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Sedaia [141]
3 years ago
10

What if a mutation caused a change in the code so the message read GAG instead of GAC? 
How would this affect the protein produc

ed?
Biology
1 answer:
Anarel [89]3 years ago
3 0

Answer: There will be no effect on the protein produced

Explanation:

According to the genetic code, the triplet codon GAG code for the amino acid, glutamate, while GAC code for aspartate.

Since GAG and GAC code for amino acids with similar properties (aspartate and glutamate are both acidic amino acids due to the extra carboxyl group present in their side chain), the change in nucleotide is still similar to the original, so will result in a different amino acid sequence, but will not alter the function of the protein at all. Thus, this kind of mutation is called silent mutation.

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What element do all the organic compounds contain?
Jlenok [28]

ALL organic compounds ALWAYS contain carbon!!

4 0
3 years ago
Read 2 more answers
In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant
Ivenika [448]

Answer and Explanation:

<em><u>Available data</u></em>:

  • Comb shape is determined by genes at two loci (R, r and P, p).
  • The walnut comb genotype is R_P_.
  • The rose comb genotype is R_pp.
  • The pea comb genotype is rrP_.
  • The single genotype is rrpp.

a. <em>Walnut crossed with single produces 1 walnut, 1 rose, 1 pea, and 1 single offspring: </em>

Parental)             RrPp       x          rrpp

Gametes)   RP   Rp   rP   rp     rp   rp   rp   rp

Punnet Square)      RP       Rp     rP        rp

                     rp   <em>RrPp    Rrpp   rrPp   rrpp</em>

                     rp    RrPp    Rrpp   rrPp   rrpp

                     rp    RrPp    Rrpp   rrPp   rrpp

                     rp    RrPp    Rrpp   rrPp   rrpp

F1 phenotype: 25% walnut, 25% rose, 25% pea, and 25% single.

F1 genotype: 4/16 RrPp, 1/16 Rrpp, 4/16 rrPp, 4/16 rrpp.

b. <em>Rose crossed with pea produces 20 walnut offspring</em>.

Parental)              RRpp       x          rrPP

Gametes)   Rp   Rp   Rp   Rp     rP   rP   rP   rP

Punnet Square)      Rp       Rp     Rp        Rp

                     rP    RrPp    RrPp   RrPp   RrPp

                     rP    RrPp    RrPp   RrPp  RrPp

                    rP    RrPp    RrPp   RrPp   RrPp

                     rP    RrPp    RrPp   RrPp   RrPp

F1 phenotype: 100% walnut.

F1 genotype: 16/16 RrPp.

c. <em>Pea crossed with single produces 1 single offspring</em>.

This is not possible, because the pea genotype involves <u>at least</u> one dominant allele P. There are two possible crosses: <em>rrPp x rrpp</em>, which must produce half of the progeny pea and the other half single, or <em>rrPP x rrpp</em> which produce a whole pea progeny with no single offspring.  

Parental)              rrPp       x          rrpp

Gametes)   rP   rp   rP   rp     rp   rp   rp   rp

Punnet Square)     rP       rp       rP      rp

                     rp   <em>rrPp    rrpp   rrPp   rrpp</em>

                    rp    rrPp    rrpp   rrPp   rrpp

                     rp    rrPp    rrpp   rrPp   rrpp

                     rp    rrPp    rrpp   rrPp   rrpp

F1 phenotype: 50% pea, and 50% single.

F1 genotype: 8/16 rrPp, 8/16 rrpp.

d. <em>Rose crossed with pea produces 2 walnut, 1 single, and 1 pea offspring</em>.

This is not possible, because having one of the parents with a rose phenotype  involves <u>at least one R allele</u>, which means that <u>there must be rose phenotype</u> in the progeny.

