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Andreyy89
3 years ago
10

Consider the diagram of an internal feedback mechanism that the body uses to help maintain homeostasis.

Biology
1 answer:
enyata [817]3 years ago
4 0

In homeostasis the body has a target set point which it tries to maintain, with the use of the negative feedback loop the body strives to stabilize the internal environment by regulating the temperature of the body.

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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.

3 0
3 years ago
Medical tests were conducted to learn about drug-resistant tuberculosis. of 142 cases tested in new jersey, 9 were found to be d
KengaRu [80]
Phat1 = 9/142
Phat 2 = 5/268
Phat 3 14/410
Test statistics = (Phat 1 - Phat 2) / (Square root of (phat × (1- phat))
times the square root of (1/n1 + 1/n2))
Test statistics = 2.373
P- value = 2× (1- normal cdf (2.373, 0, 1)) 
= 0.0176
At the level of significance, there is difference between the proportion of drug-resistant wihtin the two states.
3 0
3 years ago
At which one of the following days in the 28-day cycle is a woman MOST likely to get pregnant?
motikmotik

Answer:

The answer is day 14 - / + 3 days

Explanation:

In a 28-day menstrual cycle the most likely days for a woman to become pregnant is from day 14 - / + 3 days, that is, days 11, 12, 13, 14, 15 and 16. She should be educated, If she is in search of pregnancy, take the vaginal temperature and realize how the viscosity of the abundant flow begins to present.

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3 years ago
What describes the conversion of ADP to ATP?
vlada-n [284]

Answer:

ATP stands for adenosine triphosphate, where three phosphoric acids attach themselves to provide high energy phosphate groups to it. This process of conversion of ADP to ATP is called as phosphorylation.

Explanation:

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Which of the following statements about biofuels is true? a. Biofuels are composed of 100% renewable materials. b. Algae is the
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C. Energy processing from corn biofuels is not energy efficient.
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