I think the answer is A) steroids
Hope this helps!
Answer:
The correct answer is -
A) BbTt x bbTt
B) .5^4 or 1/16 or 0.0625
Explanation:
As it is given that Brachydactylus and PTC tasters both traits are autosomal dominant conditions which mean only one allele would be enough.
For Branchydactylus and For tasting : man and women will be heterogeneous.
Hence,
The Genotype of man = BbTt
The Genotype of wife = bbTt
b. Answer = 0.0625
For Branchydactylus:
Bb X bb
Possible genotypes:
B b
b Bb( Brachydactylus) bb(normal)
b Bb (Brachydactylus) bb (normal)
The probability of a single child being Branchydactylus = 2/4 = 0.5
So, Probability of all 4 child being Branchydactylus = .5 x .5 x .5 x .5 = 0.0625
Answer:
Pls attach an image so I can do it, then I can answer
Explanation:
Answer:
0.25
Explanation:
The population of panthers consists of 60 black panthers with the dominanat allele PP. 30 are pink panthers with the genotype Pp and 10 panthers are white with the phenotype pp.
The frequency of p allele in the population can be determined by
the number of individual with pp genotype + 1/2the number of individual with Pp genotype divided by the total number of individual.
Frequency = 10+1/2*30/100
Frequency = 0.25.
Thus, the frequency of p genotype is 0.25.
