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3 years ago

A series of choices between two characteristics that is used to identify organisms is called a

Biology

2 answers:

Tomtit [17]3 years ago

5 0

A dischotomous key which is answer D

Dmitriy789 [7]3 years ago

5 0

Answer:

The correct answer is option D, dichotomous key

Explanation:

Dichotomous keys are used to identify plant and animals and also it is an indicative parameter of relationship between two species. The dichotomous key consists of series of choices associated with questions one after the other which helps in identifying an organism. For example for identification of any plant, the first question would be “does the plant have leaves” with option either “yes” or “No”. If the answer to this question is yes, the next question will be related to some distinguishing characteristic of leaves. If the answer to this question is “no” a new set of question will come related to some other part of plant.

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The researchers set up experimental and control microcosms, or small artificial ecosystems, to measure the release of minerals f

Rudik [331]

Answer:

The correct answer is - Control microcosms did not contain living moss, while experimental microcosms did contain living moss.

Explanation:

The difference between the control microcosms and experiment Microcosms is, the presence of the living moss in the experimental group whereas the control group does not contain living moss.

The independent variable in this research setting is the presence or absence of the living moss and for the control group, the mosses are filtered out and only added the water.

7 0

3 years ago

Which of the following best describes meiosis?

Art [367]

Meiosis is when the cells divide so I’m going. To say it is b

5 0

3 years ago

Asexual reproduction can be advantageous for species that colonize new areas.

max2010maxim [7]

Answer:

True

Explanation:

It can rapidly increase if the environment is good for them

6 0

3 years ago

The false spider mite, Brevipalpus phoenicis, has only two chromosomes. Which of the following pieces of evidence would allow yo

Reptile [31]

Answer:

The correct answer is - a. If these two mite chromosomes have different genes at different loci.

Explanation:

If it is 2n= 2, it means that it is diploid and has two sets of chromosomes in which one set comes from mother and the other from father which means parent's genes contribute to diploid equally. Both sets of chromosomes form homologous chromosome pair. Each homolog of the pair has the same gene at the same loci in diploid and if it has not the same homologous gene at the same loci these are haploid.

8 0

3 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).

5 0

3 years ago