<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>
- <em>These proteins are gathered in the endoplasmic reticulum. </em>
- <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>
- <em>Finally it is transported to all part of cells in our body.</em>
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.
Answer:
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
8) released as organic carbon (some)
9) tree leaf is eaten by an animal, which then releases carbon either from respiration or when it dies
Answer:
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%