The provided question is incomplete as it lacks the combinations of parent as metion in the question, however the complete question is attached as the image.
Answer:
There are two parent combinations that are :
1. heterozygous black Ff cross with heterozygous black Ff
2. heterozygous black Ff cross with white ff
Explanation:
1. in first combination the each parent will produce F and f gametes, as it is shown as Capital F so it will be dominant over f as per the rule. so the cross can be shown as :
F f
F FF Ff
f Ff ff
where, FF are black Ff black and ff white offspring. As there is clearly 3 black offspring over one white spring so the ratio is 3:1 or 75% chances of black and 25% chances of white offspring (phenotype).
2. in second combination there would be F and f gametes for black parent and f and f for white parent, thus the offsprig would be :
F f
f Ff ff
f Ff ff
As, there is two black and two white springs are produced than the ratio would be 2:2 or 50% chances of each white offspring and black offspring (phenotype)
Answer:
The correct answer is option d) age of either or both parents.
Explanation:
Genetic variation is the difference in nucleotide sequences between individuals organisms within a particular population. Variation can occur either in germ cells that are sperm and egg or in somatic cells or in both.
Random fertilization of egg and sperm can lead to genetic variation is every germ cell has a different set of chromosomes. Independent assortment of chromosomes is also the same as random fertilization as it is also random. Crossing over of homologous chromosomes is one of a major source of the genetic variation as it forms a new set of chromosomes.
Thus, the correct answer is option D) age of either or both parents.
Answer:
148
Explanation:
According to Hardy-Weinberg equilibrium,
p + q = 1
p² + 2pq + q² = 1 where,
p = frequency of dominant allele
q = frequency of recessive allele
p² = frequency of homozygous dominant genotype
2pq = frequency of heterozygous genotype
q² = frequency of homozygous recessive genotype
Here,
Total population = 592
Number of NN people = 148
Frequency of N blood group or NN genotype (q²) = 148/592 = 0.25
q = √0.25 = 0.5
p = 1 - q
= 1 - 0.5 = 0.5
Hence, p = 0.5
Frequency of MM genotype = p² = 0.25
Number of people with MM genotype = 0.25*592 = 148
Hence, 148 people will have MM genotype or M blood group.
1. Phosphates and sugars (deoxyribose)
2. yes
3. the order of the bases aren't but what they pair with are
4. five
5. ten
6. the sequences vary between all organisms
