For a type of pea plant, round seeds (R) are dominant to wrinkled seeds (r). A pea plant
with the genotype Rr is crossed with another pea plant with the genotype Rr. What
percent of the offspring are expected to have wrinkled seeds?
O 100%
O 75%
O 50%
O 25%

Answer:

B) Mutation

Explanation:

Mutation is a mistake in the Gene's I doesnt mean it's a bad thing or a good thing. It could be both

Answer:

1)  UUC-AUG-AUC-AAG-UAA.

2) 48

Explanation:

1) Determine the possible antisense oligo nucleotides possible from this amino acid sequence

Given that the peptide sequence is = Glu-Glu-His-Asp-Val-Leu 1.

we will use a codon table to show the codons and Anticodons for each of the Amino acids  

Amino Acid   Codons                                                         Anticodons

 Glu                GAA, GAG                                                    UUC, CUC

 His                 CAU, CAC                                                    AUG, GUG

 Asp         GAU, GAC                                                    AUC, GUC

 Val                 GUU, GUC, GUA, GUG                             AAG, GAC, UAC,CAC

 Leu   UUA, UUG, CUU, CUC, CUA, CUG  | UAA, CAA, AAG, GAG,UAG,CAG

From the table above

The possible antisense oligo nucleotide sequence is : UUC-AUG-AUC-AAG-UAA.

2)  How many total unique different oligonucleotide would need to be produced to cover all possibilities

A desired peptide sequence requires 16 anticodons

Hence the total unique different oligonucleotide needed to be produced to cover all possibilities = 16 *3 = 48

The ADG1(ADP glucose pyrophosphorylase is involved in flowering timing in plants and a DNA probe is made by using partial protein sequence information. Part of the amino acid sequence is Glu-Glu-His-Asp-Val-Leu.

1. To determine the possible antisense oligonucleotides possible from this amino acid sequence:

Since, for every amino acid, there is a specific codon and anticodons are present so according to the given information, the possible antisense oligonucleotides for Glu-Glu-His-Asp-Val-Leu is UUC-AUG-AUC-AAG-UAA.

2) The total number of unique different oligonucleotides that would need to be produced to cover all possibilities is 48

since the amino acid sequence is Glu-Glu-His-Asp-Val-Leu has 16 anticodons and each anticodon have 3 nucleotides so the possible number of oligonucleotides are  :

16×3 = 48.

A codon is a trinucleotide sequence of RNA that corresponds to a specific amino acid.

An anticodon is a trinucleotide sequence complementary to that of a corresponding codon in a messenger RNA sequence.

Thus, according to codon and anticodon, the possible antisense oligonucleotides for Glu-Glu-His-Asp-Val-Leu is UUC-AUG-AUC-AAG-UAA and a number of unique different oligonucleotides that would need to be produced to cover all possibilities is 48.

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Answer:

Wilting of leaves occur.

Explanation:

if the leaf tissue is kept at high temperature during the process of cell isolation the leaf will shrink because of loss of all water. Due to high, temperature the stomata will be open and water is removed in order to remove heat from the body. A time comes that all water that is present in the cell will be removed and as a result wilting of leaves occur.

Answer:

you got it right

Explanation:

Answer:

Therefore, 98% of the US population have no alleles for PKU

Explanation:

The Hardy-Weinberg equilibrium states that the amount of genetic variation in a population will remain constant from one generation to the next in the absence of disturbing factors.

The Hardy-Weinberg equilibrium is expressed quantitatively using a mathematical equation known as the Hardy-Weinberg equation. the equation is given below:

p² + 2pq + q² = 1

also, p + q = 1

Given a pair of alleles, S and s with A dominant and a recessive

where p is the frequency of the dominant allele in the population,

q is the frequency of the recessive allele in the population,

p² represents the frequency of the (SS)  dominant genotype,

q² represents the frequency of the (ss) recessive genotype,

2pq represents the frequency of the heterozygous genotype

From the given question,

q² = 1/10000 = 0.0001

q = 0.01

from p + q = 1

p = 1 - 0.01 = 0.99

p² = 0.98

Therefore, 98% of the US population have no alleles for PKU

Answer:

sporopollenin

While the exine is composed of sporopollenin, a complex and highly resistant biopolymer containing fatty acids, phenylpropanoids, phenolics and carotenoids, the intine is largely composed of pectin and cellulose.

Explanation:

Answer:

c

Explanation:

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Answer:

i live in Terrestrial ecosystems ( land-based) &,  i shared it with frogs,fishes,birds,snakes,insects,turtles....

Explanation:

Answer:

c. that is double stranded

Answer:

biodiversity is the basis of agriculture and our food systems. agricultural biodiversity also performs ecosystem services such as soil and water conservation, maintenance of soil fertility, conservation of biota and pollination of plants, all of which are essential for food production and for human survival.

good luck :)

hopefully, this helps

have a great day !!

Answer:

The sky.

(Purposely wrong.)

Answer:

negative feedback

Explanation:

This is an example of negative feedback system.

Negative feedback occurs when some function of the output of a system, process, or mechanism is fed back in a manner that tends to reduce the fluctuations in the output, whether caused by changes in the input or by other disturbances.

