If you spilled a bottle of bleach in the laundry room and your mother smelled it in the living room, what would have taken place?osmosis
active transportation
cyrolysis
diffusion

Answers

Answer 1
Answer:

Answer:

Diffusion

Explanation:

The strong smell is diffusing from one part of the home to the other.


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(c) Based on the data in Table 1, describe why the dominant alleles for body color and wing shape are the alleles that produce a gray body and long wings, respectively. Based on the data, describe why the two genes are most likely on different chromosomes or why they are most likely on the same chromosome. Calculate the probability of producing flies that have gray bodies and vestigial wings if a cross is performed between one of the F1 flies from the first analysis and a fly that is homozygous for a gray body and vestigial wings.

Answers

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.

To know more about Mendelian ratio, refer to the following link:

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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.

What is the volume of a block charger (must be in centimeters)

Answers

Answer:\lim_(n \to \infty) a_n \neq √(x) x^(2) \n

Explanation:

Transposons need to __________________ in order to limit their negative impact on the genome of the host cell. A. control their nucleotide length B. regulate their copy number C. control their target-site choice D. avoid transposing into their own genome

Answers

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.

Learn more about Transposons here:

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Plasmodesmata in plant cells are most similar in function to which of the following structures in animal cells?A. tight junctions
B. desmosomes
C. peroxisomes
D. extracellular matrix
E. gap junctions

Answers

Answer:

E

Explanation:

Gap junctions in animals, like plasmodesmata in plants, connect the cytoplasm of two cells through dedicated and regulated channels. A gap junction is made of two hemichannels connected together leaving a channel that is 2-4 nm across -tranversing the two cells' cell membranes. The gap junctions allow for the cells to signal each other through allowing exchange of ions and other moelcules. One significance of gap junction is in enhancing communication between cells in the development of tissues and embryo.

Final answer:

Plasmodesmata in plant cells are most similar in function to gap junctions in animal cells.

Explanation:

Plasmodesmata are channels that connect plant cells and allow for the exchange of various materials, including nutrients and signaling molecules. They are most similar in function to gap junctions in animal cells. Gap junctions are specialized channels that connect neighboring animal cells, allowing for the direct transfer of small molecules and ions.

Like plasmodesmata, gap junctions play a crucial role in cell-cell communication and coordination within tissues. They facilitate the rapid passage of substances, such as ions and small metabolites, between adjacent cells, allowing for coordinated responses and sharing of resources.

Therefore, the answer to this question is E. gap junctions.

Learn more about Plasmodesmata here:

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What characteristics make a species vulnerable to extinction?

Answers

Answer:

Many rare and/or endemic species exhibit one or more of the following attributes which make them especially prone to extinction: (1) narrow (and single) geographical range, (2) only one or a few populations, (3) small population size and little genetic variability, (4) over-exploitation by people

What type of sound is particularly effective for many marine mammals because it allows sound to travel for very long distances in water?​?

Answers

They produce narrow-band high-frequency sound to help them navigate in water. These high-frequency sounds are used in echolocation.  Echolocation is used to identify and locate prey and also avoid the predators. Marine mammals like whales and dolphin use echolocation in different manners.