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Zolol [24]
3 years ago
6

Find amino acid for the mRNA code AAG

Biology
1 answer:
laila [671]3 years ago
5 0

Answer:

phenylalanine

Explanation:

A base set of triplet called codon codes for a particular amino acid of protein molecule. Protein synthesis takes place in ribosomes found in the cytoplasm. Information on the mRNA is translated by ribosomes and is used to assemble amino acids into specific protein molecules.

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Zigmanuir [339]

Answer:

the first one

Explanation:

The animals above it can no longer eat it, and the animals below it will no longer be eaten by it.

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3 years ago
The only cellular respiration process that can be considered anaerobic is _____.
Nesterboy [21]

The correct answer is fermentation.

The production of energy needs oxygen. The electron transport chain, where the most of ATP is produced, needs a huge input of oxygen. However, there are various species, which have produced methods to continue the metabolism in the absence of oxygen, or can move from aerobic to anaerobic cell respiration when oxygen is enough.

At the time of cellular respiration, some of the living species utilize an organic substance as the eventual electron acceptor. The procedures that utilize an organic molecule to reproduce NAD+ from NADH are in a combination known as fermentation.

In comparison, some of the living species utilize an inorganic substance as an ultimate electron acceptor. Both the procedures are known as anaerobic cellular respiration, where the species are obtaining energy for their application in the absence of oxygen.

7 0
3 years ago
Heat transfer by conduction occur when molecules
Reika [66]
I think the answer is B.
6 0
3 years ago
Pleas help me I’m currently in class so fast pleeeeeeeaaqwwwwse
netineya [11]

*this is based on the assumption that parent two is homo_zygous dominant.

If a hetero_zygous and a homo_zygous dominant parent are crossed:

Genotypic Ratio

WW: Ww

1 : 1

Phenotypic Ratio

white : purple

[1 : 0] or [100% : 0%]

100 will have a white colour while 0 will have a purple colour.

6 0
3 years ago
Three linked autosomal loci were studied in smurfs.
cupoosta [38]

Answer:

height -------- color --------- mood

           (13.2cM)      (14.5cM)

C=0.421

I = 0.579

Explanation:

We have the number of descendants of each phenotype product of the tri-hybrid cross.

Phenotype Number

  • pink, tall, happy            580
  • blue, dwarf, gloomy     601
  • pink, tall, gloomy         113
  • blue, dwarf, happy      107
  • blue, tall, happy              8
  • pink, dwarf, gloomy        6
  • blue, tall, gloomy          98
  • pink, dwarf, happy      101

Total number of individuals = 1614 = N

Knowing that the genes are linked, we can calculate genetic distances between them. First, we need to know their order in the chromosome, and to do so, we need to compare the phenotypes of the parental with the ones of the double recombinants. We can recognize the parental in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent. So:

Parental)

  • Pink, tall, happy            580 individuals
  • Blue, dwarf, gloomy      601 individuals

Simple recombinant)

  • Pink Tall Gloomy           113 individuals
  • Blue, Dwarf, Happy       107 individuals
  • Blue Tall Gloomy             98 individuals
  • Pink Dwarf Happy          101 individuals

Double Recombinant)  

  • Blue Tall Happy                 8 individuals
  • Pink  Dwarf Gloomy           6 individuals  

Comparing them we realize that parental and double recombinant individuals differ in the position of the gene codifying for <u>color</u><u>.</u> They only change in the position of Blue and Pink. This suggests that the position of the color gene is in the middle of the other two genes, height and mood, because in a double recombinant only the central gene changes position in the chromatid.  

So, the alphabetic order of the genes is:

---- height ---- color ----- mood ----

Now we will call Region I to the area between Height and Color, and Region II to the area between Color and Mood.

Once established the order of the genes we can calculate distances between them, and we will do it from the central gene to the genes on each side. First We will calculate the recombination frequencies, and we will do it by region. We will call P1 to the recombination frequency between Height and color genes, and P2 to the recombination frequency between color and mood.

P1 = (R + DR) / N

P2 = (R + DR)/ N

Where: R is the number of recombinants in each region (the ones that have an intermediate phenotypic frequency), DR is the number of double recombinants in each region, and N is the total number of individuals.  So:

Region I

Tall------ Pink--------happy  (Parental) 580 individuals

Dwarf ---Pink------- Happy (Simple Recombinant) 101 individuals

Dwarf--- Pink-------Gloomy (Double Recombinant) 6 individuals

Dwarf----Blue-------Gloomy (Parental) 601 individuals

Tall ------Blue------- Gloomy (Simple Recombinant)  98 individuals

Tall ----- Blue------- Happy   (Double Recombinant) 8 individuals  

Region II

Tall------ Pink--------happy (Parental) 580 individuals

Tall-------Pink------- Gloomy (Simple Recombinant) 113 individuals

Dwarf----Pink------- Gloomy (Double Recombinant) 6 individuals

Dwarf----Blue-------Gloomy (Parental) 601 individuals

Dwarf ----Blue-------Happy (Simple Recombinant) 107 individuals

Tall ----- Blue------- Happy   (Double Recombinant) 8 individuals

In each region, the highlighted traits are the ones that suffered recombination.

  • P1 = (R + DR) / N

P1 = (101+6+98+8)/1614

P1 = 213/1614

P1 = 0.132    

  • P2= = (R + DR) / N

P2 = (113+6+107+8)/1614

P1 = 234/1614

P1 = 0.145

Now, to calculate the recombination frequency between the two extreme genes, height and mood, we can just perform addition or a sum:

  • P1 + P2= Pt

0.132 + 0.145 = Pt

0.277=Pt

The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU).  

The map unit is the distance between the pair of genes for which every 100 meiotic products, one results in a recombinant product.  

Now we must multiply each recombination frequency by 100 to get the genetic distance in map units:

GD1= P1 x 100 = 0.132 x 100 = 13.2 MU = 13.2 cM

GD2= P2 x 100 = 0.145 x 100 = 14.5 MU = 14.5 cM

GD3=Pt x 100 = 0.277 x 100 = 27.7 MU = 27.7 cM

To calculate the coefficient of coincidence, CC, we must use the next formula:

CC= observed double recombinant frequency/expected double recombinant frequency

Note:  

-observed double recombinant frequency=total number of observed double recombinant individuals/total number of individuals

-expected double recombinant frequency: recombination frequency in region I x recombination frequency in region II.

  • CC= ((6 + 8)/1614)/0.132x0.145

        CC=0.008/0.019

        CC=0.421

The coefficient of interference, I, is complementary with CC.

I = 1 - CC

I = 1 - 0.421

I = 0.579

8 0
3 years ago
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