Negative feedback (or balancing feedback) occurs when some function of the output of a system, process, or mechanism is fed back in a manner that tends to reduce the fluctuations in the output, whether caused by changes in the input or by other disturbances.

A negative feedback is a self-regulatory system in which it feeds back to the input a part of a system's output so as to reverse the direction of change of the output.

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Answer:

B

Explanation:

Ive studied rhis before

Answer:

Plant cell

Explanation:

If you are trying to observe a cell wall, you should look at a plant cell. There are only two different types of cells, which are plant and animal cells. The plant cell is the only one which contains a cell wall. Animal cells actually only have a cell membrane.

Answer:

The correct answer is c.

Explanation:

Polymerase Chain Reaction (PCR) technology is an in vitro enzymatic method that allows the amplification of a specific DNA sequence, that is, it is the DNA amplification technique by means of the polymerase chain reaction (PCR) is a technique that consists of the in vitro amplification of a specific DNA fragment. To carry out the amplification experiment, it is necessary to know, at least partially, the sequence of the fragment to be amplified (a gene, a part of a gene, a non-coding region, ...). Basically, it is about replicating the same DNA fragment over and over again and, for this, we must do in vitro what cells do in vivo to replicate their DNA. Generally, PCR begins with the denaturation or separation of the DNA double helix by heating the sample to a temperature between 94 and 96 ° C to break the hydrogen bonds that joined them, in this way each chain remains as a template for the synthesis of a new complementary strand of DNA.

Answer:

The sucrose is transported around the plant in phloem vessels. It needs to be able to reach all cells in the plant so that the sucrose can be converted back into glucose for respiration. The movement of sucrose and other substances like amino acids around a plant is called translocation .

Belowground organs of plants (e.g. roots and rhizomes) are sinks during plant growth since they cannot perform photosynthesis. Some organs are both a source and sink. Leaves are sinks when growing and sources when photosynthesizing

Answer:

true

Explanation:

I've done this it's TRUE

Answer:

Arteries carry blood away from the heart and veins carry blood back to the heart. The circulatory system carries oxygen, nutrients, and hormones to cells, and removes waste products, like carbon dioxide. These roadways travel in one direction only, to keep things going where they should.

Explanation: dont be men to me i have autism and savant syndrom

a = lil b = p c = a  d = pi add it up lil papi

Explanation:

Answer:

C

Explanation:

0.01kg*(10m/s)^2/2=0.5

Answer:

box 4, because it has more atoms in the same amount of space

[tex]\huge\text{Hey there!}[/tex]

[tex]\large\text{What is \underline{Homeostasis}?}[/tex]

[tex]\large\text{Homeostasis is the self-processing in the biotic structured which}\\\large\text{tended to perpetuate (maintain) rationally while it is modified}\\\large\text{its conditions that can be prototypical for surviving}[/tex]

[tex]\large\text{Homeostasis has.... }\\\bullet\large\text{ Controllable center(s)}\\\bullet\large\text{ Effectors}\\\bullet\large\text{Variables (aka Homeostasis)}\\\bullet\large\text{ Receptors}[/tex]

[tex]\text{Good luck on your assignment and enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]

[tex]\large\text{Here is what each of your features of homeostasis means}\downarrow[/tex]

Answer:

Correct / true

Explanation:

Answer:

True

Explanation:

Chlorophyll produces a green pigment, which is why leaves look green

Answer:

true the green stuff is chlorophyll

Explanation:

The cell multiplies and creates a copy of its DNA during interphase. In the mitotic (M) phase, the cell divides into two new cells by dividing its cytoplasm and DNA into two sets.

A cell consumes the bulk of its time in what is known as interphase, where it creates, duplicates its chromosomes, and acquires available to split. The cell then moves interphase, goes direct mitosis, and completes dividing.

The cell multiplies and the nuclear DNA is duplicated during interphase. The mitotic phase immediately follows interphase.

Chromosomes that have been duplicated are separated and distributed across daughter nuclei during the mitotic phase. Usually, the cytoplasm is divided as well, producing two daughter cells.

Therefore, The first stage and the stage the cell stays in most is Interphase.

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Answer:

Approximately 6944 glucose residues are added enzymatically per second

Explanation:

Cellulose is the main structural polysaccharides in plants. It is composed of unbranched glucose monomer units linked to each other by beta 1-4 glycosidic bonds.

The cell wall and stem of plants cells are composed of cellulose fibers. They provide rigidity and support to the plant.

In the given bamboo plant, the enzymatic addition of glucose units to the growing cellulose fiber chains results in the phenomenal growth rate of the bamboo stem.

Since each glucose unit contributes ~0.5 nm to the length of a cellulose molecule, number of glucose units required for daily growth is calculated as follows:

0.5 nm = 10⁻⁹

0.3 m/0.5 x 10⁻⁹ m = 600000000 units of glucose per day

Number of seconds in a day = 24 * 60 * 60 = 86400 seconds

Number of glucose residues added per second = 600000000/86400

Number of glucose residues added per second = 6944.4 glucose molecules per second

Therefore, approximately 6944 glucose residues are added per second

Answer:

condiction

Explanation:

conduction is the transfer of heat between substances that are in direct contact with each other.