Answer:
Frequency of heterozygous carriers of the allele in the entire herd is 
i.e 
out of 
Explanation:
It is given that a recessive allele pair produces a dwarf cow.
Let this recessive allele be represented by "t" and the dominant allele for tall height trait in cows be "T".
Given
Total population of randomly mating cows 
Total number of dwarf calves in the population 
Frequency of genotype "tt" producing dwarf calves is represented as 
in HW equilibrium equation.
Here
Frequency of recessive allele is equal to "q"
As per first equilibrium equation of HW,
Frequency of genotype "TT" producing dwarf calves is represented as 
in HW equilibrium equation.
As per HW's second equilibrium equation
Substituting the given values in above equation, we get -
Frequency of heterozygous carriers of the allele in the entire herd is i.e out of
Answer:
Chi square value = 0.3
In order to find the expected value, an assumption of purple flower colour dominance over white flower colour has to be made.
Explanation:
<em>First, we will assume that the purple flower trait is dominant over white flower trait, and that the alleles segregate according to Mendelian standard. Hence, the expected F2 phenotypic ratio will be 3:1.</em>
In order to get the Chi square, the expected phenotype frequency will be calculated thus;
Expected frequency (E) for purple flowered plants = 3/4 x 160 = 120
Expected frequency (E) for white flowered plants = 1/4 x 160 = 40
Observed frequency (O) for purple flowered plants = 123
Observed frequency (O) for white flowered plants = 37
<em>Chi square 
= 
</em>
For the purple colour flower
, which gives 0.075
For the white colour flower
, which gives 0.225
<em>Total = 0.075 + 0.225, which gives 0.3</em>
Answer:
b
Explanation:
water's hydrogen bonds allow it to move upwards against gravity through plant veins.
