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slega [8]
3 years ago
7

Create a table in which you compare the components and functions of the following.

Biology
1 answer:
Sauron [17]3 years ago
5 0

Nucleic acids, proteins, carbohydrates, and lipids are the four major types of biomolecules that form all living things. These biomolecules consists of monomers linked together by covalent bonds to form polymers.

  • Nucleic acids, proteins, carbohydrates and lipids can be classified according to their basic elements, monomer constituents, and functions.

Basic elements:

  1. Nucleic acids: Hydrogen (H); Carbon (C); Oxygen (O); Nitrogen (N); Phosphorous (P)
  2. Proteins: Hydrogen (H); Carbon (C); Oxygen (O); Nitrogen (N); Sulfur (Z)
  3. Carbohydrates: Hydrogen (H); Carbon (C); Oxygen (O)
  4. Lipids: Hydrogen (H); Carbon (C); Oxygen (O); Phosphorous (P)

Monomer constituents:

  1. Nucleic acids: nucleotides
  2. Proteins: amino acids
  3. Carbohydrates: monosaccharides
  4. Lipids: fatty acids and glycerol

Functions:

  1. Nucleic acids: contains the hereditary information to synthesize proteins
  2. Proteins: regulate metabolic processes (enzymes), the main biomolecule of cellular structures
  3. Carbohydrates: store energy (short term); form cellular structures
  4. Lipids: store energy (long term); the main component of biological membranes

Examples:

  1. Nucleic acids: DNA and RNA
  2. Proteins: lactase; collagen
  3. Carbohydrates: starch (polysaccharide); glucose (monosacharide)
  4. Lipids: phospholipids; cholesterol

Learn more in:

brainly.com/question/736132?referrer=searchResults

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What is fever and explain why it is considered to be a nonspecific immune response?
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Answer:

Body of an organisms have a range of the normal temperature that is maintained by the body by homeostasis. When the range of the body temperature rise above that range it it known as fever. The normal temperature of human body is normally 36-37 degree centigrade. Mild fever is not need any assistance to bring it back to normal range it is doe by the body to fight against infection or change takes place. Fever can be measured in mouth, rectum, and underarms.

It is considered as nonspecific response mechanism as it is take place due to different type of reasons and its affect or mechanisms is same for all, not specific as the cell mediated or antibody mediated responses of specific defense mechanism.

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Place the filling in order for smallest to largest :chromatin,chromosomes ,DNA ,nucleoside ,sister chromatid
GalinKa [24]

DNA will be your answer thanks alot

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3 years ago
A cross is made between homozygous wild-type female Drosophila (a^+ a^+ b^+ b^+ c^+ c^+) and triple-mutant males (aa bb cc) (the
marusya05 [52]

Answer:

a is the middle gene.

Distance [b-a]= 24.7 mu

Distance [a-c]= 15.8 mu

Distance [b-a} = 40.5 mu

Explanation:

A homozygous wild-type female drosophila (a⁺b⁺c⁺/a⁺b⁺c⁺) is crossed with a homozygous recessive male (abc/abc). <u>The order of the genes here is arbitrary. </u>

The F1 is heterozygous for the three genes (a⁺b⁺c⁺/abc). The F1 females were test crossed (crossed with abc/abc males).

The F2 shows the following phenotypic ratios:

  • 320 a⁺b⁺c⁺
  • 308 a b c
  • 102 a⁺ b c⁺
  • 112 a b⁺ c
  • 66  a⁺ b⁺ c
  • 59 a b c⁺
  • 18 a⁺ b c
  • 15 a b⁺ c⁺

Total = 1000

The male parent is homozygous recessive for the 3 genes, so the observed phenotypes of the offspring correspond to the gametes received from the mother.

Recombination during meiosis is a rare event, so the most abundant gametes are always the parentals:  a⁺b⁺c⁺ and abc.

The least abundant gametes, following the same logic, are the double crossovers (DCO): a⁺bc and ab⁺c⁺.

<h3><u>1st. Determine the gene order</u></h3>

Compare the parental and the DCO gametes. The allele that is switched corresponds to the middle gene. In this case, gene a is in the middle of the other two.

<h3><u>2nd Determine the single crossover gametes</u></h3>

The F1 mother that generated all 8 types of gametes had the genotype b⁺a⁺c⁺/bac (correct order of genes).

  • The single crossover (SCO) gametes resulting from recombination between genes b and a are b⁺ac and ba⁺c⁺.
  • The single crossover (SCO) gametes resulting from recombination between genes a and c are b⁺a⁺c and bac⁺.
<h3><u>3) Calculate the recombination frequencies between genes </u></h3>

Recombination frequency (RF) = #Recombinants/Total progeny

  • RF [b-a]= (102+112+18+15)/1000= 0.247
  • RF [f-br]= (66+59+18+15)/1000= 0.158
<h3><u /></h3><h3><u>4) Calculate the distance in map units </u></h3>

Distance (mu) = RF x 100

Distance [b-a]= 0.247 × 100 = 24.7 mu

Distance [a-c]= 0.158 × 100 = 15.8 mu

Distance [b-a} = 24.7 mu + 15.8 mu = 40.5 mu

<h3><u>The gene map therefore looks like: </u></h3>

b------------24.7 mu--------------------------a---------15.8 mu-----------c

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Answer: see below.

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