1. 20% and yz = 10%. The distance between

1.

    It’sthe only one that is recessive and double crossed.  2.    Physicalmap – the distance between A and C would be the equal.

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 3.    Linkagemap – less than the sum of recombination frequencies due to multiple crossovers. 4.    Becauseit would be 50% parental and 50% new combination. 5.    Inspite of the distance between the genes, they will assort a minimum of 50% ofthe time due to multiple crossovers. Even numbers bring the genes together,while odd numbers bring them apart.

 6.    12½ markers – the best loci placement is at every 10 million base pairs.  7.

     xy = 20% and yz = 10%. The distance between x and z is recombination percentage.  8.    Thegenotypes with the lowest recombination % are the ones who went through doublecrossovers.

Reverse the double crossover and you end up with gene A in themiddle. The phenotype represents the parental.   9.     a.     Proved that one gene =one enzyme through the use of four different strains of arginine-dependentNeurospora. All of the Neurospora had lost the ability to use a unique genethat an enzyme needed to produce arginine.  b.

     Experiment with Strep. pneumoniae  that used rough strains thatwere non-pathogenic and smooth strains that were pathogenic. He injected amouse with living smooth strains in which the mouse died.

Another mouse wasinjected with rough strains and the mouse was still healthy. He also injected amouse with a heat killed smooth strain, in which the mouse lived. Finally, hedid a new combination with a mix of heat-killed smooth strain and rough strain inwhich the mouse died.

This proved that DNA was a genetic material and that therough strain had been “transformed” into a virulent strain by the smoothstrain. c.             Proved that DNA carriedgenetic specificity and that it was the reason behind the transformation ofnon-virulent strains into virulent ones by heat-killed strains. DNA wasisolated and purified from the smooth strains and were added to the rough cellsto transform them. When they added an enzyme, deoxyribonuclease, thetransformation was interrupted. They disproved that RNA carried geneticspecificity by adding ribonuclease or a protease, but the transformation keptgoing. d.

             They wanted to prove if nucleic acid was a carrier of viral genes.They studied a T2 bacteriophage that has a DNA core and a capsid. Phageproteins were labeled 35S and the DNA was labeled 32P. These labels were usedto follow where the protein and the nucleic acid would end up in the progenyphage. They found that the 32P labeled nucleic acid was still prominent in theprogeny and that a very small amount of the 35S labeled protein had actuallyentered the bacteria.

This led them to prove that all viruses contain nucleicacid. e.             Found that there was an exact ratio of adenine, thymine, cytosineand guanine that varied through species and that the bases had a complementaryrelationship with each other. He calculated the nucleotide percentages in DNAthrough paper chromatography and UV spectrophotometry.

 ?f.              Used Atsbury’s X-ray diffraction of DNA to prove the helical, doublestranded structure of DNA and she noted that the strands had a major and minorgroove along the helical axis. g.             Watson anUsed Franklin’s X-ray diffraction and used Chargaff’srule to conclude that DNA chains are helical and complementary. This showedthat DNA is a template for its own replication and eventually became a part ofthe Central Dogma. h.             Worked with extracts of bacteria to study DNA synthesis. 32P—cytosinewas added to unlabeled nucleotides and he used a 3′ phosphodiesterase to locatethe 32P-cytosine at a 3′ end.

With this experiment he was able to isolate DNApolymerase I and that it links 3′ hydroxy to a 5′ phosphate, therefore growingin the 5′ à3′ direction. Thisproved that DNA strands were anti-parallel. i.              A density gradient centrifugation was used to separate DNA into a heavy15N15N DNA, a light 14N14N DNA, and a hybrid 14N15N DNA. The heavy DNA made aband high in density near the bottom of the density gradient, the light DNAstayed near the top of the gradient where the density equaled the density of thesalt solution, and the hybrid chain was intermediate. The 15N heavy was grownin the 14N and so on. The results showed that after three generations, all ofthe bands were different, and it was concluded that DNA replication issemi-conservative.

j.              Concluded that sickle cells had an amino acid change in B-globinby doing a protein fingerprint on a wild type of hemoglobin and a sickle celledhemoglobin. The hemoglobin was cleaved, and it ran in two directions. The electricfield caused it to move from the right to the left and the cells were chromatographicallyseparated. k.             Showed that DNA served as a template for its own replication.

Heshowed arrows from DNA to RNA and stated that this was called “transcription”and after that RNA would “translate” the information into proteins. l.              Reticulocytes were used to measure protein synthesis. As theprotein synthesis started, a radiolabeled leucine was added, and the synthesiswas stopped before a new globin chain was synthesized. The proteins were thenseparated through chromatography and then trypsin was used to make peptidefragments. Then the globin chains that had completed synthesis were separatedfrom incomplete chains using gel electrophoresis. A plot that showed the lengthof each peptide along with the label was made to show the position of the peptidein the completed chain. The peptides on the B-globin regions had a small amountof radiolabel and would be the N terminus and the end with the highest was nearthe C terminus.

