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%
B. CO2 levels in the atmosphere could increase and contribute to global warming problems.
<h2>Development of Plant Needles</h2>
Explanation:
- Seed of pitch pine treated with colchicine delivered tetraploid seedlings which had thick and sporadic needles and less fortunate tallness and diameter growth than ordinary seedlings.
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In test of colchicine-initiated polyploidy in pines, researcher found that a significant number of the polyploid plants returned to a diploid development in light of the fact that the polyploid cells partitioned at a more slow rate and were overwhelmed by the more quickly developing diploid cells which encompassed them.
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The primary needles of both diploid and polyploid plants were more effective at low light intensity than secondary needles, and they had lower compensation points.
C. Webbed feet. I'm pretty sure that's the answer lol
Genetic diseases is one of the topics that will be covered in biology
