Answer: Bb/bb (let's take letter B this time, just for no reason since it does not matter :) )
Explanation:
if the ryan's genotype is Bb, alexa's genoype is BB, then the genotype ryan's biological father is either Bb, or bb
punnet square evidence (option 1):
B b
B BB Bb
B BB Bb
punnet square evidence (option 2):
b b
B Bb Bb
B Bb Bb
I believe the answer is 3.Cellulose
Fiber is carbohydrate that can't be digested by human. Consumption of fiber will not give calorie like other food do, but it has other benefit for the digestive system such as helping regulate blood sugar level and satiety.
Cellulose is fiber that could be found in cell wall of green plants. Cellulose is the most common insoluble fiber.
<u>Full question:</u>
For a moment after hearing his dog's high-pitched bark, Mr. Silvers has a vivid auditory impression of the dog's yelp. His experience most clearly illustrates ________ memory.
echoic
short-term
iconic
procedural
<u>Answer:</u>
His experience most clearly illustrates echoic
memory.
<u>Explanation:</u>
Echoic memory is a portion of sensory memory and relates to auditory memories. The sensory memory that conveys into record sounds that you’ve recently confronted is a kind of this memory type. When you listen to a sound, your ears convey that sound to the brain and it is saved by echoic memory for an aggregate of four seconds.
While that concise time, your mind forms and holds an accurate reflection of the sound that you listened, so that if you stayed in a tranquil room you could yet "listen" it subsequent the sound has ended.
I would say man made but not to sure with out answer choices..
The answer is 0.42.
According to the Hardy-Weinberg equilibrium:
p + q = 1
p² + 2pq + q² = 1
where p is a frequency of a1 allele, q is a frequency of a2 allele, p² is a frequency of a1a1 genotype, 2pq is a frequency of a1a2 genotype, q² is a frequency of a2a2 genotype.
If <span>70% of the gametes produced in the population contain the a1 allele, then:
p = 70% = 0.7.
If p = 0.7 and p + q = 1, then q = 1 - p = 1 - 0.7 = 0.3
So, the proportion of </span>the flies that carry both a1 and a2 (a1a2 genotype) is:
2pq = 2 * 0.7 * 0.3 = 0.42
