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Harlamova29_29 [7]
3 years ago
6

Assume that a cross is made between aabb and aabb plants and that all the offspring are either aabb or aabb. these results are c

onsistent with ________.
Biology
1 answer:
Iteru [2.4K]3 years ago
6 0
~Hello there!

Your question: <span>Assume that a cross is made between aabb and aabb plants and that all the offspring are either aabb or aabb. these results are consistent with ________.

Your answer: </span>Assume that a cross is made between aabb and aabb plants and that all the offspring are either aabb or aabb. these results are consistent with absolute linkage in cis.

Any queries ^?

Happy Studying!
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What does a S-shaped curve demonstrate for population growth?
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A design of growth in which (occurs only in a different situation), the population frequency of an organism rises gradually originally, in a good acceleration time, later progresses swiftly nearing an exponential growth pace just like the J-shaped curve, however, it later declines (meaning rejects) in a bad acceleration time till at zero growth rate the population maintains. For example, yeast is a tiny fungus, so tiny, a microscope is required to view it, furthermore, it is used to produce bread and alcoholic drinks, displays the traditional S-shaped curve if raised in a test tube. Its maturity levels off as the population deplete the nutrients that are essential for its growth.

7 0
3 years ago
A population of 150 individuals has an allele frequency of 0.2 for the dominant allele (H) and a frequency of 0.8 for the recess
Amanda [17]
Hardy-Weinberg Equation (HW) states that following certain biological tenets or requirements, the total frequency of all homozygous dominant alleles (p) and the total frequency of all homozygous recessive alleles (q) for a gene, account for the total # of alleles for that gene in that HW population, which is 100% or 1.00 as a decimel. So in short: p + q = 1, and additionally (p+q)^2 = 1^2, or 1
So (p+q)(p+q) algebraically works out to p^2 + 2pq + q^2 = 1, where p^2 = genotype frequency of homozygous dominant individuals, 2pq = genotype frequency of heterozygous individuals, and q^2 = genotype frequency of homozygous recessive individuals.
The problem states that Ptotal = 150 individuals, H frequency (p) = 0.2, and h frequency (q) = 0.8.
So homozygous dominant individuals (HH) = p^2 = (0.2)^2 = 0.04 or 4% of 150 --> 6 people
Heterozygous individuals (Hh) = 2pq = 2(0.2)(0.8) = 0.32 or 32% of 150
--> 48 people
And homozygous recessive individuals (hh) = q^2 = (0.8)^2 = 0.64 = 64% of 150 --> 96 people
Hope that helps you to understand how to solve these types of population genetics problems!
6 0
2 years ago
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