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%
Answer:
the messenger RNA (mRNA) is decoded in a ribosome, outside the nucleus, to produce a specific amino acid chain, or polypeptide. The polypeptide later folds into an active protein and performs its functions in the cell
Explanation:
Both human and bacteria cells contain DNA and both have a outer membrane that controls how materials come into and leave the cell.
Predominantly multicellular not in bacteria but in human body
*Cell contains a nucleus and other membrane bound organelles not in bacteria but in human body..
*DNA occurs in a circular form* in bacteria only..
* Ribosomes size in bacteria 70s and in human body 80s
*Capable of growth at temperatures greater than 80 C only in bacteria
*His tone proteins present in cell only in human body cells
* Operons present in DNA only in bacteria..
