The correct answer is option A, It will decrease in number
Reason -
Since the red-cockaded woodpecker, depends for its fooding on insects which are found in the trees. As the trees are removed, the insects get simultaneously removed. Now the woodpecker will not get food for feeding itself so the first and foremost affects that will take place is the depletion of woodpeckers due to non-availability of foods. However, after certain period of time the woodpecker population may shift to a new location with appropriate food for survival.
Answer:
It will bring about a stop to the translation process
Explanation:
The mutant tRNA despite still being charged with Glu, since it's anticodon is mutated and then recognizes and reads the codon UAA which is one of the stop codons (UAA, UAG and UGA) on the mRNA strand causes the translation process of that particular mRNA strand to stop. And the growing polypeptide chain to be released if any from the ribosomes.
Answer:
The anticodon will be unable to recognize the mRNA codon that is GAA, and the translation of this protein/polypeptide will be abruptly stopped. This may result in a truncated protein which is defected and hence, will be degraded by the relevant mechanisms in place. Since, UAA is actually one of the stop codons, the ribosome will not continue onward with the translation and fall off the mRNA.
Hope that answers the question, have a great day!
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:
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:
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:
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.
Monosaccharides
Fatty acids and glycerol
Amino acids
Water
Ca2+ , Na+ , Ca2+
Answer:
The absorption of monosaccharides takes place via facilitated and cotransport mechanism and is transported through blood capillaries present in the villi.
The absorption of glycerol and fatty acids takes place through diffusion and the majority of them are transported via lymph capillaries, while some are transported through blood capillaries.
The absorption of amino acids takes place via a cotransport mechanism with sodium ions and is transported through blood capillaries.
The absorption of water takes place by the process of osmosis via diffusion and is transported with the help of blood capillaries.
The absorption of calcium and sodium ions takes place via an active transport mechanism, while the absorption of chlorine takes place via the process of diffusion. The transportation of all these ions takes place with the help of blood capillaries.
Monosaccharides, amino acids and ions are generally absorbed through active transport into the blood capillaries, while fatty acids and glycerol are transported into the lymphatic capillaries. Water is absorbed via osmosis into both the blood and lymph capillaries.
The absorption mechanisms of the various food breakdown products are different and determine whether they move into the blood capillaries or into the lymphatic capillaries. This absorption is either through passive or active transport.
#SPJ3
Answer:
"The main role of plasma is to take nutrients, hormones, and proteins to the parts of the body that need it. Cells also put their waste products into the plasma."
towards; away from
B.
away from; towards
C.
side by side to; away from
D.
away from; above
Answer:
A!
Explanation:
Hope this Helps! :)