Answer:
a) The genotype of her father = "aa".
The genotype of her mother = "Aa".
The genotype of Sally is "Aa".
b) 1/2
c) 1/2
Explanation:
a) Sally's father has alkaptonuria. So, the genotype of her father is "aa". Her mother is normal. Since her brother is affected by the disease, he should be homozygous recessive (aa). This means that the mother also has one copy of "a" allele and is heterozygous dominant (Aa). A cross between Aa (mother) and aa (father) would give 50% affected progeny with "Aa" genotype. So, the genotype of Sally is "Aa".
b) The genotype of her mother= Aa
The genotype of her father = aa
Aa x aa = 1/2 Aa (affected): 1/2 aa (normal)
So, if Sally's parents have another child, there is a 1/2 probability that this child will have alkaptonuria.
c) The genotype of Sally = Aa
Genotype of the man Sally marries= aa (since he is affected with alkaptonuria)
Aa x aa= 1/2 Aa (affected): 1/2 aa (normal)
There is a 1/2 probability that their child will have alkaptonuria.
Answer: drink water
Explanation:
water keeps the body cold and hydrated.
Answer:
150.
Explanation:
150 flies are expected to have normal body color and to be eyeless in F2 generation. The reason for that population is the ratio of the species produced. The ratio of the species are 9:3:3:1. Ratio 9 refers to first parent while 1 refers to second parent and the middle 3's represent the hybrid formed. So the population of first parent is 450 and population of second parent is 50 while the population of two hybrids are 150 each.
Answer: Foods with calcium
The amount of bone tissue in the skeleton or bone mass can keep growing until the age of 30. It is recommended to eat food with calcium prior to this age for maximum strength and density of the bones. Examples of food rich in calcium are dairy foods, green leafy vegetables, beans, nuts, fish and a lot more.
