Given
E dominant allele for wet
e recessive allele for dry
that means
for phenotype wet, the possible genotypes are EE, or Ee, and
for phenotype dry, the only possible genotype is ee.
Therefore we also know that the child who has dry earwax has genotype ee.
Since the child inherits one allele from each parent, therefore each parent must have a recessive allele "e".
If both parents have phenotype wet earwax, they both must be heterozygous for wet/dry earwax, namely Ee.
Answer:
Maltose is a disaccharide sugar made up of two units of glucose.
In cyclic structure, the glucose exists in two anomeric forms; alpha and beta.
These glucose units can either joined by α (1→4) glycosidic bond or by β (1→4) glycosidic bond.
Thus, the maltose exists in two anomeric form alpha and beta.
Answer:
a. Fluid intelligence
Explanation:
Cattell-Horn's theory of fluid and crystallized intelligence suggests that intelligence is composed of a series of different skills that interact and work together to produce general individual intelligence.
<u>
Fluid intelligence</u> implies being able to think and reason abstractly and solve problems. This ability is considered independent of learning, experience and education. It is basically configured by primary skills such as induction, deduction, relationships and figurative classifications, breadth of associative memory and intellectual speed, among others. It reaches its maximum splendor in the early twenties and tends to decrease in parallel to the aging and deterioration of neurons. This ability can be measured from tests that measure the biological potential of the individual to learn or acquire knowledge.
Answer:
1/8
Explanation:
Given that the trihybrid parents have AaBbCc genotype for fruit color. The trait is a quantitative trait i.e. each dominant allele will have an additive effect on it. In this case, AaBbCc and AABBCC will not produce same fruit color because AaBbCc has only three loci contributing to the color while in AABBCC all the six loci are contributing to the color. For an offspring to be exactly similar to the AaBbCc parents it should have the same genotype of AaBbCc.
The probability of Aa to come from a cross between Aa and Aa is 2/4 or 1/2
The probability of Bb to come from a cross between Bb and Bb is 2/4 or 1/2
The probability of Cc to come from a cross between Cc and Cc is 2/4 or 1/2
So the collective probability of AaBbCc offspring from a cross between AaBbCc and AaBbCc parents would be=
1/2 * 1/2 * 1/2 = 1/8
Hence, assuming no effects of the environment, 1/8 of the offspring will have the same fruit color phenotype as the trihybrid parent.
