Answer:
Consider the heterozygous oval, thick cell walled bacteria to have the alleles OoTT and the thin cell walled bacteria to have alleles oott. Results will be 50% oval, thick walled bacteria and 50% round, thick walled bacteria. This will be the F1 progeny.
When the oval, thick walled bacteria from the F1 progeny is cross bred with round, thick walled bacteria then 25 percent of the bacteria will be heterozygous oval, thick walled. 25 percent will be heterozygous oval and heterozygous thick walled. 25 percent will be round and thick walled. 25 percent will be round and thin walled.
Answer: 4%
Out of 5000kcal, the owl loses 2300kcal which mean it could only process 2700kcal. Out of 2700kcal processed, 2500 kcal is used for cellular respiration so there is 200 kcal used to make body cells. The production efficiency would be: 200kcal/5000kcal= 4%
Full question found from other source
The F2 generation phenotypes for each cross are shown in Table 1. (See attachment) Which of the following is the mean number per cross of F2 generation offspring that are the result of crossing over?
Answer:
B, 2.2
Explanation:
The parental genotypes are long and black vs short and white. Therefore the phenotypes that result from crossing over are long and white, and short and black. (middle two rows of the table). If we add up the total number of offspring with these genotypes we get 6 long whites, and 5 short blacks.
The total is 11 from 5 crosses, so the mean is 11/5 = 2.2
Answer:
The overview is defined in the clarification section elsewhere here, and according to the particular circumstance.
Explanation:
Including genetic mapping as well as tracing the characteristics of hereditary disorders, genetic markers are valuable.
Basic DNA sequence determination for chromosomes.
- Healthy Marker Properties.
- This needs to have been Polymorphic.
- It is indeed meant to be non-epistatic.
- Throughout the genome, these are distributed uniformly.
