Genetics Problems...?

Ectrodactyly (lobster claw) is an inherited physical defect in humans occurring in homozygous recessive individuals.

a. If 2 normal parents had a daughter affected with this condition %26amp; a normal son, what is the probability that the son will be a carrier of the recessive allele?

b. If this son married a normal woman whose mother was afflicted with lobster claw, what is the probability that a 2nd child would be afflicted if the 1st child born to these 2 ppl had this physical defect?



A cross between 2 snapdragon plants produced 83 plants with pink flowers, 35 with red, and 36 with white.

a. What is the genotype %26amp; phenotype of each parent?

What phenotypes %26amp; in what proportions would you expect among progeny of the following crosses:

b. pink x pink

c. red x red

d. red x white

e. pink x white





I am still confused on how to do these
Genetics Problems...?
snapdragons



parents:

phenotype:both are pink

genothype: both are Ww (W=white allel, w=red allele)



ratios of offspring phenotypes:

b: pink x pink: pink:white:red = 2:1:1

c: red x red: all red

d: red x white: all pink

e: pink x white: pink:white = 1:1





Note: answer above: the probability that a child is born with the affliction under these circumstances is ALWAYS 1/4. This is not changed by the fact that there was an earlier child already born with the disease. The 1/16 probability is only right if you were asking "what is the probability that 2 children are born with the disease". This is different from the question above.
Reply:Use Punnett Square to determine your answer.
Reply:a. Well if two normal parents have this, and the disease only shows up with homozygous recessive, and they have a daughter affected by the condition (xx) then they both must be Xx. Otherwise they wouldn't be able to genetically produce a child that is xx. So the probability they would have a NORMAL son who is a carrier of the recessive allele would be 50%.

Because - (Xx)(Xx) produces the following genotype percentages:

1/4 XX; 1/2 Xx; 1/4 xx



b. This would be yet another (Xx)(Xx) cross - since the mother would have passed the recessive x to the daughter (even though she's normal, since it requires homozygosity to show up phenotypically). So the percentage of having TWO children that have the disease is 1/4*1/4 which is 1/16.

(First child probability * Second Child probability)



The second one I'm too lazy to read, sorry.