Answer:
The correct answer is-
F1 - AaBb (lacerate)
F2 - A_B_; A_bb; and aaB_ (lacerate)
- aabb (normal)
2. Two genes, with a dominant allele at either or both loci.
Explanation:
The given information gives THE following data:
Dominant: Lancerate leaves - AABB
Recessive: normal leaves - aabb
F1 has - all Lacerated leaves - AaBb
F2 by selfing F1:
AB Ab aB ab
AB AABB AABb AaBB AaBb
Ab AABb AAbb AaBb Aabb
aB AaBB AaBb aaBB aaBb
ab AaBb Aabb aaBb aabb
Here, 15/16 = lacerate which is 0.94 which is equal to the value of lacerate given in the question - 249/265 = 0.94
And normal 1/16 = 0.062 is almost same as 16/265 = 0.060
Thus, the genotypes of -
F1 - AaBb (lacerate)
F2 - A_B_; A_bb; and aaB_ (lacerate)
- aabb (normal)
B if it’s not correct sorry
Answer:
The minimum mass of dissolved cholesterol that the new system must be able to detect is 0.43 milligrams.
Note: The question is incomplete. The complete question is:
<em>GenAlex Medical, a leading manufacturer of medical laboratory equipment, is designing a new automated system that can detect borderline high levels of dissolved cholesterol (170, to 200. mg/dL), using a blood sample that is as small as 250 μL Calculate the minimum mass in milligrams of cholesterol that the new system must be able to detect Be sure your answer has the correct number of significant digits.</em>
Explanation:
Range of the equipment to detect cholesterol in blood = 170.0 to 200.0 mg/dL
Minimum amount of cholesterol that can be detected by instrument = 170 mg/dL
Volume of blood sample = 250 <em>μ</em>L= 0.0025 dL (1 dL = 100000 <em>μL)</em>
Amount of dissolved cholesterol in 250 <em>μL</em> sample of blood =
170 mg/dL * 0.0025 dL = 0.425 mg
The minimum mass of cholesterol that the new system must be able to detect is 0.43 milligrams.
Answer:
1/1024 is the proportion of the F2 genotypes will be recessive for all five loci
Explanation:
When crossing between both parents, all genotypes will give us 100% AaBcCcDdEe. When a self-fertilization is performed this means that it can have a cross for example of two flowers of the same plant with the genotype AaBcCcDdEe In this way, given the law of independent segregation which states that the alleles of two or more different genes are distributed in the gametes independently of each other. The proportion that at this junction the alleles are aa (1/4) bb (1/4) cc (1/4) dd (1/4) and ee (1/4). The proportions (1/4)* (1/4)*(1/4)*(1/4)*(1/4) are multiplied, obtaining a value of 1/1024
