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2 years ago

Use the scenario to answer the question.

Biology

1 answer:

DochEvi [55]2 years ago

3 0

Answer:

All the planets besides Earth lack a breathable atmosphere for terrestrial beings, but also, many of them are too hot or too cold to sustain life. A “habitable zone” which exists within every system of planets orbiting a star. Those planets that are too close to their sun are molten and toxic, while those that are too far outside it are icy and frozen.

But at the same time, forces other than position relative to our Sun can affect surface temperatures. For example, some planets are tidally locked, which means that they have one of their sides constantly facing towards the Sun. Others are warmed by internal geological forces and achieve some warmth that does not depend on exposure to the Sun’s rays.

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3 years ago

If a hemoglobin molecule comes into close contact with a respiring cell, the presence of CO2 in the environment will cause hemog

lapo4ka [179]

Answer:

No, when the concentration of carbon dioxide is high, such as in peripheral tissues, CO2 binds to hemoglobin and the affinity for O2 decreases, causing it to release.

Explanation:

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3 years ago

In Drosophila, the genes for withered wings (whd), smooth abdomen (sm) and speck body (sp) are located on chromosome 2 and are s

lesya692 [45]

Answer:

A) 47; B) 33; C) 272; D) 122

Explanation:

The three genes are linked.

The female with withered wings and a smooth abdomen has the genotype whd sm sp+/whd sm sp+.

The male with a speck body has the genotype whd+ sm+ sp/whd+ sm+ sp.

Both individuals are homozygous for all genes, so each of them only produces one type of gamete. The resulting F1 therefore has the genotype whd sm sp+/ whd+ sm+ sp, heterozygous for all genes and with a wild-type phenotype.

The females of the F1 were mated with homozygous recessive males (test cross): whd sm sp/whd sm sp.

If we assume interference is 0, the probability of crossing over happening between the genes whd and sm is independent from the probability of crossing over happening between sm and sp.

The distance = frequency of recombination × 100, so the frequency of recombination (RF) between genes whd and sm is 0.305 and the RF between genes sm and sp is 0.155.

<u>The expected double crossover progeny among the 1000 offspring will be:</u>

RF whd-sm × RF sm-sp × 1000 =

0.305 × 0.155 × 1000 = 47 individuals will be double crossover.

Interference is 0.3

The interference is calculated as 1- coefficient of coincidence (cc).

cc = observed double crossover/expected double crossover

Therefore:

I = 1 - cc

cc = 1 - I

Observed DCO / 47 = 0.7

Observed DCO = 0.7 × 47

Observed DCO ≅ 33

The parental gametes are whd sm sp+ and whd+ sm+ sp (the genotype of the F1 female is known).

Looking at them and at the gene map we can tell that the gametes that give rise to withered wings, speck body (whd sm+ sp) and smooth abdomen (whd+ sm sp+) phenotypes are the result of recombination occurring between genes whd and sm.

To calculate the expected number of individuals with those phenotypes among the 1000 progeny we need to determine the frequency of recombination between the genes whd and sm considering there's interference.

The distance whd-sm = RF x 100

The recombination frequency is the sum of the single crossover between whd and sm and the double crossovers.

The frequency of DCO is 33/1000=0.033.

Distance whd-sm/ 100 = SCOwhd-sm + DCO

0.305 - 0.033 = SCO whd-sm

<u>Frequency of SCO whd-sm= 0.272</u>

And the expected number of individuals with those phenotypes will be 0.272 x 1000 = 272.

The gametes that originate the phenotypes withered wings, speck body, smooth abdomen (whd sm sp) and wild type (whd+ sm+ sp+) are the result of recombination between genes sm and sp.

Distance sm-sp/ 100 = SCOsm-sp + DCO

0.155 - 0.033 = SCOsm-sp

<u>Frequency of SCO sm-sp= 0.122</u>

And the expected number of individuals with those phenotypes will be 0.122 x 1000 = 122.

6 0

3 years ago