10.  It was believed that RNA did nothave the ability to differentiate between side groups of amino acids andsolubility so there had to be “something” that translated it.  11.  Since rRNA is about the same inevery species, it cannot explain the genetic variations between the species.

 12.  5′ AUGCCAUUCGAA 3′ – from the 3′ – 5’template strand.  13.

  Each amino acid has multiple butspecific codons and each of those codons codes for only one amino acid.  14.  Inosine helps with t-RNA stabilityat the wobble position and allows to read the codons as an A, U or C in thisposition. This enhances variation and also follows the degenerate but not ambiguousrule.  15.

  Purification removes anynon-functional material from the polypeptide. If they’re not removed, theresults may be unclear. 16.  Weak bonds are further apart whilestrong bonds are closer together. Strong bonds are hard to break but they releasea high amount of energy when they are broken whereas weak bonds are reversiblebut release a smaller amount of energy when they’re broken. 17.

  Peptide bonds are rigid and planarbecause of resonance interactions.  18.  Strong bonds are harder to breakdue to activation energy, and since all reactions must move towards higherentropy, strong bonds need to be coupled with an energy releasing source thatallows them to reach their activation energy without violating the laws ofthermodynamics.  30. At Keq = 100, the reaction willfavor the products if G is negative, and it will favor the reactants if G is positive.

 31. Free energy is noted as G and it is the energy available for a reaction tooccur. When G is negative, the reaction is considered spontaneous andit will favor the products. When G is positive, it is considered less organized and notspontaneous and it will favor the reactants unless it is positive at a hightemperature.  32.

At a high temperature, a positive G will be spontaneous but it will not be spontaneousat a low temperature.  33. Although the given G value is negative, H2 and O2 will not reactspontaneously to form water because energy needs to be added to the system forit to overcome activation energy, become less orderly, and for the molecules tocollide.

This is due to the second law of thermodynamics, which states that reactionsmove towards a more disordered system.  34.Weak bonds can be broken through a small amount of energy because kineticenergy is dispersed.

  35.             a) The smaller the van der Waalsbonds radius is, the stronger the van der Waal interaction is.               b) Hydrogen bonds are strongestwhen they are linearly aligned with other molecules. Any bond angle greaterthan 30° will weaken the hydrogen bond.               c) Ionic bonds occur betweenpartially charged molecules.

If an ionically bound group is submerged in water,the ionic forces would weaken.              d) Hydrophobic bonds strengthen theprimary structure of proteins and they promote coiling and folding in alphahelices and beta sheets.                e) Nonpolar molecules will cluster ifsubmerged in water due to the van der Waals forces and hydrophobic groups.               f) Polar molecules are soluble inwater and may participate in hydrogen bonds and interact with the latticestructure of water.   36.In water, the nonpolar molecules are kept together by the hydrophobic effect. Thehydrogen atoms will break apart from the water molecules, releasing energy intothe system.  The water molecules then formmore hydrogen bonds and encapsulate the nonpolar molecule, therefore decreasingthe entropy in the system.

 37.Hydrophobic proteins would be within the protein because the hydrophobicinteractions would help with the stability and further folding and coiling ofthe protein.   38.When two molecules fit each other according to their shape, they are able tohave multiple interactions and therefore, stronger binding energies.  39.They mediate reactions between enzymes and substrates because of their abilityto break and form again under physiological conditions.

Weak bonds also givemacromolecules their shape. 40.The formation of peptide bonds is achieved through the breakdown of ATP to AMPand pyrophosphate, which yields a net of – G. Activation energy keeps proteins from easily comingapart. 41.

Weak bonds are thermodynamically unstable and can give off more energy than astronger bond.  42.G is the actual energy change in the cell, while G° is measured under standard-state conditions.

G°’ is measured under standard conditions and also wherethe pH is at 7. Biochemists prefer G°’ because it closely resembles cellular conditions.  43.The overall pathway has a -G because citrate synthase keeps oxaloacetate concentrationslow.  44.Covalent bonds are strong and cannot be broken or rearranged because theactivation energy is too large to overcome without the help of enzymes.  45.

Enzymes lower activation energies and speed up the rate of reactions. They donot affect equilibrium.  46.

The phosphoanhydride bonds are negatively charged and constantly fighting overresonance. Hydrolysis is favored because ADP is higher in -G than ATP is.  47.Molecules exchange functional groups through group transfer and this allowshigh energy bonds to be transferred between molecules. 48.ATP activates amino acids by using group transfer to give amino acids an AMP,which gives the overall pathway of the reaction more entropy.

 49.ATP uses group transfer to activate nucleotides and then pyrophosphate isreleased and then split to make the overall reaction – G.  50.ATP hydrolysis is used to help proteins make the conformational changes neededto carry out their specific functions.