Answer:
AA is labels would correctly model photosynthesis
Its stored in the chloroplasts
Answer:
<em>Backyard bird feeding is a well known </em><em>type of human–untamed life </em><em>communication in specific regions of the northern and southern side of the equator.</em>
Explanation:
In territories where sustaining is mainstream in the Southern Hemisphere, supplemental encouraging may excessively profit acquainted winged creature species while driving with decreases of local species, possibly due to hetero specific competition.
However, Backyard bird feeders can bring about constructive outcomes on some feathered creature species, for example, improved overwinter endurance, expanded populace sizes and geographic range extension.
Answer:
we will know that the allelic frequencies are for R 0.95 and r 0.05
Explanation:
We know that the population is in Hardy-Winberg equilibrium, we deduce the following formula:
p + q = 1
p2 + 2pq + q2 = 1
data
R: red flower allele
r: allele blor blanca
p would be equal to the allelic frequency R
q will be equal to the frequency allelic r
2p = RR
2q = rr
2pq = Rr
If there are 25 white flowers in 1000 plants, their frequency will be:
2pq frequency of the Rr genotype
white flower = 25/10000 = 0.0025 = rr = 2q = 0.0025
we deduce that q is equal to 0.05
we replace the data with the previous formula
p + q = 1
p = 1-0.05
we get as a result
p = 0.95
if p = 0.95 and q = 0.05
we will know that the allelic frequencies are for R 0.95 and r 0.05
Answer:
The correct answer would be:
- Genotype ratio: 1 (PP) : 2 (Pp) : 1 (pp)
- Phenotypic ratio: 3 (polka-dot tails) : solid colored tail
The genotype of both parents is Pp (heterozygous).
Thus, both the parent would produce two types of gametes which are P and p.
The cross would result in the production of offspring with three types of genotype PP, Pp, and pp in 1:2:1.
Offspring with PP and Pp will have a polka-dotted tail as it is the dominant trait.
Thus, the phenotype ratio would be 3 (polka-dotted tail) : 1 (solid colored tail).
