Answer:
B. As the distance between loci increases, some multiple crossovers go undetected such that the relationship between recombination frequency and map distance ceases to be linear.
Explanation:
When calculating recombination frequencies, and hence, map distances, we might notice that these distances are not completely additive. They might vary. <em>For example, let us say that we have three genes, A, B, and C, in that order. </em>We calculated that the <em>distance between A and B equals 5.9</em> MU and that <em>B and C are 19.5 MU apart.</em> According to this, we might say that the <em>total distance between A and C is 25.4 MU (5.9 + 19.5). </em>However, after a<em> two-point calculation between A and C, the value equals 23.7 MU. </em>
The recombination frequency between these two genes located in the extremes and far apart underestimate the actual genetic distances between them because there might occur other crossing-overs that were not detected. This is <em>when calculating the distance between A and C, we probably will not detect the occurrence of a double recombinant between them, and hence, we might sub-estimate the real distance.</em>
The relationship between the actual map distance (number of crossing overs) and the recombination frequency between two loci, is not lineal. The farther apart are the two genes, the worse is the distance estimation.
Answer:
C)
Explanation:
Meiosis creates sex cells (gametes) for sexual reproduction. And the resulting daughter cells have have the number of chromosomes as the parent cell.
Answer:
number 1 its 25 number 2 its 10000
Explanation:
There are 4 nitrogenous bases in DNA. Adenine pairs with thymine, and Guanine pairs with cytosine. A Dna Nucleotide consists of a 5 carbon sugar, a phosphate group and a nitrogenous base. These bases are paired up by weak hydrogen bonds, so when they have have to replicate, the enzyme helicase can easily unzip the Dna. <span />
