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Anni [7]
3 years ago
10

Gin Edji | AICE Week 1 r... - Wolf CER

Biology
1 answer:
Elden [556K]3 years ago
4 0
Math 32 wolf CER gin EDJI AICE week 1
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If a doctor prescribes tylenol iii with codeine, he or she has prescribed a __________.
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The right option is c. narcotic

Tylenol iii with codeine belongs to the group of medications called narcotic analgesics (pain relievers). Narcotics are drugs that are administered to provide relief from severe pain. Narcotics are sometimes prescribed when other forms of pain relievers are not working. Examples of narcotics are hydrocodone, codeine, tramadol and opium.






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I need help with these two questions
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What’s a geographic variation in fossil record?
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Geographical variation refers to differences among populations in genetically based traits across the natural geographic range of a species. ... In the simplest case, we divide these factors into purely genetic versus environmental components to tease apart their relative contributions to observed phenotypic variation.
6 0
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The ventral rami of spinal nerves ________ form the major nerve plexuses.
Mrrafil [7]

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C1 through T1 and L1 through S4

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6 0
3 years ago
You are a botanist looking at the plant color in a new breed of Snapdragon. The two parent plants are Blue and Green and when yo
irakobra [83]

Answer:

Incomplete dominance is the inheritance pattern where the dominant allele did not mask the recessive allele completely and form a mix of both alleles. Here the inheritance is the incomplete inheritance. The ratio of F2 generation is 1:2:1.

Given:

R1R1 = 42

R2R2 = 39

R1R2 = 86

Total R1 alleles = 2*42+86 = 170

Total R2 alleles = 2*39+86 = 164

Total alleles = 334

Frequency of allele R1 = 170/334 = 0.51

Frequency of allele R2 = 164 / 334 = 0.49

Expected number of each phenotype:

Total population = 167

Blue = R1R1 = 0.51 * 0.51 * 167 = 43.44

Green = R2R2 = 0.49 * 0.49 * 167 = 40.10

Cyan = 2*R1*R2 = 2*0.51*0.49*167 = 83.46

Phenotype     Observed(O)    Expected (E)    O-E      (O-E)2     (O-E)2/E

Blue                 42                   43.44               -1.44       2.0736     0.0477

cyan                86                   83.46                 2.54     6.4516     0.0773

green              39                    40.1                    -1.1       1.2100      0.0302

Total              167                     167                                               0.1552

Chi-square value = 0.155

Degrees of freedom = no. of phenotypes – 1

Df = 3-1 = 2

Critical value = 5.99

Chi-square value of 0.155 is less than the critical value of 5.99. So we accept the null hypothesis.

4 0
3 years ago
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