Which of the following statements about homologous chromosomes is likely true

Which of the following statements correctly describes some aspect of ATP hydrolysis being used to drive the active transport of

saw5 [17]

Answer:

B) This is an example of energy coupling

Explanation:

Energy coupling is a process by which cells carry out thermodynamically unfavorable, endergonic (energy-requiring) reactions to other exergonic (reactions that liberate free energy) reactions.

Reactions which are endergonic do not occur spontaneously as they result in a decrease in entropy and a positive free energy change, ∆G.

Exergonic reactions occur spontaneously as they result in an increase in entropy and have a negative free energy change, ∆G.

For example, the active transport of ions against their concentration gradients in cells is a non-spontaneous endergonic process which is coupled to the hydrolysis of ATP, an exergonic process in other for the reaction to proceed. ATP hydrolysis phosphorylates an amino acid side chain in the transport protein which then drives the forward process of transport of the ion.

Considering the above,

Option A: The hydrolysis of ATP is endergonic, and the active transport is exergonic is false because ATP hydrolysis is exergonic while active transport is endergonic.

Option B: This is an example of energy coupling is true.

Option C: Both ATP hydrolysis and active transport are spontaneous because they result in an increase in entropy of the system is false because only ATP hydrolysis is spontaneous

Option D: Neither ATP hydrolysis nor active transport is spontaneous is false because ATP hydrolysis is spontaneous.

Option E: ATP is acting as a transport protein to facilitate the movement of the ion across the plasma membrane is false because ATP is not a protein, rather it serves to activate the transport protein by phosphorylating it.

8 0

3 years ago

1. Ai là người đã đặt nền móng cho di truyền học

makvit [3.9K]

Explanation:

here your answer

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7 0

2 years ago

4) A homozygous groucho fly ( gro, bristles clumped above the eyes) is crossed with a homozygous rough fly (ro, eye abnormality)

docker41 [41]

Answer and Explanation:

A homozygous groucho fly ( gro, bristles clumped above the eyes) is crossed with a homozygous rough fly (ro, eye abnormality).

The F1 females are testcrossed, producing these offspring: groucho 518 rough 471 groucho, rough 6 wild-type 5 1000 a) What is the linkage distance between the two genes? B) Plot the genes on a map c) If the genes were unlinked and the F1 females were mated with the F1 males, what would be the offspring in the F2 generation?

1st cross:

Parental) grogro ro+ro+ x gro+gro+ roro

F1) gro+gro ro+ro

2nd cross:

Parental) gro+gro ro+ro x grogro roro

Gametes) gro+ro+ gro ro

gro+ro gro ro

gro ro+ gro ro

gro ro gro ro

Punnet square)

gro+ro+ gro+ro gro ro+ gro ro

gro ro gro+gro ro+ro gro+gro roro grogro ro+ro grogro roro

F2)

0.518 grogro ro+ro (518 individuals)

0.471 gro+gro roro (471 individuals)

0.006 grogro roro (6 individuals)

0.005 gro+gro ro+ro (5 individuals)

Total number of individuals 1000

Note: These frequencies were calculated dividing the number of individuals belonging to each genotype by the total number of individuals in the F2.

To know if two genes are linked, we must observe the progeny distribution. If individuals, whos genes assort independently, are test crossed, they produce a progeny with equal phenotypic frequencies 1:1:1:1. If we observe a different distribution, that is that phenotypes appear in different proportions, we can assume that genes are linked in the double heterozygote parent.

In the exposed example we might verify which are the recombinant gametes produced by the F1 di-hybrid, and we can recognize them by looking at the phenotypes with lower frequencies in the progeny.

By performing this cross we know that the phenotypes with lower frequencies in the progeny are groucho, rough and wild-type. So the recombinant gametes are gro+ro+ and gro ro, while the parental gametes are gro+ro and gro ro+.

So, the genotype, in linked gene format, of the double heterozygote individual in the F1 is gro+ro/gro ro+.

To calculate the recombination frequency we will make use of the next formula: P = Recombinant number / Total of individuals. The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU).

The map unit is the distance between the pair of genes for which one of every 100 meiotic products results in a recombinant product.

The recombination frequency is:

P = Recombinant number / Total of individuals

P = 6 + 5 / 1000

P = 11 / 1000

P = 0.011

The genetic distance between genes, is 0.011 x 100= 1.1 MU.

Genetic Linkage Map:

Parental Phenotypes)

-----gro+------ro---- -----gro------ro+----

----- gro ------ro---- ---- gro------ ro ----

Recombinant phenotypes)

-----gro+------ro+---- -----gro------ro----

----- gro ------ ro---- -----gro------ro----

If the genes were unlinked and the F1 females were mated with the F1 males, the offspring in the F2 generation would have been

4/16 = 1/4 gro+gro ro+ro

4/16 = 1/4 gro+gro roro

4/16 = 1/4 grogro ro+ro

4/16 = 1/4 grogro roro

Their phenotypic frequencies would be 1:1:1:1 related.

7 0

3 years ago

The large number of alveoli in lungs improve diffusion by ____.

nekit [7.7K]

Answer: a. Increasing the surface area for diffusion

Explanation:

When the air is inhaled through the nostrils, the air containing the oxygen the air diffuses inside the alveoli. The alveoli are the tiny sacs where air enters and exchange of gases takes place. From the alveoli the oxygen enters into the blood stream. The oxygen is utilized by the cells for respiration. The increase in network and number of the alveoli will increase the surface area for the diffusion of oxygen and hence, will speed up the respiration process.

6 0

2 years ago

Two organisms, with the genotypes FFEe and FfEE, are mated. Assuming independent assortment of the F/f and E/e genes: 1) Provide

IgorC [24]

Answer:

<h2> FFEE, FfEE, FFEe, and FfEe</h2><h2> All the progeny show same 1/4 or 25% probability.</h2>

Explanation:

Given; parents are with genotype FFFe and FfEE

F/f and E/e are showing independent assortment,

1. Genotype of all possible offspring are;

gamete from parent FFEe are; FE and Fe ,

gamete from parent FfEE are; FE and fE.

So genotype of offspring are: FFEE, FfEE, FFEe, and FfEe

2. All the progeny show same 1/4 or 25% probability.

7 0

3 years ago