Physics topics for freshmen Students, please help

Set the drip interval time to t

t4=t,t3=2t

1 2)*g*(2t*2t)-(1 2)*g*(t*t)=1 get t=1 15 under the change sign, height h=(1 2)*g*(4t*4t)=

g takes 10

The answer upstairs is not right.

This process can be seen as the free-fall process of a drop of water.

And t is equal.

s2-s1=1m

s2=s1=gt2/2

t。=2t

Here the total height can be set to h and the drip interval to t.

Then there are 4 time intervals for 5 drops of water, and the total time is 4t, h=(1 2)g(4t*4t).

The third and fourth drops are located on the upper and lower edges of the window at a height of one meter.

h3-h4=(1 2)g(2t*2t)-(1 2)g(t*t)=1m, the height h is 16 3, and the time t is 1 15 s under the root number

1/2)g（4t*4t）

The third and fourth drops are located on the upper and lower edges of the window at a height of one meter.

h3-h4=(1 2)g(2t*2t)-(1 2)g(t*t)=1m, the height h is 16 3, and the time t is 1 15 s under the root number

Grafting is a method of propagation that takes advantage of the regenerative power of plants. During grafting, the cambium layers of the two wounded surfaces are brought close together and tied together, and as a result, due to cell proliferation, they heal each other and become a whole connected by vascular tissue.

Therefore, the grafting process does not involve the activity of genes.

In other words, the AA buds are grafted onto AA and grow by transporting water and nutrients through vascular bundles. His genotype is still AA.

So three generations of inbred, the genotype remains the same, and it is still AA.

aa。This is because the shoots and flowers that bloom on this bud are all AA. Therefore, for three consecutive generations of inbred, the genotype of the offspring is AA.

A A Because grafting only uses the nutrients and water uploaded by AA, it will definitely not change the genetic material of AA.

So AA's descendants can only remain AA.

So the genetic factor is AA

Because it is grafted, the genotype of the bud does not change, and the plant AA actually acts as a nutrient provider for the bud AA. And because it is a bud AA self-breeding, the homozygous will not have trait separation, no matter how many generations it has been crossed, the gene is AA.

An amino acid requires at least 65 codons to be translated, and each codon requires 3 base pairs on the DNA for transcription. The number of bases in the gene for this protein is at least 65*3*2=390

1 amino acid corresponds to 3 codons corresponds to 6 bases, or 390 if the stop codon is omitted).

2.The peel of the peach fruit is developed from the ovary wall cells of the female parent, so the fruit developed from the pistil should be hairy when homozygous light peach pollen is given to the hairy plant.

1. At least 65*6=390, the least base is that the protein is composed of 65 homoamino acids.

Although a codon has only three bases, the question asks about the base number of genes, because DNA is double-stranded, so the base number should be six.

2. It should be hairy, because the pulp is developed from the female parent, and the female parent is hairy. After the seeds germinate, they will grow and bear fruit, and they will show trait separation.

1 amino acid:messenger RNA:DNA = 1:3:6 (without regard for termination code).

2 The fruit should be hairy, and the phenotype of the fruit is determined by its own genes of the female parent, as it develops from the ovary wall.

1 codon is needed for 1 amino acid. , transcribed RNA is a codon for every 3 bases, and the RNA and DNA bases are 1:2, so an amino acid needs 6 bases, and the gene is at least 390 bases.

2.Light peach, because it is homozygous, is a dominant pollen, and it has a dominant light peach gene no matter what.

65*6=390

There is hair, the fruit follows its mother, and the seed follows its father.

At the end of the question, the question "What is the proportion of individuals whose offspring do not have trait separation to the yellow round grains of F2", which means that the yellow round grains in F2 do not have trait separation, in fact, it is to ask what is the proportion of homozygous in the yellow round grains in F2.

The genotype of F1 is YYRR

The ratio of progeny trait segregation: 9:3:3:1

Then the ratio of the yellow circle in F2 is 9 16, and the homozygous is naturally only 1 part, and the ratio is 1 9, so it should be that you think about the topic too complicated, but this topic is also ambiguous. But genetically the proportions are generally compared with contemporary comparisons. It is seldom that the number of the next generation is compared to the number of the previous generation.

Personal opinion, I hope it has reference value.

25/36

Yellow round-grained peas in F2 account for 9 16 of F2, then it is assumed that there are only 9 yellow round-grained peas in F2 as follows:

1AABB (no trait separation).

