Assorted Flower

Assorted Flower
genetics problems?

I have a set of genetics problems that I have no clue what I'm doing. Can someone please explain....

Instructions: In humans "unattached" earlobes are dominant over "attached" earlobes. Use E and e for he earlobe phenotype allees.

1. a female and a male, both with unattached earlobes, have a child with attached earlobes. What is the probability their next child will have attached earlobes.

instructions:
In Mendel's pea plants, yellow seeds are dominant to green seeds. Purple fowers are dominant to white flowers. Use Y and y for the seed color allele and P and p for the flower color alleles. Flower color and seed color assort independently.

5. If a YyPp plant is crossed to a Yypp plant, what is the probability that the resulting plant will have genotype Yypp?

THANK YOU!!!
Nice comment! I just want an explanation not an answer because as I stated in my question I don't know what I'm doing.

1. Unattached earlobes are dominant, so the genotype for both parents is Ee, otherwise there can't be an attached earlobe child (crossing EE with EE or Ee with EE all produce unattached earlobed children as E is dominant).

Hence the child's genotype is ee (must be ee as attached is recessive)

Genetic cross of Ee x Ee gives 1/4 chance of EE, 1/2 chance of Ee and 1/4 chance of ee.

Hence the chance for their next child will still be 1/4 (25%).

5. Damn long dihybrid crosses....oh well here goes:

Each gamete (reproduction cells) must have one gene for seed colour and one gene for flower colour.

So gametes for YyPp = YP, Yp, yP, yp
& gametes for Yypp = Yp, yp

Now we carry out the cross: (can't really draw it here, but draw 5 columns, and 3 rows, put each gamete for YyPp across the top and Yypp down on the left, then cross them and see what you get.)

We should get: 1/8 for YYPp, 1/8 for YYpp, 2/8 for YyPp, 2/8 for Yypp, 1/8 for yyPp, 1/8 for yypp.

Hence the probability for Yypp = 2/8 = 1/4 = 25%

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