<span>While some areas of the U.S. have a high population density, others must have extremely low population densities.</span>
Answer:
There are a couple different possible answers, since there are multiple different structures that are present in both types of cells. So you have to make the choice of what to put.
The structures that prokaryotic and eukaryotic cells have in common are: plasma membrane, ribosomes, and cytoplasm
Answer:
Lamproptera meges, the green dragontail, is a species of swallowtail butterfly (family Papilionidae) found in parts of South Asia and Southeast Asia. There are ten subspecies. A specimen from Java is the type species of the genus.
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:
It enters fish through the gills and the female attaches to the tongue, with the male attaching on the gill arches beneath and behind the female. Females are 8–29 millimetres (0.3–1.1 in) long and 4–14 mm (0.16–0.55 in) wide.
Explanation:
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