Answer:
The active smokers and those getting exposed, that is, passive smokers are at enhanced threat of bacterial infections. Getting exposed to tobacco smoke enhances vulnerability to respiratory tract infections, comprising pneumonia, tuberculosis, and Legionnaires disease.
Smoking compromises the anti-bacterial activities of leukocytes incorporating monocytes, neutrophils, B cells, and T cells, thus, demonstrating the mechanism for enhanced risk of infections.
I believe that Calpurnia finds the food on the back porch.
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.
Answer:
It would be b. 1 TT: 2Tt: 1 tt. Because crossing two heterozygous parents means the genotypes for them are Tt and Tt. Which means that there is only one way to get TT or tt and 2 ways to get Tt
Explanation: