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DedPeter [7]
3 years ago
7

Which statement about RNA is not true? Uracil is found in RNA as one of the two pyrimidine nitrogenous bases. RNA possesses cata

lytic activity, which earned it the name "ribozyme." RNA is thought to have dominated early life on earth, serving as both genetic information and as a catalyst. RNA is a more stable molecule than DNA. RNA typically consists of a single polynucleotide strand with distinct secondary structures.
Biology
2 answers:
Alla [95]3 years ago
6 0

Answer:

RNA is a more stable molecule than DNA

Explanation:

This is a false statement. DNA is far more stable than RNA.

This is  because of  the stability provided by the presence of double helix  strands held by  hydrogen bonds holding the purine and pyrimidine bases  is shape,which releases free energy for   thermodynamic  stability.

it is also stable because of the  interaction of the  hydrogen bonds with the surrounding  water molecules in the  cells.

The presence of Thymine instead of Uracil in DNA is additional factor to DNA stability, because Thymine has resistance to photochemical mutations compare to U in RNA. Thus  genetic code encoded in DNA molecule  is more stable against mutation  compare to RNA.

Furthermore, the phosphodiester  bonds holding the alternate backbone of phosphate and sugar further provided needed stability.

RNA is made up of single strands, made up of Uracil base  of low resistance to photochemical mutations, OH-groups in position 2 of its ribose sugar which makes it to easily hydrolysed in alkaline solution. These factors makes it less stable compare to DNA.

mash [69]3 years ago
4 0

Answer: RNA is a more stable molecule than DNA

Explanation: DNA (deoxyribonucleic acid) is much stable than RNA (ribonucleic acid) for many reasons; Frist of all its because of the double strand of the DNA versus the single-stranded RNA. This is thanks to the Hydrogen bonding between bases since those bondings release free energy that allows the molecule to be provided by more thermodynamic stability.

Also the lack of 2′ OH groups makes DNA more stable since the OH groups are very reactive by themselves. In the DNA sugar molecule there is not a spare OH group that could start reacting on its own. So this is why DNA is immune to chemical reactions, while RNA (having an extra OH group) is more chemically "promiscuous".

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Myosin i think  not sure

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PLEASE HELP ASAP!!!!
marin [14]

Answer:

B. Smaller fragments travel faster and farther than larger fragments.

Explanation: Shorter molecules move faster and migrate farther than longer ones because shorter molecules migrate more easily through the pores of the gel. This phenomenon is called sieving. [2] Proteins are separated by charge in agarose because the pores of the gel are too large to sieve proteins.

6 0
3 years ago
How do global systems interact to affect ecosystems?
miskamm [114]

is a functional unit that results from interactions of abiotic, biotic, and anthropogenic components and are a combination of interacting, interrelating parts that form a unitary whole.  Ecosystems vary in size.  They can be as small as a puddle, or as large as the earth itself.  Basically, any living and non-living things interacting together can be considered an ecosystem.  Within each ecosystem, there are habitats that vary in size.  A habitat is a place where a population lives.  A population is a group of living organisms of the same kind living in the same place at the same time.  Natural ecosystems are made up of abiotic factors such as air, water, rocks, and energy and biotic factors such as plants, animals, and microorganisms.

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6 0
3 years ago
In Drosophila, the genes for withered wings (whd), smooth abdomen (sm) and speck body (sp) are located on chromosome 2 and are s
lesya692 [45]

Answer:

A) 47; B) 33; C) 272; D) 122

Explanation:

The three genes are linked.

The female with withered wings and a smooth abdomen has the genotype whd sm sp+/whd sm sp+.

The male with a speck body has the genotype whd+ sm+ sp/whd+ sm+ sp.

Both individuals are homozygous for all genes, so each of them only produces one type of gamete. The resulting F1 therefore has the genotype whd sm sp+/ whd+ sm+ sp, heterozygous for all genes and with a wild-type phenotype.

The females of the F1 were mated with homozygous recessive males (test cross): whd sm sp/whd sm sp.

<h3>A)</h3>

If we assume interference is 0, the probability of crossing over happening between the genes whd and sm is independent from the probability of crossing over happening between sm and sp.

The distance = frequency of recombination × 100, so the frequency of recombination (RF) between genes whd and sm is 0.305 and the RF between genes sm and sp is 0.155.

<u>The expected double crossover progeny among the 1000 offspring will be:</u>

RF whd-sm × RF sm-sp  × 1000 =

0.305  × 0.155 × 1000 = 47 individuals will be double crossover.

<h3>B)</h3>

Interference is 0.3

The interference is calculated as 1- coefficient of coincidence (cc).

cc = observed double crossover/expected double crossover

Therefore:

I = 1 - cc

cc = 1 - I

<u>cc = 0.7</u>

Observed DCO / 47 = 0.7

Observed DCO = 0.7  × 47

Observed DCO ≅ 33

<h3>C)</h3>

The parental gametes are whd sm sp+ and whd+ sm+ sp (the genotype of the F1 female is known).

Looking at them and at the gene map we can tell that the gametes that give rise to withered wings, speck body (whd sm+ sp) and smooth abdomen (whd+ sm sp+) phenotypes are the result of recombination occurring between genes whd and sm.

To calculate the expected number of individuals with those phenotypes among the 1000 progeny we need to determine the frequency of recombination between the genes whd and sm considering there's interference.

The distance whd-sm = RF x 100

The recombination frequency is the sum of the single crossover between whd and sm and the double crossovers.

The frequency of DCO is 33/1000=0.033.

Distance whd-sm/ 100 = SCOwhd-sm + DCO

0.305 - 0.033 = SCO whd-sm

<u>Frequency of SCO whd-sm= 0.272</u>

And the expected number of individuals with those phenotypes will be 0.272 x 1000 = 272.

<h3>D)</h3>

The gametes that originate the phenotypes withered wings, speck body, smooth abdomen (whd sm sp) and wild type (whd+ sm+ sp+) are the result of recombination between genes sm and sp.

Distance sm-sp/ 100 = SCOsm-sp + DCO

0.155 - 0.033 = SCOsm-sp

<u>Frequency of SCO sm-sp= 0.122</u>

And the expected number of individuals with those phenotypes will be 0.122 x 1000 = 122.

6 0
3 years ago
In chickens, comb shape is determined by alleles at two loci (R, r and P, p). A walnut comb is produced when at least one domina
frutty [35]

Answer:

a) All Walnut , Genotype - RrPp

b) Walnut : Rose: Single

1:1:2

Explanation:

Phenotype and their Genotype -

Walnut comb - R_ P_

Rose Comb - R_ pp

Pea Comb - rr P_

Single Comb  - rr pp

a) RR PP *  rr pp

All progenies will have genotype RrPp (Walnut Comb)

b) Rr Pp *  rr pp

RrPp (4), Rrpp (4), rrpp (8)

Walnut : Rose: Single

1:1:2

4 0
2 years ago
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