Answer:
B. 0.42
Explanation:
Since the two alleles are at Hardy-Weinberg equilibrium, the sum of the frequency of two alleles will be one. If the frequency of a dominant allele is "p" and that of the recessive allele is "q", then p+q=1
According to the given information, the frequency of one allele in the population is= 0.7
This means that the frequency of the other allele would be= 1-0.7 = 0.3
Frequency of heterozygote in the population =2pq = 2 x 0.7 x 0.3 = 0.42
<u>Answer</u>: A) It is a step in glycolysis.
<u>Explanation</u>: <em>Glycolysis</em> is the metabolic process through which <em>glucose molecules</em> are broken down, whereas <em>β-oxidation</em> is the catabolic process through which <em>fatty acid molecules</em> are broken down.
Avoid intramuscular injection,Examine the skin for ecchymotic areas
About the question:
I failed to find the complete question with the Punnett square, so in the answer and explanation I will propose five options for crosses between dogs with different genotypes, and for each option, I will remark the percent of the offspring that is expected to have wirehair. You just need to look for the option that matches your Punnett Square.
Answer:
- Option 1: HH x HH ----> 100% of the progeny is wire-haired, HH
- Option 2: HH x Hh---> 100% of the progeny is wire-haired, HH + Hh
- Option 3: Hh x Hh --->75% of the progeny is wire-haired, 25% HH + 75% Hh
- Option 4: Hh x hh ---> 50% of the progeny is wire-haired, Hh
- Option 5: hh x hh ----> 0% of the progeny is wire-haired
Explanation:
You will find the complete explanation in the attached files due to technical problems
