Answer:
Osteoblasts penetrate the disintegrating cartilage and replace it with spongy bone. This forms a primary ossification center. Ossification continues from this center toward the ends of the bones. After spongy bone is formed in the diaphysis, osteoclasts break down the newly formed bone to open up the medullary cavity.
que contiene el condón?
Answer:
plss translate it in English so i Can easyly answer it.
Explanation:
Thank you.
What are three techniques that can be used to show that the electron transport chain is found on the matrix side of the inner mitochondrial membrane. Select all that apply.
Answer:
Explanation:
Step 1: Generating a Proton Motive Force
The hydrogen carriers (NADH and FADH2) are oxidised and release high energy electrons and protons
The electrons are transferred to the electron transport chain, which consists of several transmembrane carrier proteins
As electrons pass through the chain, they lose energy – which is used by the chain to pump protons (H+ ions) from the matrix
The accumulation of H+ ions within the intermembrane space creates an electrochemical gradient (or a proton motive force)
Step Two: ATP Synthesis via Chemiosmosis
The proton motive force will cause H+ ions to move down their electrochemical gradient and diffuse back into matrix
This diffusion of protons is called chemiosmosis and is facilitated by the transmembrane enzyme ATP synthase
As the H+ ions move through ATP synthase they trigger the molecular rotation of the enzyme, synthesising ATP
Step Three: Reduction of Oxygen
In order for the electron transport chain to continue functioning, the de-energised electrons must be removed
Oxygen acts as the final electron acceptor, removing the de-energised electrons to prevent the chain from becoming blocked
Oxygen also binds with free protons in the matrix to form water – removing matrix protons maintains the hydrogen gradient
In the absence of oxygen, hydrogen carriers cannot transfer energised electrons to the chain and ATP production is halted
Below is a mature eukaryotic mRNA transcript. Translate this mRNA into a protein, also showing the tRNA anticodons involved. Make sure you start and end translation in the right place! Label the ends of the polypeptide chain as N and C terminus.
mRNA: 5'GMUUACAUGCGGCUCAGUUGAGGCGAAAAAA 3'
tRNA:
amino acids:
Answer:
mRNA ⇒ 5'GMU UAC AUG CGG CUC AGU UGA 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 codons 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 anticodon. 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 AUG CGG CUC AGU UGA 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.
the flowers of the ____ bloom with sunrise and close with sunset.
Cystic fibrosis is most common in individuals of Northern European descent, affecting 1 in 3200 newborns. Assuming that these alleles are at Hardy-Weinberg equilibrium, what is the frequency of the disease-causing CFTR alleles in this population
Answer:
0.0177
Explanation:
Cystic fibrosis is an autosomal recessive disease, thereby an individual must have both copies of the CFTR mutant alleles to have this disease. The Hardy-Weinberg equilibrium states that p² + 2pq + q² = 1, where p² represents the frequency of the homo-zygous dominant genotype (normal phenotype), q² represents the frequency of the homo-zygous recessive genotype (cystic fibrosis phenotype), and 2pq represents the frequency of the heterozygous genotype (individuals that carry one copy of the CFTR mutant allele). Moreover, under Hardy-Weinberg equilibrium, the sum of the dominant 'p' allele frequency and the recessive 'q' allele frequency is equal to 1. In this case, we can observe that the frequency of the homo-zygous recessive condition for cystic fibrosis (q²) is 1/3200. In consequence, the frequency of the recessive allele for cystic fibrosis can be calculated as follows:
1/3200 = q² (have two CFTR mutant alleles) >>
q = √ (1/3200) = 1/56.57 >>
- Frequency of the CFTR allele q = 1/56.57 = 0.0177
- Frequency of the dominant 'normal' allele p = 1 - q = 1 - 0.0177 = 0.9823
Which is true if energy in ecosystem
Explanation:
energy flows in only one direction through an ecosystem.
The energy source for active transport is ________ , while the force driving facilitated diffusion is ________.
Answer:
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Explanation:
