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Vika [28.1K]
3 years ago
5

Describe how a change in dna sequence could result in a changed sequence of amino acids in a protein.

Biology
1 answer:
egoroff_w [7]3 years ago
8 0
A change in DNA could result in a changed sequence by the amount of atoms and chemicals in the DNA
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To better understand the processes of cellular respiration and photosynthesis, answer the following questions for all steps of b
Readme [11.4K]

Cellular respiration: Reagents are oxygen and sugar. The products are water and carbon dioxide.

Each glucose molecule undergoes a slow breakdown into carbon dioxide and water during cellular respiration. In the course of the processes that change glucose, some ATP is created directly. However, a process known as oxidative phosphorylation produces a lot more ATP later.

It occurs in mitochondria.

Cellular respiration can occur both aerobically (using oxygen), or anaerobically (without oxygen).

Photosynthesis: The reagents are carbon dioxide and water. The products are oxygen and glucose.

In the process of turning water and carbon dioxide into glucose, photosynthesis harnesses solar energy as chemical energy. As a byproduct, oxygen is released. As glucose is broken down by oxygen during cellular respiration, chemical energy and heat are released.

Photosynthesis occurs in Chloroplasts.

conditions required: carbon dioxide, water, and sunlight.

To know more about cellular respiration and photosynthesis visit:

brainly.com/question/19255074

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4 0
1 year ago
Please answer question below
Sever21 [200]

Answer:

i will edit in the answer if i get to understand it better

if its 1 choice only then its comet

Explanation:

3 0
3 years ago
Read 2 more answers
Fill out the Alien Periodic table
Oliga [24]

Refer to the attachment for periodic table

7 0
2 years ago
Below is a mature eukaryotic mRNA transcript. Translate this mRNA into a protein, also showing the tRNA anticodons involved. Mak
diamong [38]

Answer:

mRNA ⇒ 5'GMU  UAC  <u>AUG  </u>CGG  CUC  AGU  <u>UGA  </u>GGC  GAA  AAA  A 3'

tRNA ⇒                           UAC  GCC  GAG  UCA  ACU

protein ⇒ N - MET   ARG   LEU   SER   Stop - C

Explanation:

In protein synthesis, the ribosome reads mRNA in the 5´ to 3´ direction, and, according to the <u>codon</u>s that are being readen, tRNA transfers the correct amino acids to build the polypeptide chain. A codon is a short sequence of three nucleotides that store the genetic information for the aminoacids´ assembly. Each tRNA has two important sites. One of them that couples with the codon of the mRNA molecule, named <u>anticodon</u>. The other site couples with an amino acid. tRNA allows amino acids to align according to the nucleotidic sequence in the mRNA molecule.  

Once the new amino acid links to the growing peptidic chain, the binding between the amino acid and the tRNA molecule breaks. The tRNA is now free to join another amino acid and repeat the cycle.  

The protein is synthesized from the amino terminus to the carboxy terminus, while the added amino acids to the chain are coded by a codon formed by three bases in the mRNA. mARNs also have a start and end codon that are the signals of the synthesis initiation and finish. When the ribosome reaches the end codon, protein synthesis is over.    

Each of the codons represents one of the 20 amino acids used to build the protein. Each amino acid can be codified by more than one codon. From the total 64 codons, 61 codify amino acids, and one of them is a start codon. The left three codons are stopping translation points.

The codons indicating the initiation or stop points during the translation process are:

• The start codon AUG is the most common sequence used by eukaryotic cells and places near the 5´extreme of the molecule.  

• The end codons are UAA, UAG, UGA.

Protein synthesis initiates in the AUG start codon -Metionin-, and ends when reaching either of the stop codons UAA, UAG, UGA.

In the exposed example we have the following mRNA.

mRNA ⇒ 5'GMU  UAC  <u>AUG  </u>CGG  CUC  AGU  <u>UGA  </u>GGC  GAA  AAA  A 3'

Codons are separated by a space left between them. AUG is the start codon placed near the 5´ extreme. UGA is the end codon near the 3´ extreme. tRNA will add amino acids from the start codon, not before.

tRNA ⇒ UAC  GCC  GAG  UCA  ACU

Anticodons are separated by a space left between them.

protein ⇒ N - MET   ARG   LEU   SER   Stop - C

Each mRNA codon codifies for an amino acid. The start codon codifies for methionine. AUG = Met, CGG = Arg, CUC = Leu, AGU = Ser, UGA = Stop codon. The amino terminus is represented as an N and the carboxy terminus is a C. The first extreme to be translated carries the amino-terminal group, while the other extreme carries the carboxy-terminus group.

5 0
3 years ago
Huntington’s disease: Huntington’s disease is a disease in which certain parts of the brain waste away; symptoms usually begin l
neonofarm [45]

Answer:

Regarding Huntington's disease, it can be said that it is an inherited disease with an autosomal dominant character, so it is enough that one of the children inherits one of these characters to manifest this neurodegenerative disease, it is considered that the pattern is usually 50% of the total of children of sick parents can suffer it, however if both parents are carriers the number rises leading to 100% of the children affected, on the contrary if no parent has the dominant gene and are homozygous recessive, neither your children will suffer the disease

The possibilities are the following:

Parent 1, inheritance pattern: Hh * hh, the possible gametes for this cross-linking are H, H, h, h, offspring; Hh, Hh, hh, hh 50% of the children could be sick

Parents 2, inheritance pattern: hh * hh, the possible gametes for this cross-linking are h, h, h, h. the offspring will be; hh, hh. no sick child

Parent 3, inheritance pattern: HH * HH, the possible gametes for this cross-linking are H, H, H, H, the offspring will be HH, HH, all children may be sick, 100%

parents 4, inheritance pattern: Hh * Hh, the possible gametes for this cross-linking are h, H, h, H. the descendants will be HH, Hh, Hh, hh 50% of sick children

Parents 5, inheritance pattern: Hh * hh, the possible gametes for this interbreeding are H, h ,, offspring will be Hh all sick children

Parent 6, inheritance pattern: Hh * Hh the possible gametes for this cross-linking are H, h, the descendants will be Hh, the heterozygous descendants will have the disease.

5 0
3 years ago
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