Humans convert ammonia to the water soluble compound urea. True O False
1 answer:
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Answer:
A) 9:3:3:1
B) All bitter fruit, yellow spotted offsprings
C) Phenotypes are bitter yellow spotted (4), bitter no spot (4), sweet yellow spot (4), and sweet no spot (4). 1:1:1:1
Explanation:
This is a typical dihybrid cross involving two genes, one coding for fruit taste and the other for spot color. The allele for bitter taste (B) and yellow spot (S) is dominant over the allele for sweet taste (b) and no spot (s) respectively.
Hence, a heterozygous F1 resulting from a cross between an homozygous dominant (bitter fruit, yellow spot) and homozygous recessive (sweet fruit, no spot) will have a BbSs genotype. The heterozygous F1 offsprings are self-crossed and produce gametes BS, Bs, bS, bs. (See punnet square). The F2 offsprings will have the following phenotypes: Bitter fruit, yellow spot (9)
Bitter fruit, no spot (3)
Sweet fruit, yellow spot (3)
Sweet fruit, no spot (1)
Back cross between a F1 offspring (BbSs) and homozygous dominant parent (BBSS) will produce all bitter fruit, yellow spot offsprings (see attached image). BBSS (4), BBSs (4), BbSS (4), and BbSs (4) are the offsprings' genotypes.
For the back cross between a F1 offspring (BbSs) and a homozygous recessive (bbss) parent, the Phenotypes with their proportions are as follows:
Bitter fruit, yellow spot (BbSs, 4)
Bitter fruit, no spot (Bbss, 4)
Sweet fruit, yellow spot (bbSs, 4)
Sweet fruit, no spot (bbss, 4).
