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%
u get a mechanical pencil
Earth quakes can happen anywhere that two tectonic plates interact because thats how earthquakes happen
Answer:
The correct answer is D: "<em>Apes have a Y-5 pattern of cusps, whereas Old World monkeys have a bilophodont pattern</em>".
Explanation:
Apes and old world monkeys only have four kinds of teeth: <em>two incisors, one canine, two premolars, and three molars</em>. This dental formula is 2.1.2.3.
The old world monkeys characterize for having molars with four cusps joined by ridges. They have bilophodont teeth with better crushing, shearing and wearing capabilities than apes.
Apes´molars have 5 cusps, a Y-shaped space between those raised points, and no transverse lophs. These characteristics are known as the Y-5 pattern.
