Answer:
All cells have these four parts in common: a plasma membrane, cytoplasm, ribosomes, and DNA.
Explanation:
So we’ll just use “R” and “r” for this example. If the mother AND father are heterozygous, then both of their genotypes are “Rr” if you work out the lumber square or use the foil method, the box would look like this: RR on top left, Rr on top right, Rr on bottom left, and rr on bottom right. So the genetic probabilities, using four as the sum would be 1:2:1
Salinity refers to the principal characteristic that determines whether an aquatic ecosystem is classified as marine or freshwater. It is the measure of all the salts dissolved in water.
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Answer:</h2>
The principle is <u>4) Archimedes' principle</u>.
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Explanation:</h2>
Archimedes principle, found by the old Greek mathematician and creator Archimedes, expressing that any object totally or incompletely submerged in a liquid (gas or fluid) very still is followed up on by an upward, power the size of which is equivalent to the heaviness of the liquid dislodged by the body.
The volume of dislodged liquid is identical to the volume of an item completely drenched in a liquid or to that portion of the volume underneath the surface for an article halfway submerged in a fluid. The heaviness of the uprooted bit of the liquid is comparable to the extent of the buoyant force.
Answer:
Map distance = Genetic distance, GD = 3.9 MU ≅ 4 MU
Explanation:
The more separated two genes are, the more chances of recombination there will be. The closer they are, the fewer chances of recombination there will be. Two genes that are very close will have a few recombination events and are strongly bounded.
To analyze the recombination frequency, we have to know that:
- 1% of recombination = 1 map unit = 1centi Morgan = 1,000,000 base pairs.
- 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.
Knowing that the genes are linked, we can calculate genetic distances between them. First, we need to identify the genotypes of the parental gametes with the ones of the recombinants. We can recognize the parental gametes in the descendants because their phenotypes are the most frequent, while the recombinants are the less frequent. So:
Parentals:
- black, curved 337
- yellow, straight 364
Recombinants:
- Black, straight 17
- yellow, curved 12
To calculate the recombination frequency we will make use of the next formula: P = Recombinant number / Total of individuals.
P = Recombinant number / Total of individuals.
P = 17 + 12 / 337 + 364 + 17 + 12
P = 29 / 730
P = 0.039
The genetic distance (GD) will result from multiplying that recombination frequency (P) by 100 and expressing it in map units (MU).
GD = P x 100
GD = 0.039 x 100
GD = 3.9 MU ≅ 4 MU