Identify the stage of the cell cycle that is represented by the letter I in orange.

Answer:

hi

Explanation:

Answer:

Explanation:

I would say A. Learning to ride a bicycle is a process of getting on and falling off. You have no idea how to do it properly without practice. That's learned.

But look out for D. I'm not sure that that isn't instinctive. Somewhere along the line I think even a baby knows not to let his/her eyes be touched. Is it learned? It's close but A is a much better answer.

The salmon is instinctive. The plant doesn't learn to turn towards the sun. It does so because more food is produced.

Yes, it's A.

A is the only option that is not instinctual. Riding a bicycle is something that has to be taught (and even if you weren't taught to ride a bike, you had to figure it out by trial and error.)

The other options are instinctual and reflexive responses. The salmon returning to the river is a behavior that is pretty much hard-wired into the salmon's brain. The plant turns to follow the sun because it is a source of light and energy.

The other user expressed uncertainty as to option D, but in fact, option D is very much a reflexive, instinctual response as well. To flinch or blink when something abruptly moves towards you is not a choice or something you have to be taught — it's a reflex, something you can't control.

V

Explanation:

1. F

2.T

3.T

4.F

5.F

6.T

7.T

Answer:

Spanish

Explanation:

Answer:

12 out of 16 have yellow leaves

4 out of 16 have green leaves

1 out of 16 has green leaves and rough seeds

Explanation:

<3

The answer to the three questions is as follows:

A dihybrid cross may be characterized as a type of cross that significantly takes place between two individuals with two observed traits that are controlled by two distinct genes. It is a breeding experiment between two organisms that are identical hybrids for two traits.

According to the context of this question, the allele Y encodes for yellow color leaves, and y codes for green color leave. Yellow is dominant over the green. Similarly, S encodes for smooth leaves while s codes for rough seeds. Smooth is dominant over rough.

yyss is the only genotype that has a green and rough phenotype. While the possibilities that have yy in their genotypes must possess a green phenotype.

To learn more about Dihybrid cross, refer to the link:

https://brainly.com/question/12540319

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Answer:

Natural selection is a factor of biological evolution, by means of which individuals who possess a certain characteristic tend to prevail over those who lack such characteristic. Thus, by reproducing, individuals with favorable characteristics end up imposing their own set of characteristics, taking control of the environment in which they develop and relegating other individuals or condemning them to their extinction or displacement.

Thus, in the case, if some mice entered the environment of others, most likely they would not be able to adapt to the conditions to which the original group of that environment had previously adapted. In this way, natural selection itself would make the new mice end up disappearing from it, in favor of the mice that already lived there and that, therefore, had adapted their characteristics to those of the environment.

Answer:

- The virus binds to a receptor on a human cell’s membrane.

- The virus releases its RNA genome into the cell.

- The virus’s RNA is translated into proteins by the cell’s ribosomes.

- Viral RNA polymerase helps transcribe more copies of the virus’s RNA.

- New viruses travel to the cell membrane of the infected cell and are released outside the cell.

Explanation:

Coronaviruses are viruses from the family Coronaviridae (Order Nidovirales), which are structurally composed of positive-sense single-stranded RNA (ssRNA) genomes with a size of approximately 25 to 33 kilobases. Since the coronavirus genome consists of an ssRNA, they can be directly translated into protein inside infected host cells. The coronavirus that has caused the recent pandemic can enter into host cells through its spike (S) glycoprotein which binds to the Angiotensin-converting enzyme 2 (ACE2) receptor. Subsequently, this virus exploits the host molecular machinery to encode an RNA polymerase which is used to generate RNA from an RNA template. Finally, these RNAs are translated into all viral components (structural proteins) and new viruses are recruited to infected cells.

Answer:

Insulin is composed of an A chain of 21 amino acids and a B chain of 30 amino acids. C-peptide is a peptide composed of 31 amino acids

Explanation:

Insulin is composed of an A chain of 21 amino acids and a B chain of 30 amino acids, the chains being held together by two disulfide bonds. A third disulfide bond is present within the A chain. Human insulin differs from porcine insulin by a single amino acid and from beef insulin by three amino acids.

Explanation:

Dashed circles around either an R or K indicate sites of peptidase cleavage that result in generation of the mature insulin that is comprised by two separate peptides: the A chain with 21 and the B chain with 30 amino acids. A third C peptide with 21 amino acids served as a linker between the A and B chains.

Answer:

(b) The interquartile range of B is greater than the interquartile range of A.

(d) The median of A is the same as the median of B.

Explanation:

Given

[tex]A = \{1, 4, 2, 2, 3, 1, 1, 2, 1\}[/tex]

[tex]10th\ run = 9[/tex]

So:

[tex]B = \{1, 4, 2, 2, 3, 1, 1, 2, 1,9\}[/tex]

Required

Select all true statements

(a) & (d) Median Comparisons

[tex]A = \{1, 4, 2, 2, 3, 1, 1, 2, 1\}[/tex]                         [tex]B = \{1, 4, 2, 2, 3, 1, 1, 2, 1,9\}[/tex]

[tex]n = 9[/tex]                                                         [tex]n = 10[/tex]

Arrange the data:

[tex]A = \{1, 1, 1, 1, 2, 2, 2,3, 4\}[/tex]               [tex]B = \{1,1,1,1,2,2,2,3,4,9\}[/tex]

                               [tex]Median = \frac{n + 1}{2}th[/tex]

[tex]Median = \frac{9 + 1}{2}th[/tex]                            [tex]Median = \frac{10 + 1}{2}th[/tex]

[tex]Median = \frac{10}{2}th[/tex]                              [tex]Median = \frac{11}{2}th[/tex]

[tex]Median = 5th[/tex]                                 [tex]Median = 5.5}th[/tex] --- average of 5th and 6th

[tex]Median = 2[/tex]                                    [tex]Median = \frac{2+2}{2} = 2[/tex]

Option (d) is correct because both have a median of: 2

(b) & (c) Interquartile Range Comparisons

[tex]A = \{1, 1, 1, 1, 2, 2, 2,3, 4\}[/tex]               [tex]B = \{1,1,1,1,2,2,2,3,4,9\}[/tex]

[tex]n = 9[/tex]                                                         [tex]n = 10[/tex]

First, calculate the lower quartile (Q1)

[tex]Q_1 = \frac{n + 1}{4}th[/tex][Odd n]             [tex]Q_1 = \frac{n}{4}th[/tex] [Even n]

[tex]Q_1 = \frac{9 + 1}{4}th[/tex]                            [tex]Q_1 = \frac{10}{4}th[/tex]

[tex]Q_1 = \frac{10}{4}th[/tex]                              [tex]Q_1 = 2.5[/tex]

[tex]Q_1 = 2.5th[/tex]                              

This means that:

[tex]Q_1 = 2nd + 0.5(3rd - 2nd)[/tex]              [tex]Q_1 = 2nd + 0.5(3rd - 2nd)[/tex]

[tex]Q_1 = 1 + 0.5(1- 1)[/tex]                   [tex]Q_1 = 1+ 0.5(1 - 1)[/tex]                      

[tex]Q_1 = 1[/tex]                                       [tex]Q_1 = 1[/tex]

Next, calculate the upper quartile (Q3)

[tex]Q_3 = \frac{3}{4}(n + 1)th[/tex] [Odd n]             [tex]Q_3 = \frac{3}{4}(n)th[/tex] [Even n]

[tex]Q_3 = \frac{3}{4}(9 + 1)th[/tex]                            [tex]Q_3 = \frac{30}{4}th[/tex]

[tex]Q_3 = \frac{30}{4}th[/tex]                                     [tex]Q_3 = 7.5th[/tex]  

[tex]Q_3 = 7.5th[/tex]                                    

This means that:

[tex]Q_3 = 7th + 0.5(8th- 7th)[/tex]           [tex]Q_3 = 7th + 0.5(8th- 7th)[/tex]

[tex]Q_3 = 2 + 0.5(3- 2)[/tex]                       [tex]Q_3 = 2+ 0.5(4 - 2)[/tex]                      

[tex]Q_3 = 2.5[/tex]                                       [tex]Q_3 = 3[/tex]

The interquartile range is  [tex]IQR = Q_3 - Q_1[/tex]

So, we have:

[tex]IQR = 2.5 - 1[/tex]                  [tex]IQR = 3 - 1[/tex]

[tex]IQR = 1.5[/tex]                       [tex]IQR =2[/tex]

(b) is true because B has a greater IQR than A

(e) This is false because some spread measures (which include quartiles and the interquartile range) changed when the 10th data is included.

The upper quartile and the interquartile range of A and B are not equal

Answer:

D

Explanation:

LKM iF i AM CORRECT

Explanation:

perform the following the rhythm patterns using the given legend use any kind of membranophones pearl shoes us

Answer:

leopard gecko

Explanation:

Answer:

you chose based off what i type

Explanation:

crested geckos need a vivarium and a uvb light but turn it off at night cause they are nocturnal they need stuff to climb on a hide and a clean cage the eat a powder mixed with water and crickets you can give them other bugs to they can jump from your hand to your arm if they want to they will drop there tail if scared they do not grow back also super cute

leopard geckos need a 20 gallon or bigger terrarium with uvb to turn off at night and a heat pad 24\7 make sure its no to hot leopard geckos need a variety of bugs like crickets and bubia roaches and other  and a clean tank and at least 2 hides they will drop there tail but they do grow back they just look a little different but are super cute

Answer:

1) option a is correct. (260+270)/1545

(Note about this option: In the statement, it is written as (260+270) 1545. Lacks the division symbol, / )

2) From these results one can conclude that the two genes are linked (option a)

Explanation:

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

We might verify which are the recombinant gametes produced by the di-hybrid, and we will be able to recognize them by looking at the phenotypes with lower frequencies in the progeny. Phenotypes with the highest frequencies represent the parental gametes.

