I need help with question #1
2 answers:
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Answer:
1
Female gametes:
ab= 0%
ab+=0%
a+b=0%
a+b+=0%
for the miniature phenotype to occur, it is necessary to contain two recessive characters and the female must have both for it to occur, if it is homozygous for b + it will have 2 alleles b +, understanding that the female is aab + b it will only occur gametes ab +
.
2. Phenotype
Miniature body + gray=25%
Large body +gray = 50%
Large body +ebony = 25%
Miniature body +ebony =0%
As the crossing cannot occur in males, the gametes it can produce are: ab + and a + b. Although the crossing can occur in women. The gametes produced are: ab +, a + b, ab and a + b +.
the crossing occurs only in the females generating the gametes ab +, a + b, ab and a + b +
The genotypes corresponding to:
Punnet graphic
large gray body: a + ab + b, a + ab + b +, a + a + b + b
big body and ebony: a + a + bb, a + abb
miniature and gray body: aab + b +, aab + b
Answer:
A monosaccharide is made of a 6-carbon cyclic ring. 5 out of the 6 form part of the ring while the other carbon acts as a side group of the 5’ carbon in the ring. When forming a disaccharide with the next monosaccharide, the OH of the 1’ C of one monomer undergoes a condensation reaction with the OH of the 5’ C of the next monomer. This results in the formation of a glycosidic bond and the loss of a water molecule.
