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Cerrena [4.2K]
2 years ago
4

Plate tectonics is the theory that the Earth’s crust is made up of plates that glide over a layer of mantle. A prediction of thi

s theory is that the plates will move relative to each other as they move over the mantle. To test this hypothesis, a scientist places a marker on two adjacent plates to measure their movement over a decade. What would be the expected result of the experiment? How would you interpret the results if the scientist sees no movement?
Biology
2 answers:
MA_775_DIABLO [31]2 years ago
8 0
The plates should have moved, and if they have not then the plate tectonic theory is incorrect
BlackZzzverrR [31]2 years ago
6 0

The expected result of the experiment will be that the markers are either further away from each other, or closer to each other, thus confirming that the tectonic plates are in fact moving and are not static.

On the other hand, the distance between the markers can remain the same, and that is a tricky situation. This may indicate that there's actually no movement, thus making the theory nit valid, but it can also mean something else, that the scientist put the markers on two tectonic plates that are moving in the same direction with pretty much the same speed, so even though there is movement, it can not be noticed by the markers. A nice example will be if there's markers set on the North American plate, and on the South American plate, they move at pretty much the same speed, and also in the same direction, so with setting up markers we can not get to valid results.

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Answer:

Cross each to a rabbit who you know is homozygous - i.e. a white rabbit. If there are any white offspring you know this is the heterozygote.

Explanation:

Heterozygous individuals are those with 2 different alleles of a gene. Homozygous individuals have have 2 alleles that are the same.

Imagine the allele for coat color is B black, or b white. If you cross 2 individuals who are homozygous for the black gene (BB), they would only produce black rabbits in the F1 (BB). If you cross the F1, they could only produce BB rabbits in the F2.

However, if you cross a heterozygous rabbit with a homozygous black rabbit (Bb x BB), you would get either BB or Bb rabbits in the F1. However, intercrossing them could produce BB, Bb, or bb rabbits. Therefore, white rabbits can be produced.

You know that white rabbits are bb. So if you are unsure about the genotype of the black rabbit, you can cross it with a white rabbit (either BB x bb or Bb x bb). If any white rabbits appear in the F1, you know there must be a b allele in the black rabbit genotype, so that rabbit must be heterozygous.

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Answer:

The correct genotypic notation for a wild type with phenotypic characteristics: red eyes, long bristle and round eyes will be any of CNRDll, CnRdl, CnRDl or CNRdl if c, n, r and d are capable of independent assortment and CNRDl if they cannot.

Explanation:

The eye color and bristle size is coordinated by either recessive cn and dominant CN or Cn genes. The wild type in questions has red color eyes, which is a dominant trait coordinated by CN or Cn if c and n can undergo assortment independently or just CN if they cannot. The second characteristics of the wild type is long bristle. We were told that reduced bristle is coordinated by recessive rd Gene. Then the long bristle in our specimen wild type will have dominant genotypic notification of RD or Rd, depending if r and d can undergo assortment independently. The last trait is a round eyes, which is a recessive trait controlled by recessive allele l. Therefore, the combined characteristics of red eye color, long bristle and round eyes shape of the wild type will be CNRDl or any of CNRDll, CnRdl, CnRDl CNRdl depending on the segregation pattern of c, n, r and d.

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