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Licemer1 [7]
1 year ago
15

the fact that the human brain is not completely developed at birth means that___greatly influence(s) human brain development.

Biology
1 answer:
Brums [2.3K]1 year ago
8 0

The fact that the human brain is not completely developed at birth means that <u>environment</u> greatly influence(s) human brain development.

Human brain development is a complex procedure and unlike other organs, the human brain is not completely developed at the time of birth.

Apart from the development inside the mother, the environment also has a major influence on the development of the human brain.

This is the reason, the brain of each person is influenced by the exposure to the environment that he gets. The development of the cognitive brain is associated with various environmental aspects of society. Hence, not all of the brain of a person has been developed at the time of birth.

To learn more about brain, click here:

brainly.com/question/1247675

#SPJ4

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Select the correct answer from each drop-down menu. what are the four types of nitrogen bases of dna nucleotides? the four types
Hitman42 [59]

The four types of nitrogen bases of dna nucleotides are:

  1. Adenine (A)
  2. Cytosine (C)
  3. Ganine (G)
  4. Thymine (T).

These bases form specific pairs (A with T, and G with C).

<h3>What is DNA nucleotides?</h3>

Nucleotides can be defined as those organic substances consisting of a nucleoside and a phosphate.

They serve as monomeric units of the nucleic acid polymers –

  • Deoxyribonucleic acid
  • Ribonucleic acid,

So therefore, the four types of nitrogen bases of dna nucleotides are:

  • Adenine (A)
  • Cytosine (C)
  • Ganine (G)
  • Thymine (T).

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brainly.com/question/16273896

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2 years ago
1. What are the key players in translation?
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MRNA, ribsomes,tRNAs, and many enzymatic factors
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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.

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