Answer:
The answer is explained below
Explanation:
The BFS program consists of training for agility, flexibility and sport-definite technique five days per week and weight training three days per week. This program is specifically for athletes and not for your average fitness clients; it still does not adhere to the Principle of Individual Differences.
CrossFit- Crossfit is a strength and conditioning program mainly the mix of aerobic exercise, calisthenics (Body weight exercise). In this type of gyms use equipment for multiple disciplines including barbells, dumbbell, pull-up bars. This type of workout makes body more flexible and capable to bear the abnormalities when necessary. Many types of workout are in it like, Chipper (squats, press ups, burpees), Ladder.
HIT – High Intensity Training refers to the one set failure type training program promoted as the most effective and scientifically based strength training program. In this type of program multiple sets of each exercise are performed. Volume is the most common form of weight training program seen in the gym and fitness centers. HIT also recommended sporting persons.
Human born with different strengths and weakness, this fact is considerable for the training program. When beginners at any sport see great improvement when he or she starting their programs. Person needs to improve their loading as possible to make them stronger.
Nowadays CrossFit is the best workouts to have to do because it makes physically stronger, flexible and smarter.
One reason is for safety! in case you get stranded in a body of water you can keep yourself afloat
Answer:
Incomplete dominance is the inheritance pattern where the dominant allele did not mask the recessive allele completely and form a mix of both alleles. Here the inheritance is the incomplete inheritance. The ratio of F2 generation is 1:2:1.
Given:
R1R1 = 42
R2R2 = 39
R1R2 = 86
Total R1 alleles = 2*42+86 = 170
Total R2 alleles = 2*39+86 = 164
Total alleles = 334
Frequency of allele R1 = 170/334 = 0.51
Frequency of allele R2 = 164 / 334 = 0.49
Expected number of each phenotype:
Total population = 167
Blue = R1R1 = 0.51 * 0.51 * 167 = 43.44
Green = R2R2 = 0.49 * 0.49 * 167 = 40.10
Cyan = 2*R1*R2 = 2*0.51*0.49*167 = 83.46
Phenotype Observed(O) Expected (E) O-E (O-E)2 (O-E)2/E
Blue 42 43.44 -1.44 2.0736 0.0477
cyan 86 83.46 2.54 6.4516 0.0773
green 39 40.1 -1.1 1.2100 0.0302
Total 167 167 0.1552
Chi-square value = 0.155
Degrees of freedom = no. of phenotypes – 1
Df = 3-1 = 2
Critical value = 5.99
Chi-square value of 0.155 is less than the critical value of 5.99. So we accept the null hypothesis.
Answer:
25%
Explanation:
When looking at a pedigree remember that:
- squares are males
- circles are females
- the solid colored figure represents an individual affected by a disease
- the empty figure represents a healthy individual
Let us assign the symbol X⁺ to represent the dominant allele linked to the X-chromosome and expressing healthiness, and X⁻ to represent the recessive allele expressing the dissease.
According to this pedigree
- I1 is a man affected by the disease, YX⁻
- I2 is a healthy woman X⁺X⁻
- we can see that among the progeny (generation II) there are two individuals affected (a boy and a girl) and one healthy girl. This means that the mother I2 is heterozygous for the trait.
So, having their genotypes we can know what are the probabilities of getting a son with DMD
Parentals) YX⁻ x X⁺X⁻
Gametes) Y X⁻ X⁺ X⁻
Punnett square)
X⁺ X⁻
X⁻ X⁺X⁻ X⁻X⁻
Y X⁺Y X⁻Y
F1)
- The probabilities of getting a healthy daughter X⁺X⁻ are 25%
- The probabilities of getting a healthy son X⁺Y are 25%
- The probabilities of getting a daughter with DMD X⁻X⁻ are 25%
- The probabilities of getting a son with DMD X⁻Y are 25%
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