To calculate the recombination frequency, we will use the following formula: P = Recombinant number / Total of individuals.  

In the present example:

Parental)        YySs                        x                 yyss

Gametes)  YS parental type                        ys, ys, ys, ys

                 ys parental type

                 Ys recombinant type

                 yS recombinante type

The total number of individuals in the offspring: 1545

Phenotypic class Number of offspring  

According to this information, the phenotypic frequencies of the progeny differ from the phenotypic ratio 1:1:1:1. We can assume then, that these genes are linked.

The recombination frequency is:

P = Recombinant number / Total of individuals

P = 260 + 270 / 1545

P = 530/1545

P = 0.343

The genetic distance between genes is 0.343 x 100= 34.4 MU.

Answer:

C. Autotrophs (plants) are organisms that can produce their own food and heterotrophs (animals) are organisms that must consume or break down other organisms for food.

Answer:

F = 13.5 N

Explanation:

Answer:

they absorb water and minerals from the soil and use the energy of the sun to synthesize sugars (carbohydrates)

Explanation:

Isotherms are lines of constant or equal temperature. They are often used on weather maps by meteorologists to give a large scale view of temperatures across the U.S. If you have ever looked at a weather map in a newspaper, the isotherms are used to divide the color-filled temperatures.

Answer:

cytosine

Explanation:

Guanine is one of the building blocks of DNA. It's the G in the A, C, G, or T. Guanine in the double helix pairs with cytosine, so you will see CG pairs; one on one strand and one on the other.

Answer:

In a DNA molecule, G pairs with C

Explanation:

Wish you luck! Have a good day~

Answer:

.0625 or 6.25%

Explanation:

Answer:

Shrink

Explanation:

The blood cell would shrink because of osmosis. The molecules move with the concentration gradient so water molecules will move outside of the cell where there is less water than inside of the cell.

Blood cells are the main constituents of the circulatory system and carry the most vital function of oxygen transportation. Blood cells have a specific concentration and osmotic pressure to remain vital.

The correct answer is:

Option D. Shrink.

This can be explained as:

Therefore when the red blood cells are placed in 0.8% salt solution they will shrink.

To learn more about hypotonic and hypertonic solutions follow the link:

https://brainly.com/question/4237735

Answer:

a. Chi-Square Value = 0.12

b. Degrees of freedom = 1

c. Reject or fail to reject the hypothesis = fail to reject the hypothesis

Explanation:

Available data:

Let us name the allele for Long L and the allele for dumpy l

The number of observed individuals:

In the F2 generation, if l was a recessive allele and L a dominant allele, exhibiting complete dominance, we would expect a 3:1 phenotypic ratio.

Cross:          Ll   x   Ll

Gametes)   L   l    L  l

Punnett square)   L     l

                     L    LL    Ll

                      l     lL     ll

F2) 3/4 LL + Ll long-winged individuals

      1/4 ll dump-winged individuals

From this ratio and the total number of individuals in the F2 generation, we can get the expected number of individuals:

4 possible genotypes ------- 890 individuals

3 LL + Ll genotype ------------X = 667.5 long-winged individuals

1 ll genotype ---------------------X = 222.5 dump-winged individuals

Chi square = ∑ (O-E)²/E

LL + Ll = (O-E)² /E

LL+ Ll = (672 - 667.5) ² / 667.5

LL + Ll = 0.03

ll= (O-E)²/E        

ll= (218 - 222.5)² / 222.5

ll= 0.09  

0.12 < 3.84 meaning that there is enough evidence to reject the null hypothesis.

Fail to reject the hypothesis.  The genotypes in the population might be in equilibrium H-W. L is the dominant allele, and l is the recessive allele. The dumpy wing is inherited as a recessive trait.

Answer:

A. Compound Light Microscope

Explanation:

USA TestPrep

Answer:

coumpound light microscope

Explanation:

Answer:

I think the answer is D

Explanation:

Hope that helps!

Answer: herbaceous plants are plants with very flexible stems. Their leaves and stems die down to soil level at the end of every growing season amd woody plants are plants with very strong and not easily bendable stems such as trees. Woody plants produce wood as a structural tissue