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2 years ago

Mitochondria

Biology

1 answer:

PilotLPTM [1.2K]2 years ago

5 0

Answer:
P

Explanation:
Mitochondria will release ATP, where the structure of ATP is:
Nitrogenous base: Adenine
Pentose sugar: Ribose sugar
Phosphate group: 3 phosphate group

ATP is form by adding one phosphate group to a ADP which only have 2 phosphate group by condensation

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In some chickens, the gene for feather color is controlled by codominance. The allele for black is B and the allele for white is

Agata [3.3K]

Answer:

The question is incomplete, it lacks the main questions, those questions are:

a. What is the genotype for black chickens?

b. What is the genotype for white chickens?

c. What is the genotype for erminette chickens?

Explanation:

This question involves a gene coding for feather colour in some birds, in which the black allele (B) and white allele (W) is involved. According to the question, the gene coding for feather colour exhibits co-dominance i.e. both the black (B) and white (W) allele are expressed in a heterozygous state (BW) to form an erminette phenotype.

Hence, if the neither alleles is dominant or recessive over the other. The black phenotype will only be expressed when the genotype is BB.

The white phenotype will only be expressed when the genotype is WW

The combination of both alleles which results in the simultaneous expression of both in the erminette phenotype, will be BW genotype.

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2 years ago

Which body cavity is further separated into other cavities?.

Tom [10]

Answer: The thoracic cavity is separated from the abdominopelvic cavity by the diaphragm. The thoracic cavity is further separated into the pleural cavity which contains the lungs and the superior mediastinum which includes the pericardial (heart) cavity.

Explanation:

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2 years ago

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The fastest growing segment of the homeless population is ______.

elixir [45]

Unemployed men maybe the answer

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2 years ago

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What is ecodiversity?

Ghella [55]

Answer:

the variety of life in the world or in a particular habitat or ecosystem

Explanation:

you silly goose

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2 years ago

Cystic fibrosis (CF) is one of most common recessive disorders among Caucasians it affects 1 in 1,700 newborns. What is the expe

Phantasy [73]

Answer: The expected frequency of carriers is P(Aa)=0.046.

The proportion of childs with CF is P(aa)=0.024.

25% of having a child with CF (aa).

Explanation:

Hardy-Weinberg's principle states that in a large enough population, in which mating occurs randomly and which is not subject to mutation, selection or migration, gene and genotype frequencies remain constant from one generation to the next one, once a state of equilibrium has been reached which in autosomal loci is reached after one generation. So, a population is said to be in balance when the alleles in polymorphic systems maintain their frequency in the population over generations.

Given the gene allele frequencies in the gene pool of a population, it is possible to calculate the expected frequencies of the progeny's genotypes and phenotypes. <u>If P = percentage of the allele A (dominant) and q = percentage of the allele a (recessive)</u>, the checkerboard method can be used to produce all possible random combinations of these gametes.

Note that p + q = 1, that is, the percentages of gametes A and a must equal 100% to include all gametes in the gene pool.

The genotypic frequencies added together should also equal 1 or 100%, and all the equations can be summarized as follows:

$p+q=1\\(p+q)^{2} = p^{2} +2pq+q^{2} = 1\\P(AA)=p^{2} \\P(aa)=q^{2} \\P(Aa)=2pq1$

So, there are 1700 individuals and only one is affected. Since it is a recessive disorder, the genotype of that individual must be aa. So the genotypic frequency of aa is 1/1700=0.000588.

Then, $P(aa)=q^{2}=0.000588$ . And with that we can calculate the value of q,

$P(a)=q=\sqrt{0.000588}=0.024$

And since we know that p+q=1, we can find out the value of p.

$p+0.024=1\\1-0.024=p\\p=0.976$

Next, we find out the genotypic frequency of the genotype AA:

$P(A)=p=0.976\\P(AA)=p^{2} = 0.976^{2}=0.95$

Now, we can find out the genotypic frequency of the genotype Aa:

$P(Aa)=2pq=2 x 0.976 x 0.024 = 0.046$

Notice than:

$p^{2} + 2pq + q^{2} = 1\\x^{2} 0.976^{2} + 2 x 0.976 x 0.024 + 0.024^{2} = 1$

Then, the expected frequency of carriers is P(Aa)=0.046

The proportion of childs with CF is P(aa)=0.024

If two parents are carriers, then their genotypes are Aa.

Gametes produced by them can only have one allele of the gene. So they can either produce A gametes, or a gametes.

In the punnett square, we can see that there genotypic ratio is 2:1:1 and the phenotypic ratio is 3:1. So, there is a probability of 25% of having an unaffected child, with both normal alleles (AA); 50% of having a carrier child (Aa) and 25% (0.25) of having a child with CF (aa).

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3 years ago