2AAB (there are 2*3 4 grains that will not appear trait separation) 2AAB (there are 2*3 4 will not appear trait separation) 4AAB (there are 4*9 16 will not appear trait separation) then a total of 25 4 grains will not have trait separation, and then the total number of 9 will be 25 36

1) AA or AA, AA, AA

2）aa，aa，aa

Analysis: Dominant is sick! That is, both aa and aa are syndacty, i.e., a is syndacty.

Grandfather A Grandmother AA Maternal Grandfather A Maternal Grandmother AA Father AA Mother AA

Daughter 1 4aa 1 2 aa

When a daughter marries a syndactyly, the male genotype may be AA or AA, so all 3 genotypes may appear.

1.Females suffer: AA or AA Father:

AA Mother: AA Analysis: If the disease is a dominant disease, then the grandmother and maternal grandmother will inherit a gene A from the parents, then the parents should not have the disease, so the disease is a dominant disease, and the genes of the grandmother and maternal grandmother are AA.

And because both grandmothers and grandmothers will pass on a gene A to their parents, and both parents are sick, the parents' genes are AA.

2.aa or aa or aa

Analysis: The genes of male patients with syndactyly type I may be AA or AA. Females can also have AA or AA, so all genotypes can be.

Analysis: The genotype of the female may be AA or AA, and the genotype of the male is AA. There are two types of marriages: AA and AA, and AA and AA.

If it is AA and AA, the offspring must be diseased, with a 50% chance, and if it is AA and AA, the offspring have a half chance of being diseased, with 25% of the disease and 25% of the disease.

1.Female: AA or AA Father: AA Mother: AA

or AA or AA

1) Female patients: AA or AAFather's is AAMother's is AA

2) AA or AA or AA.

3) One-third. Explain the third question: the female patient's parents are both aa, so the probability ratio of the female patient's genotype is aa:

aa=1/3 ：2/3.The genotype of a normal male can only be AA.

So what a normal male gives must be the A gene, so 2 3 1 2 = 1 3

1) The possible genotype of the female patient: AA or AA analysis: because her grandfather and maternal grandfather are both patients, their genotype is A (whether the disease is controlled by the autosomal dominant gene), and her grandmother and maternal grandmother are normal, then their genotype is AA, and because both parents are patients, the genotype of both parents is AA, and it cannot be AA, so it can be seen that (2) the father is AA and the mother is AA.

3) The possible genotype of the offspring is AA, AA, AA(4)1 4 Analysis: The genotype of the female patient is (AA or AA) AA (normal male), and the genotype of the offspring may be 2AA, 4AA, 2AA, and the normal is AA, so the probability of normal offspring is 2 8, that is, 1 4.

The following is copied elsewhere.

Climbing plants such as morning glory and honeysuckle have an extraordinary ability to cling to scaffolds and use the "movement" of the stem tip to climb upward. Take the morning glory as an example, the top of its stem is 10 15 cm long, due to the inconsistent growth rate of the surface in all directions, it can constantly change its position in the air, and always rotate in a certain direction, and with this as the radius, after encountering the attachment in a circle, it will wrap the attachment. In this way, climb to the heights to fight for the sun and rain.

Interestingly, the "head turning movement" of most plants has a certain direction, such as honeysuckle, dodder flower, blood vine, etc., always rotate to the right, morning glory, lentils, aristolochia, yam, etc., rotate to the left and wrap upwards, while Polygonum multiflorum is"Do whatever you want"Turn your head, sometimes left, sometimes right. So, why do these winding stem plants have a fixed winding direction?

Recent research by scientists has shown that the directional nature of the rotation and winding of plants is an instinct inherited from their respective ancestors. Billions of years ago, there were two kinds of climbing plants that were the ancestors, one in the southern hemisphere and one in the northern hemisphere. In order to get more sunlight and space for them to grow and develop better, the tips of their stems are tightly oriented towards the rising sun in the east and setting in the west.

In this way, the stems of plants growing in the southern hemisphere rotate to the right, and the stems of plants growing in the northern hemisphere rotate to the left. After a long process of adaptation and evolution, they gradually formed their own rotational and winding fixed directions. Later, although they were transplanted to different geographical locations, the characteristics of their rotation direction were inherited and fixed.

Climbing plants that originate near the equator, because the sun is in the sky, they do not need to rotate with the sun, so their winding direction is not fixed, and they can rotate and wrap at will.

The winding growth of carob is caused by the uneven distribution of auxin, and the influencing factor is mainly single metering.

Phototropism and backsoil are the growth characteristics of things.

The outer side has a high concentration of auxin. The concentration inside is low, so the outer side grows fast, the inner side grows slowly, and the winding growth is exhibited.