AA X aa
= 4 Aa
======the only possible cross between a true breeding homozygous dominant and a true breeding homozygous recessive.
2:1:2 answer: 1
The alleles of the f1 offspring will depend on the alleles of the parents. In theory all of the alleles in the parental genotypes could be present in the f1 generation.To work out which combinations of alleles will be present in the f1 generation/the proportion with one allele etc. you would need to draw some kind of cross.AA x AaA AA AA AAa aA aASo the f1 offspring have both the A and a alleles, because the two alleles from each parent are separated into the gametesAA gives two gametes both with 'A' alleleAa gives on gamete with 'A' and one with 'a'
2 genes control each characteristic of an offspring. x
If the parents are both AA, which results in the cross AA X AA, then the offspring will all be AA. If both parents are AA, resulting in the cross AA X AA, then all offspring will be AA. If BOTH parents are Aa, resulting in the cross Aa X Aa, then the offspring will be 25% AA, 50% Aa, and 25% AA. This is only true if the alleles are not sex-linked.
They determine the sex of the offspring in Humans.
In humans, it is the father's chromosomes that determine the sex of offspring. The father contributes either an X or a Y chromosome, while the mother always contributes an X chromosome. If the combination is XX, the offspring is female, and if it's XY, the offspring is male.
25%
F1 offspring obtained by monohybrid cross of AA and AA will be Aa.
A dihybrid cross results in 16 boxes for the offspring. For example, the cross RrDd X RrDd is shown below:RDRdrDrdRDRRDDRRDdRrDDRrDdRdRRDdRRddRrDdRrddrDRrDDRrDdrrDDrrDdrdRrDdRrddrrDdrrdd
offspring will be produced in following AA :Aa:aa and is in the ratio of 1:2:1 so the probabiltiy is 50%
The genotypic ratio of a cross of Aa and Aa is: one AA, one aa, and two Aa. Or 1:2:1
And the boxes are how big ????
6 x 6 = 36 boxes
There are 9 boxes in a 3x3 grid.
80 eight boxes wide by ten boxes long
That depends on the size of the little boxes. If they're all 1-cubic-foot boxes, then you can fit (7 x 7 x 8) = 392 of them in there. And by the way ... 7 x 7 x 8 is not a "cube".
None - if the little boxes are taller than the big box!
17 x 13 = 221 221/23.97 = 9.22 9.22 boxes or 10 whole boxes.