Answer:
7%.
Explanation:
The recombinant progeny might occur due to the exchange of genetic material between non sister chromatids known as crossing over during the meiosis.
The chromosome sequence is DABC. A and B distance is 4 map units, B and C is 2 map units, B and D is 5 map units. The distance between A and D is 1 map unit that can be calculated by subtracting the distance between BD and AB ( 5-4). The CD map distance can be calculated by adding the distance between AD, AB and BC ( 1 + 4+ 2). The recombination frequency is equal to distance between them is 7%.
Thus, the answer is 7%.
Explanation:the haemoglobin of lamprey is different from that of humans as it consists of characteriatics partly of vertebrates and partly of invertebrates.
It has a molecular weight equal to that of muscle haemoglobin but it is found to be very efficient in oxygen transport.
The haemoglobin in the oxygenated form is monomeric and undergoes association with diamers and Tetramers when they undergo disassociation.
Fixation or volatilization, mineralization, nitrification, immobilization, and denitrification.
1.What is the probability that a sperm from the father will contain the PKU allele?
Probability = ½
2.What is the probability that an egg from the mother will contain the PKU allele?
Probability = ½
3.What is the probability that their next child will have PKU?
Probability = ¼ (because each parent has ½ chance ½ X ½ = ¼)
4.What is the probability that their next child will be heterozygous for the PKU gene?
Probability = ½ (because each parent has ½ chance of donating the ‘P’ allele and ½ chance of donating the ‘p’ allele (½ x ½) + (½ x ½) = ½
Answer:
No B is the answer....................
