<span>The barn owl's right ear opens slightly upward while its left ear opens slightly downward. this difference enables the owl to detect the location of a sound. It would actually depend on the location of where it would be at, and based on this, depending where it would then be located, this would then be able to differentiate where it would be located.</span>
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:
CGTATC - DNA complementary strand- GCATG
CTTTCAAG- DNA complementary strand-GAAAGTTG
GAGACTTAC-DNA complementary strand-CTCTGAATG
Explanation:
Remember that the complementary sequence will be based on base pairing for DNA, so the A will pair with T, C with G
Answer:
The cell membrane acts as a barrier that protects the cell from the external environment and it regulates the movement of the particles or substances in and out of the cell.
<span>they both have a question, make a hypothesis, test the hypothesis, and both collect and analyze data and come to a conclusion </span>