Answer:
<h2>Helicase, topoisomerase ii /gyrase, single strand binding proteins.</h2>
Explanation:
DNA replication is the process in which DNA is replicated with the help of various enzymes and proteins..
Helicase is the enzyme which unwind the DNA strands, After unwinding, topoisomerase removes these twists. Single strand binding proteins stabilize the single strands of DNA during replication.
Answer:
The claim by the researcher is supported because of the known functions of insulin and growth factors. Both hormones have a tendency to respond to the presence of glucose. The growth hormone initiates STAT5 signaling for producing the Igf-1 factor that is important for utilizing glucose. The insulin hormone is well known for utilizing glucose.
This would be the organ. The order of all would be organelle, cell, tissue, organ, organ system, organism.
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%