As recently as 1996, the black-footed ferret, North America's only native ferret species, was listed as "extinct in the wild", with only captive individuals surviving in zoos and captive rearing programs. In 2002, 223 ferrets were released in Wyoming in an attempt to re-establish a wild population. Of these only 68 individuals survived the winter into 2003. By 2010, conservation biologists counted 760 ferrets, with an estimated annual growth rate of 0.35, one of the highest of any endangered species managed under the Endangered Species Act. The status of black-footed ferrets will be reevaluated and potentially downgraded from "critically endangered" to "endangered" once their population size exceeds 2000 individuals. Assuming no change to their annual growth rate in 2010, in what calendar year (in AD) would this occur?

Answer:

The correct answer is B

Explanation:

Transposons need to regulate their copy number to avoid errors with chromosomal pairing during meiosis and mitosis such as unequal crossover.

A typical example of this error is called the Alu Sequence or Elements. Alu elements contain more than one million copies found everywhere in the genome of human beings.

Many inherited human diseases such as cancer are related to Alu insertions.

Cheers!

Final answer:

To minimize their negative impact on a host cell's genome, transposons need to regulate their copy number. Unregulated replication could lead to harmful mutations.

Explanation:

Transposons, also known as jumping genes, are sequences of DNA that can move around to different positions within the genome of a single cell. Their movement can cause mutations, which can have negative impacts on the host cell. In order to limit their negative impact on the host cell's genome, transposons need to B. regulate their copy number. If they did not manage this, over-replication of transposons could overload the cell with unnecessary genetic material, leading to harmful mutations or even cancer.

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Answer:

subsoil or horizon b I hope it will help you please follow me

Answer:

When viewing a specimen through a light microscope, scientists use special stains to distinguish the individual components of cells.

Explanation:

Microscope cell staining -

Microscope cell staining is a method that is used to for the better visualization of the cells and the parts of the cell , when are studied under microscope .

Stains are basically a type of biological tissues .

By the use of different type of stains , a nucleus of the cell or the cell wall of the cell are viewed easily , cells are sometimes stained in order to highlight metabolic process and can be used to find the difference between a dead or a live cell .

Some type of stains are specific for living cells , but some stains can work both for living as well as the non living cells .

Mendel was the first to explain the transmission of phenotypic characters and the independent assortment of the genes.

In the given data, F2 progeny of phenotypic traits are shown.

In the above observations, the gray body long wings and ebony body vestigial wing are parental combinations.

Also, the gray body and vestigial wings, and ebony body long wings are the recombinants.

The given ratio in the progeny indicates that gray bodies and long wings are expressed.

The genes for the two traits are independently assorted, which means that genes are unlinked present on the samechromosome.

Now,

For the F1 progeny:

• Gg Ll - G for Grey and L for Long
• The cross between F1 progeny and true gray vestigial will be:
• GGll x GgLl
• Gametes:  GL Gl gL gl Gl

The cross between Gl and hybrid will result in a 50% chance of flies having the gray body and vestigial wings.

b) In the above given F2 progeny, the cross between true gray body and long wings with true ebony body and vestigial wings, will result in the independent assortment of the genes.

Given:

• Parental gametes - GGLL x ggll
• Gametes produced - GL and gl

For the F1 progeny, all the offspring will have genotype GgLl  (Gray body and long wings but in heterozygous condition)

The above cross can be shown in the Punnett square, which is given in the attachment below.

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Answer:

Check the explanation

Explanation:

Consider the below table of F2 data of phenotypic flies: we can reconstruct the table as seen in the second attached image below:

From the above observations it is clear the Gray body long wings and Ebony body vestigial wings are the parental combinations and that the Gray body vestigial wings and Ebony body long wings are the recombinants.

Also the ratio indicates that Gray body is dominant over ebony body and that long wings is dominant over vestigial wings.

here the genes for the two characters have shown independent assortment which means that the genes are unlinked if located on the same chromosome or are located  on different chromosomes.

Now F1 hybrid= GgLl (G for Grey and L for Long)

Cross between F1 hybrid and true breeding Gray vestigial (GGll)

GgLl x GG ll

Gametes-----------> GL Gl gL gl Gl

        GL                          Gl                      gL                  gl

Gl    GGLl                      GGll                  GgLl               Ggll

   (Gray long)      (Gray vestigial)     (gray Long)     (Gray vestigial)

Therefore the probability of getting the flies with gray body vestigial wings= 2/4= 50%

b) The reason why the students obtained the above given F2 results involving a cross between true breeding Gray body and long wings with true breeding Ebony body and vestigial wings is that the genes for the two charcters asssort independently in F2 generation and that the genes are not linked as:

Parents------------------> GGLL x ggll

Gametes -----------------> GL gl

F1---------------------> GgLl (Gray long but in heterozous condition)

Now GgLl x GgLl

Gametes GL Gl gL gl   GL Gl gL gl

Here gametes assorted independently and hence in F2 generation we got the above results (U can show the results in the form of punnett square.

Answer:

0.549 is the frequency of the F allele.

0.495 is the frequency of the Ff genotype.

Explanation:

FF or Ff genotypes determine freckles, ff determines lack of freckels.

In this class of 123 students, 98 have freckles (and 123-98= 25 do not).

If the class is in Hardy-Weinberg equilibrium for this trait, then the genotypic frequency of the ff genotype is:

q²= 25/123

q²=0.203

q=

q= 0.451

q is the frequency of the recessive f allele.

Given p the frequency of the dominant F allele, we know that:

p+q=1, therefore p=1-q

p=0.549 is the frequency of the F allele.

The frequency of the Ff genotype is 2pq. Therefore:

2pq=2×0.549×0.451

2pq=0.495 is the frequency of the Ff genotype.

Final answer:

The frequency of the dominant allele, F, in this class is 0.55. The frequency of the heterozygous genotype, Ff, is 0.495. This is calculated using Hardy-Weinberg equilibrium and observed phenotype frequencies.

Explanation:

To start, we need to calculate the frequency of the recessive allele, f, which is easily calculated as those who do not have freckles. From a total of 123 students, 98 have freckles, leaving 25 students with no freckles, which represents individuals who are homozygous for the recessive trait (ff). As these are the only individuals we can be sure of, we take the square root of their frequency to get the frequency of the recessive allele, q. In this case, q = sqrt(25/123) = 0.45. To find the frequency of the dominant allele, p, we subtract q from 1 (because p + q = 1), so p = 1 - q = 0.55.

Next, we'll calculate the frequency of the heterozygous genotype Ff.

Using Hardy-Weinberg equilibrium, we know this is represented by 2pq. Hence, the frequency of genotype Ff would be 2 × 0.55 × 0.45 = 0.495.

This process offers an example of applying the principles of population genetics and Hardy-Weinberg equilibrium to determine the likely genotype frequencies in a given group of individuals based on observed phenotype frequencies.

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