Question

16. Part B Convert the expected phenotypic ratio from Part A into the each of the four phenotypes and record them in the table belo il. Calculate the probability for each of the four phenotypes o rividing the numbho from the data presented at the beginning of the question by dividing the number in the ch class by the total number of progeny and record thes table below Probability- Number of Progeny in Phenotype Class+ Phenotypes Red and Round Number of Offspring Expected Red and Oval Yellow and Round Yellow and Oval 17. Part C Compare the expected probabilities of each phenotype to the observed probabilities. Are the gene for fruit colour and the gene for fruit shape assorting independently? Explain.

Answer 1

19 a- haemophiliac male Genotype: X^hY ; phenotypically normal female Genotype: X^hX (she is normal but a carrier of haemophiliac gene) ; haemophiliac daughter genotype: X^hX^​h

	X​​​​​​h	Y
X​​​​h	XhXh (haemophiliac daughter)	XhY (haemophiliac son)
X	XhX (phenotypically normal daughter, carrier)	XY (normal son)

b- it is more common for male to be haemophiliac than a female, since haemophilia is a X- linked recessive trait (major characteristic of X linked inheritance being that the father's cannot pass X linked traits to their son's chromosome). In males (who have only one X chromosome), one altered copy of the gene in each cell is enough to cause haemophilia; whereas females have two X chromosomes, therefore a mutation must be present in both copies of the gene to cause the hemophilia.As a result, Males are affected by X-linked recessive disorders much more frequently than females.

Since 16 part A not provided, part B cannot be answered.

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