The answer is; glycolysis
This process converts glucose molecule to pyruvate. It is an oxygen-independent pathway, unlike the Krebs cycle. Glycolysis occurs in the cell cytoplasm while the Krebs cycle (aerobic pathway) occurs in the mitochondria. In the presence of oxygen, the product of glycolysis, i.e pyruvate, is fed to the Krebs cycle. If oxygen is unavailable the pyruvate is converted to lactate.
It is the metric unit of measurement for temperature.
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%
Spermatogenesis produces<span> four </span>gametes<span>, while </span>oogenesis produces one with two to three polar bodies left over. because oocytes contain not only genetic material but also mitochondria.
The man responsible for developing the modern scientific classification system, that scientists still use and follow today is Carolus Linnaeus.
