Answer:
Explanation:
Spongy bone is also called cancellous or trabecular bone. It is found in the long bones and it is surrounded by compact bone. ... Trabeculae are spaces created in the tissue by thin areas of osteoblast cells. As a result, trabecular bone has about 10 times the surface area of compact bone.
When there is density it controls the high and low pressure the answer is A
Development and growth
cell reproduction
asexual reproduction
The appropriate response is aponeurorrhaphy. Aponeurorrhaphy alludes to the stutured of an aponeurosis, which is the more profound and thicker band of stringy connective tissue appending muscles to bones. It is a strategy in which the solid sheet of tissue that fortified the patient's muscle to close-by bone.
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%
