D is the answer hopw this helps
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:
They concluded that these antibodies were a) Globular proteins
Explanation:
Globular proteins are known to be water-soluble, spherical in shape and to be polypeptides as well. Compared to the other options provided, the selected one is the most correct.
63/64
<span>all you have to do is add up all the possibilities of drawing at least 1 yellow seed </span>
<span>YYY=.75x.75x.75=27/64 </span>
<span>YYG=.75x.75x.25=9/64 </span>
<span>YGY=.75x.75x.25=9/64 </span>
<span>GYY=.75x.75x.25=9/64 </span>
<span>totaling up to 63/64!!!!!</span>