Answer:
C. DNA ligase adds nucleotides to the lagging strand
Explanation:
DNA replication is the process during cell division in which DNA copies itself. DNA strands unwind with the help of helicase to initiate the process. DNA Polymerase III is responsible for prokaryotic replication and adds nucleotides in 5' to 3' direction. Since both the strands of DNA run in opposite direction their replication is slightly different. The lagging strand is also formed by DNA Polymerase III in discontinuous manner leading to formation of Okazaki fragments. DNA ligase joins these fragments once the replication process is completed.
Answer: The pathology described in the question is "testicular torsion".
The most sensitive physical exam that is specific to this pathology is testing for reflexes. The inner thigh of the affected side when rubbed should cause the testicle to contract; but in this pathology, it does not.
Explanation:
Testicular torsion is a condition caused by the twisting of the spermatic cord, which causes a loss of blood flow to the testicle. It is the leading cause of testicular loss in adolescent boys, if no surgical intervenvention is carried out as soon as possible.
Rhinoviruses are transmitted through the air or via contact. We might expect this sort of transmission to require a fairly healthy host (one who gets out and comes into contact with others) and, hence, to select against virulent strains. If we take precautions and try to stay away from people or avoid any contact with people, more hand-washing stations, increased attention to sanitization, and isolation of patients will help to reduce the transmission of the disease and, in the process, may favor the evolution of less virulent strains of the virus.
Gymnosperms:- non-flowering,exposed seeds,pine or fir fruit.
angiosperms:- flowering,seeds enclosed in fruit,garden flowers and tomatoes
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%
