In bone growth, the medullary cavity is enlarged because of the activity of:
1 answer:
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Answer:
1/2
Explanation:
Let's assume that the allele "p" is responsible for PKU in the homozygous state. According to the given information, both the parents are unaffected by the disorder but have an affected child. This means that both the parents carry at least one copy of the recessive allele responsible for the disease. Therefore, the genotype of each of the parents is Pp. A cross between Pp and Pp would produce progeny in following ratio=
Pp x Pp = 1/4 PP: 1/2/ Pp: 1/4 pp
Therefore, the probability that their next child will carry just one recessive allele is 1/2.
Answer:
Explanation:
Given that:
The abundance of three algal species in Lake A is now represented by the vectors:
[97, 84, 43] and [100, 80, 50]
Now if we look at Lake A, the change occurring in the vector of algae species can be determined as:
a = [100 -97, 80 - 84. 50 - 43]
a = [3, -4, 7]
Thus, the magnitude of that change is:
The abundance of three algal species in Lake B is now represented by the vectors:
[25, 59, 22] and [20, 63, 15]
At Lake B, the change occurring in the vector of algae species can be determined as:
b = [20 -23, 63 - 59. 15 - 22]
b = [-3, -4, -7]
Thus, the magnitude of that change is:
Hence, for both Lake A and B, the magnitude of change is the same.
Answer:
50%
Explanation:
This question involves a single gene with two alleles W and w in puppies. According to the question, the genotype of the parent organisms are Ww. In a cross between the two parents i.e. Ww × Ww, the following gametes will be produced by each parent: W and w.
Using these gametes in a punnet square (see attached image), the following genotype of offsprings will be produced: WW, Ww, Ww, ww. Hence, based on this question, the percentage of puppies that would have the same genotype as the parents, Ww are 1/2 × 100% = 50%.
