Answer:
producer rely for photosynthesis and consumers for vision and growth
Answer:
(a) 1/2; (b) no
Explanation:
Glucose-6-phosphate dehydrogenase deficiency (G6PD) is an X-linked recessive disorder and the woman's father was diseased so it means that woman is a carrier of the allele but has normal phenotype. It means that she will have XXᵇ genotype.
In contrast to this, her husband is diseased so his genotype will be XᵇY.
The Punnett square diagram related to the cross is attached.
(a) Proportion of their sons expected to be G6PD is 1/2:
They both may give birth to 4 progeny with genotypes XXᵇ, XᵇXᵇ, XY and XᵇY. It means they both may have 2 sons out of which one with genotype XᵇY will be diseased while the one with genotype XY will be healthy. So the proportion of their sons having G6PD is 1/2 or 50%.
(b) If the husband were G6PD deficient, the answer will not change.
The reason behind this is that this disease is caused by an allele located in X chromosome. But father contributes only Y chromosome to his son not X chromosome. The X chromosome will affect the genotype of his daughter not son that is why answer will not change. It means they will still have 1/2 of their sons diseased.
It’s to little I can not see it
1. All cells are similar in structure and function.
2. Cell contains hereditary information that is passed from cell to cell during cell division.
3. Living cells can be created from dead cells. All cells perform similar metabolic activities.
hope this helps
Answer:
2.275% of this population has a diastolic blood pressure less than 60 mmHg
Explanation:
Hello!
Yo have the distribution of the diastolic blood pressure in a certain population. Be X: diastolic blood pressure of an individual, X~N(μ;δ²)
Where
μ= 82mmHg
δ=11 mmHg
You need to calculate the probability of an individual of this population having less than 60mmHg diastolic blood pressure.
Symbolically:
P(X<60)
To obtain the value of probability you need to standardize the value of diastolic pressure so that you can obtain it from the standard normal distribution. The way to standardize the value is to subtract the mean and divide by the standard deviation
Z= (X-μ)/δ~N(0;1)
P(Z<(60-82)/11)
P(Z<-2)= 0.02275
I hope it helps!