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grandymaker [24]
3 years ago
7

E is a dominant allele for wet earwax and e is a recessive allele for dry earwax. if two individuals with the phenotype of wet e

arwax marry, and one of their children has the phenotype of dry earwax, what do we know about the parents?
Biology
1 answer:
MrRissso [65]3 years ago
8 0
Given 
E dominant allele for wet
e recessive allele for dry

that means
for phenotype wet, the possible genotypes are EE, or Ee, and
for phenotype dry, the only possible genotype is ee.

Therefore we also know that the child who has dry earwax has genotype ee.
Since the child inherits one allele from each parent, therefore each parent must have a recessive allele "e".

If both parents have phenotype wet earwax, they both must be heterozygous for wet/dry earwax, namely Ee.

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Describe the relationship between fatty acid chain length, saturation, and the melting point or fluidity of fatty acids.
Advocard [28]

Answer:

Because double bonds cause the hydrocarbon chain to bend. Therefore, the fatty acids cannot compact tightly together, reducing the van der Waals interaction between the fatty acids. The melting point of fatty acids is also affected by chain length. The longer the hydrocarbon chain is, the higher the melting point.

Explanation:

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A diseased cell is no longer able to produce proteins. Which cell structure is most likely malfunctioning?
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The ribosomes

Explanation:

Ribosomes are the organelle that produce proteins within eukaryotic and prokaryotic cells- taking information from the DNA in order to produce proteins.

Other organelles may be held responsible for the lack of protein production, but in this case it is most likely the ribosome.

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What causes the dark urine associated with acute poststreptococcal glomerulonephritis?
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often the first clinical manifestation of APSGN. Dark urine is caused by hemolysis of red blood cells that have penetrated the glomerular basement membrane and have passed into the tubular system. Periorbital edema is typical.

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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.

<h3>A)</h3>

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.

<h3>B)</h3>

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

<u>cc = 0.7</u>

Observed DCO / 47 = 0.7

Observed DCO = 0.7  × 47

Observed DCO ≅ 33

<h3>C)</h3>

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.

<h3>D)</h3>

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.

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