The probability of the offspring of a heterozygous father and homzygous mother having five fingers is 50%.
<h3>How to calculate genotype of a cross?</h3>
According to this question, a gene coding for the number of fingers in humans is involved. The allele for six fingers (F) is the dominant trait while the allele for five fingers (f) is the recessive trait.
If a cross between a heterozygous father that posseses a genotype of Ff and a homzygous mother that posseses a genotype of ff, the following offsprings will be produced:
This shows that the probability of the offspring of a heterozygous father and homzygous mother having five fingers is ½ (50%).
Learn more about genotype at: brainly.com/question/12116830
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Answer: Lk for this plasmid = 47
Explanation:
The formula to compute Lk is given by :-
Topological Linking Number (Lk): Twist (Tw) + Writhe (Wr)
Given: A plasmid 41 twists and 6 right handed writhes.
i.e. Twist (Tw) = 41 , Writhe (Wr) = 6
Then, Topological Linking Number (Lk) for this plasmid : = 41+6
= 47
Hence, the Lk for this plasmid = 47
It’s C new seeds
Hope this helps
