Answer:
C. 1/4
Explanation:
Let's assume that the allele for the spotted coat is "S" and the one for the even coat is "s". The allele "L" gives short horns while the recessive allele "l" imparts long horns. The genotype of the cattle heterozygous for both traits would be SsLl. A cross between two heterozygous cattle would produce progeny in following phenotype ratio=
9 spotted coat and short horns: 3 even coat and short horns: 3 spotted coat and long horns: 1 even coat and long horns.
Therefore, the proportion of the progeny with long horns = 4/16= 1/4
You have also the total power generated is 200W
To figure out the kind of exercises, you need to master the definition of each unit. for example
power: watt (W) is
joule/secWork: is Newton. meter =
Joule
To obtain the work done by the adult, we have to convert the power (watt) into a work by multiplying by the time (
* seconds = Joule)
10 minutes is 60 * 10 =
600 seconds200Watt * 600 second = 120000 joules =
120 Kj
So the work is 120Kj
Answer:
d) passage of a solute against its concentration gradient
Explanation:
When a solute is transported against its concentration gradient, the cells use metabolic energy. To move a substance from the region of its lower concentration to that of its higher concentration, the energy of ATP hydrolysis is utilized. These types of transport mechanisms are called active transports. If ATP hydrolysis is inhibited in a cell, it would not be able to perform the uphill movement of solutes due to the lack of any source of energy to drive the process.
Answer: Workplace electrocusion
Explanation:
The human body is a good conductor of electricity. The workplace electrocution occurs due to electric shock. This leads to workplace injuries that may often lead to disabilities and can also be fetal. The workplace electrocution can also lead to cardiac arrest, and tissue damage due to thermal burns. The severity of injuries may depend upon the intensity of electricity, health of the victim and how the current flows in the body.
According to the given situation, the freshly mopped floor is a source of water and water, as well as metal, which is a good source of electricity. The electric wire is an energized source of electrical energy that will allow the flow of electric current in the body, as well as contact with the floor containing water, which will enhance the effect. Thus this is an example of workplace electrocution.
Answer:
Frequency of B allele is 0.6681
Explanation:
If p represents the frequency of dominant allele and q represents the frequency of recessive allele, according to Hardy-Weinberg equilibrium:
p + q = 1
p² + 2pq + q² = 1
where p² = frequency of homozygous dominant genotype
q² = frequency of homozygous recessive genotype
2pq = frequency of heterozygous genotype
Given that number of recessive chestnut horse = 28
Total horses = 226 + 28 = 254
frequency of b² genotype = 28/254 = 0.1102
frequency of recessive b allele = √0.1102 = 0.3319
So, frequency of B allele =
1 - 0.3319 = 0.6681
Hence frequency of B allele is 0.6681
