Answer:
Let "L" be the dominant allele for little finger shape and "l" be the recessive allele for the same.
It is given that a person is homozygous dominant for blood group B and little finger shape. Hence, the genotype would be 
<em>LL</em>
The other person is heterozygous for blood group A and homozygous recessive for little finger shape. Hence, the genotype would be 
The type of gametes formed by the first parent would be 
<em> </em>only.
The type of gametes formed by second parent would be 
and <em>il.</em>
The cross is shown in the attached figure.
According to the cross, we get two phenotype:
- offspring with blood group AB and dominant little finger shape, and
- offspring with blood group group B and dominant little finger shape.
- They are present in 4 : 4 = 1 : 1
According to the cross, we get two genotype:
- They are also present in 4 : 4 = 1 : 1.
The answer is; C
Rhizoids are analogous to roots in vascular plants. They anchor the gametophyte of the moss into the substrate. The rhizoid is also significant in absorbing nutrient from the substrate just like roots. While roots of vascular plants are multicellular, rhizoids may be unicellular or multicellular. Unlike in roots where water is absorbed and passes through the roots up the plant, in rhizoids, water is carried by capillary action in between the rhizoid threads.
Answer: Hello your question is incomplete below is the complete question
Let’s look at a different gene locus, we’ll call it B. The normal, wild type, the population was 100% BB. There are 150 Pakicetus in the population at present. Assume Charles had another mutation; this one at the B locus, to produce allele b. What is the gene frequency of allele b in the population if the population of Pakicetus reaches 100,000
answer : ≈ 1
Explanation:
<u>Determine the Gene frequency of allele b </u>
Population = 100,000
There are 150 Pakicetus in population i.e. 300 alleles at locus B
hence the frequency of b = 1/300 = 0.33%
Increase in population does not affect frequency of b ( HW equilibrium )
<em>therefore the Gene frequency of allele b </em>
= 0.33% * 0.33% * 100000 = 1.08
≈ 1
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