago.

Explanation:

  • Uniformitarianism is the doctrine of uniformity that assumes that same environmental laws and process operate till date that has also occurred on the past but not with the same uniformity. Unlike the atmospheric that is short-lived and sudden changes over the period of time.
  • William Smith was a geologist who has noticed a certain type of fossil in a certain layer of sedimentary and igneous rock. A similar layer of strata could be found in the areas that were further apart.
  • The Permian period was from 229 to 251 million years and was the last period of the paleozoic era and is recorded for the last major extinction in the history of life in earth and it lasted for 48 million years.
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<u>Answer:</u>

<em>The steps in making packaging and exporting a protein from a cell are listed below in the following points:</em>

  • <em>Protein is made from the Ribosomes.</em>
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  • <em>From ER the proteins are exported to the Golgi bodies. These Golgi apparatus is found in the vesicles.</em>
  • <em>The Golgi bodies modify the protein to suitable forms that can be absorbed.</em>
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Lipogenesis is common for organic molecules because the intermediate ________ is formed in most nutrient metabolic processes. gl
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I think, Lipogenesis is a common for organic molecules because the intermediate Acetyl-CoA is formed in most metabolic processes. Lipogenesis involves the formation of fatty acids from Acetyl-CoA. Acetyl-CoA is an intermediate stage in metabolism of simple sugars, such as glucose, which is the preferred source of energy for most living organisms.  
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Lydia hypothesizes that warmer temperatures promote faster cell division. Which experimental design would best allow Lydia to te
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4 0
2 years ago
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you are a molecule of carbon. choose a starting point in the carbon cycle and describe the process you would go through to move
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1) start as a carbon molecule in the atmosphere
2) taken in by trees through photosynthesis
3) carbon is taken into decayed organism
4) then it is taken into dead organisms and waste products underground
5) millions of years later, it is stored in a fossil
6) fossil fuels used by factories then emit carbon dioxide back into the atmosphere (back to starting position

if you want the whole cycle then..
7) used again by a tree
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1 year ago
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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%                                

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