Answer:
b
Explanation:
because an object is in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
Homeostasis is the attempt to maintain stable conditions (particularly in an organism). It is done by using feedback loops (positive and negative).
There are four bases found in DNA: adenine (A), cytosine (C), guanine (G), and thymine (T). Adenine forms a base pair with thymine, and cytosine forms a base pair with guanine. There is a one-to-one relationship in these base pairings (Chargaff’s rule), which means that if you know the percentage of any one of them within a given DNA sample, you can calculate the percentages of the other three. In this case, you're given the percentage of guanine, and you want to find out the percentage of adenine.
Since guanine base-pairs with cytosine and since there must be as much cytosine as there is guanine, 41% of the bases in this gene are cytosine as well. That means that adenine and thymine <em>together </em>make up the remaining 18% (100% − 41% G − 41% C) of the base pairs. If there must be an equivalence in the number of thymine and adenine bases per Chargaff's rule, then half of the remaining base pairs must comprise adenine and the other half comprise thymine. Half of 18% is 9%.
Thus, adenine makes up 9% of the bases in this gene.
Answer: The probability of producing offspring with genotype Rr is 100%.
Explanation-
Round seed is a dominant trait ( depicted by dominant R allele) whereas wrinkled seed is a recessive trait ( depicted by recessive allele r).
According to the law of segregation of genes, the alleles will separate from one another during gamete formation.
Pea plant with round seeds will produce two gametes that is R and R whereas plant with wrinkled seeds will produce r and r gametes.
When the parents are crossed, they will result in the offspring with genotype Rr ( that will exhibit round seed phenotype).
Refer punnett square.
In order for a recessive trait to be apparent in an organism, the organism must receive a copy of the allele for that trait from both parents.
Let the:
black coat: B
brown coat: b
Trotter: T
pacer: t
The offspring received both of the recessive alleles because it demonstrates the recessive phenotypes. Therefore, they genotype of the offspring is:
bbtt
The black horse that is a pacer that can produce this offspring will be:
Bbtt
Therefore, the brown horse's genotype must be:
bbTt
The offspring will receive b from both parents and t from both parents.
