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Kisachek [45]
3 years ago
9

If too many chloride ions are built up outside of the cell, with which type of transport is the carrier protein associated???

Biology
2 answers:
Archy [21]3 years ago
7 0

Answer:

<h2>Facilitated diffusion.</h2>

Explanation:

It's also called passive transport, this refers to the process of transporting molecules or ions across a membrane. The passive name is given to it, because for this process chemical energy, like ATP, is not required. That's the difference between active and passive transportation.

Also, ions are only involves in a facilitated diffusion.

Lerok [7]3 years ago
3 0
Ion transport or active transport depending on which situation you have.
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Due to hunting, the population of Northern elephant seals was reduced to 20 individuals at the end of the 19th century. Since th
anygoal [31]

Answer:

Option C

Explanation:

Long term hunting has led to following  

a) Reduction in population size and hence the gene pool

b) Causation of Bottleneck effect

Due to above factors, the remaining population of Northern elephant seals does not represent the original genetic problem.  

Due to this restriction in mating and genetic variation shall prevail pertaining to reduced gene pool.  

Due to continuous hunting, the gene pool will further shrink, leading to genetic similarity among the organism

Option C is correct

8 0
3 years ago
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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