Question

Please help and include punnett squares!

David Beckham can roll his tongue but Sandra (his mother) cannot He also has attached earlobes. He marries Victoria, who cannot roll her tongue and has free earlobes. Anthony (Victoria's father), however, has attached earlobes. What are the chanees that their first child had free earlobes and was able to roll his tongue? 13. Huntingon's Disease is caused by a dominant allele. Roger is homozygous tor Huntington's and has cystic fibrosis. Elvira s homozygous normal for Huntington's but is a carrier of cystic fibrosis. Cystic fibrosis is inherited via a recessive allele. What are the chances that their child has neither Huntington's nor cystic fibrosis? 14. Homer is type A and Grandpa (his father) is type O. Marge is type B and her father was type O. The Rh blood group is not linked to the ABO system. It is a two allele group, with positive dominant to negative. Homer is positive, but his mother is negative. Marge is also negative. What are the chances that Lisa, Homer and Marge's daughter, will be AB negative? 15. Both William and Kate are heterozygous for arm placement and also for thumb placement. What are their changes of having a child with...... both left arm and left thumb on top?... both right arm and right thumb on top?

Answer 1

12) rolling tongue and free earlobes are dominant traits

Let the alleles be

R- Rolling tongue

r- cannot roll tongue

F- Free earlobes

f- attached earlobes

Man is Rrff ( since his mother cannot roll tongue she is rr so she passed r allele to son) and the woman is rrFf ( since her father has attached earlobes he has passed f allele to his daughter)

Rrff* rrFf

	Rf	rf
rF	RrFf (free earlobes and rolling tongue)	rrFf
rf	Rrff	rrff

so the probability of getting Free earlobes and rolling tongues= number of free earlobes and rolling tongue/ total number

=1/4

13) let the alleles be D- dominant allele causing Huntington's disease, d- recessive allele normal allele

C- normal allele c- recessive allele causing cystic fibrosis

so DDcc* ddCc

	dC	dc
Dc	DdCc(Huntington's disease, not affected with cystic fibrosis)	Ddcc ( affected with both Huntington's disease and cystic fibrosis)

the probability that the child has no cystic fibrosis and Huntington's disease= number of children without both diseases/ total number

=0/4

=0