Parental)             Rrpp       x          rrPp

Gametes)   Rp   Rp   rp   rp     rP   rP   rp   rp

Punnet Square)     Rp       Rp       rp      rp

                     rP  <em> RrPp </em>   RrPp  <em> rrPp</em>   rrPp

                     rP   RrPp    RrPp   rrPp   rrPp

                     rp    <em>Rrpp</em>    Rrpp   <em>rrpp </em>  rrpp

                     rp   Rrpp    Rrpp   rrpp   rrpp

F1 phenotype: 25% walnut, 25% rose, 25% pea, and 25% single.

F1 genotype: 4/16 RrPp, 1/16 Rrpp, 4/16 rrPp, 4/16 rrpp.

e. <em>Rose crossed with single produces 31 rose offspring</em>.

Parental)              RRpp       x          rrpp

Gametes)   Rp   Rp   Rp   Rp     rp   rp   rp   rp

Punnet Square)     Rp       Rp       Rp      Rp

                     rp    Rrpp    Rrpp   Rrpp   Rrpp

                     rp    Rrpp    Rrpp   Rrpp   Rrpp

                     rp    Rrpp    Rrpp   Rrpp   Rrpp

                     rp    Rrpp    Rrpp   Rrpp   Rrpp

F1 phenotype: 100% rose (31 individuals equal 100% of the progeny).

F1 genotype: 16/16 Rrpp.

f. <em>Rose crossed with single produces 10 rose and 11 single offspring.</em>

Parental)              Rrpp       x          rrpp

Gametes)   Rp   Rp   rp   rp     rP   rP   rp   rp

Punnet Square)      Rp       Rp       rp      rp

                     rp    Rrpp    Rrpp   rrpp   rrpp

                     rp    Rrpp    Rrpp   rrpp   rrpp

                     rp    Rrpp    Rrpp   rrpp   rrpp

                     rp    Rrpp    Rrpp   rrpp   rrpp

F1 phenotype: 50% rose, 50% single.

F1 genotype: 8/16 Rrpp, 8/16 rrpp.

3 0
3 years ago
Name the animal used as a beast of burden in the andes
satela [25.4K]

Answer;

The llama and the alpaca were used as beasts of burden in the Andes.

Explanation;

-A beast of burden is drought animal that carries or pulls heavy loads, such as a donkey, mule, llama, camel etc.  

-The Andean animal used as a beast of burden is an Alpaca and the Llamas. They are docile, fluffy animals that can withstand the cold of the mountains. As such, they make excellent work animals.

-Male llamas have been been used as beasts of burden in the Peruvian and Bolivian Andes ranges for more than 4000 years.

8 0
3 years ago
What is the volume of 0.25 mol of ammonia gas at 1.00 atm and 273 K? (R= 0.0821 Latm/molK)What is the volume of 0.25 mol of ammo
ICE Princess25 [194]

Answer:

5.6L

Explanation:

Given parameters:

number of moles  = 0.25mol

pressure on gas  = 1atm

temperature  = 273K

Gas constant R = 0.0821Latm/molK

Unknown:

Volume of gas  = ?

Solution:

Using the ideal gas equation, we can solve this problem. The equation is a combination of the three gas laws: Boyle's law, Charles's law and Avogadro's law.

It is mathematically expressed as;

             PV = nRT

where P is the pressure

           V is the volume

          R is the gas constant

          T is the temperature

          n is the number of moles

All the parameters are in the appropriate units and we simply solve for the volume of the gas;

                1 x V = 0.25 x 0.0821 x 273

                      V  = 5.6L

4 0
3 years ago
In eukaryotes, the size of the primary transcript is generally ________ the gene in the template dna strand.
Hunter-Best [27]
Had to look for the options and here is my answer. 
Basically, when we say eukaryotes, these are kinds of organisms which consists of cells that has DNA and other organelles that are enclosed by a membrane. In this case, the size of the primary transcript is generally THE SAME LENGTH the gene in the template of the DNA strand. Hope this helps.
3 0
3 years ago
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