Which of the following are characteristics of ultramafic igneous rocks

Oth seagulls and wild geese are large birds. they're also similar in that both kinds of birds can fly over great distances. on t

rusak2 [61]

The correct option is A, FEEDING HABIT AND MIGRATION. From the passage given above, it can be seen that seagulls are scavengers, they usually eat from garbage dumps while wild geese eat seeds and insects, thus they differ in feeding habit. In term of migration, seagulls does not follow seasonal pattern of annual migration while wild geese do. These are the contrasts between the two birds.

6 0

3 years ago

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Which of the following best describes the structure of a mushroom that you observe above the ground

Norma-Jean [14]

The right answer is Fruiting type

The sporophore (literally "spore carrier"), also called Fruitbodies, is the reproductive system of the so-called superior mushrooms. It is, in popular language, the organ of the "fructification" of the mushroom mycelium. It contains sporocysts (basid and asci) that differentiate in the hymenium and produce spores in various forms. It is present in the mushroom's cap.

7 0

3 years ago

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A mutation may cause a change in the genotype of a trait? <br><br> True or false

AnnyKZ [126]

Answer:

true

Explanation:

mutations deal with chromosomal changes, and while they might not always be reflected in the phenotype.. the genotype does change

6 0

3 years ago

Meat is not a good source of protein because it does not provide the body with all the essential amino acids.

tankabanditka [31]

Answer: False.

Meat is a good source of protein. There are nine essential amino acids which are required under special conditions like illness.

These are- Histidine, Leucine, Isoleucine, Valine, Threonine, Methionine, Lysine, Phenylalanine and Tryptophan. They cannot be synthesized by the body and therefore need to be taken from external protein sources.

Since meat contains all the essential amino acids, therefore it is considered as a good source of protein. Hence, the given statement is false.

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3 years ago

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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).

5 0

3 years ago