Genetics Punnett Square problem?

Question

I'm stuck on this genetics problem. I have the answers, but I need to know how to figure out each question. The question is:

In sesame plants, the one-pod conidition (P) is dominant to the three-pod condition (p), and a normal leaf (L) is dominant to wrinkled leaf (l). Pod type and leaf type are inherited independently. Determine the genotypes for the two parents for all possible matings producing the following offspring:

a. 318 one-pod, normal leaf : 98 one-pod, wrinkled leaf
b. 323 three-pod, normal lead : 106 three-pod, wrinkled leaf
c. 401 one-pod, normal leaf
d. 150 one-pod, normal leaf : 147 one-pod, wrinkled leaf : 51 three-pod, normal leaf : 48 three-pod, wrinkled leaf.
e. 223 one-pod, normal leaf : 72 one-pod, wrinkled leaf : 76 three-pod, normal leaf : 27 three-pod, wrinkled leaf.

Please show work and explain. =/ My teacher failed to explain this, so thanks! =]

Answer 1

On all of these reduce the ratios to whole numbers. Divide the larger numbers by the smallest in each ratio. The patterns tell you what kind of cross this is.

a & b) Show a ratio of 3:1 (This is a genotype of 1:2:1 seen in a hybrid cross. Therefore this is heterozygous for one trait only, the other must be homozygous.)

a) The only trait with two versions is the leaf. Figure the cross by drawing the punnet square and filling in the known offspring details. You can fill in the phenotypes beginning with heterozygous recessive ll. All are one pod, so no recessives there, just put PP for both parents. The rest are normal leafed but there are wrinkled ones present so the parents are heterozygous for leaves. Ll x Ll => 1:2:1 LL, (lL, Ll,) ll
A standard pattern to memorize for a hybrid cross.
PPLl x PPLl => 3:1

b) Again a single hybrid pattern 3:1 phenotype (1:2:1 genotype is implied) The only gene that changes is the leaf so it is the heterozygous trait again. All three pod so both parents are pp.
ppLl x ppLl => 3:1

c) PPLL only one population self crossing.

d) 3:3:1:1
3 of P_L_:3 of P_ll: 1 of ppL_:1 of ppll
The grid is four by four instead of 2x2. This cross is distinct by having two of the three pod forms. One is double recessive ppll while one is heterozygous for leaves ppLl. This is not a dihybrid cross but a variant. Look at e.

Look at the ppLl x ppll portion => 1:1 ratio of a classic Ll x ll
Pp x Pp => 1:2:1
Together the (1:2:1) x (1:1) always =>1:3:3:1
PpLl x Ppll 1:3:3:1 is another pattern to memorize.

e) 9:3:3:1 This number ratio is always seen with a dihybrid cross!
9 of P_L_:3 of P_ll:3 of ppL_:1 of ppll again in a 4x4 square

This one is dihybrid PpLl x PpLl with all possible gametes against all possible gametes PL, Pl, pL, pl x PL, Pl, pL, pl

Answer 2

For this question, forget the numerical value for a while. Concentrate on the genotype of the offspring. Ok. Now represent the offsprings genotype
e.g: a) one-pod, normal leaf = PpLl or PPLL or PPLl PpLL ; one-pods, wrinkle leaf = PPll or Ppll
Okay, now you know that each allele comes from each parent, right? both of the off-springs have dominant allele for no. of pod, therefore the possible combinations for both parent is PP and Pp. pp is not relevant because non of the off-spring have the allele, p . Then, lets look at the type of leaf allele of the off-spring. Since wrinkle leaf phenotype is only shown if the allele is ll, then, the parents allele must have atleast 1 l. From the one-pod, normal leaf, we can conclude that either one of the parents or both have the dominant alelle L. Therefore, the possible combination for this are Ll or ll. Thus! Combine the two genes! PPLl or PpLl or PPll or Ppll. You can draw the punnet square to check. you have to get the same off-spring to consider my prediction is correct.

Answer 3

You will have to put each of your allele variables into a different "box" on the punnett square so you will have a total of 16 different outcomes. DH dH Dh dh dh DdHh ddHh Ddhh ddhh dh " " " " " " " " dh dh