Genetic problem!! help!!?

Question

in rabbits black is due to a dominant gene,B, brown to its recessive allele,b. short hair is due to a dominant gene, L, long hair is to its recessive allele, l. in a cross between a homozygous black short-haired male and a homozygous brown long-haired female wahat would be the genetic constitution and the appearance of the F1 generation? of the F2 generation?

how would the F1 and F2 of the problem above compare with the F1 and F2 from a cross between a homozygous vlack long-haired male and a homozygous brown short-haired female?

Answer 1

Crosses with two traits are a little more complex than when just one trait is involved.  You could do this by making a large Punnett square, but instead of having two columns and two rows, you'd need to have 4 because of the number of potential gametes the parents could produce - two traits with two alleles (genes) each would give you 2^2 or 2 x 2 possible combinations for each parent.

For example, if you had a rabbit that was heterozygous for the two traits you have(BbLl), you could use "FOIL" (First letter of each set of letters, Outside, Inside, and Last) to get possible combinations of BL (first "B" and first "L"), Bl, bL, and bl.  But for some reason some students have trouble figuring out which letters to put together.  It helps if you think of each letter of the alphabet as a gene, and all gametes have to have one of each unique letter.  Otherise you could have a gamete with two genes for coat color and none for hair length.  You need to have one of each.

To set this up, you would first figure out what the genotype for each parent would be.  A homozygous black short-haired male would have BBLL (homozygous means the alleles for the traits are the same - either capital or lower case).  The homozygous brown long-haired female would be bbll.  Using "FOIL" you'll see the possibilites would be all BL for the male parent and bl for the female.

So putting the possible gametes from the male across the top and the female down the side, you'd have a Punnett square that woul look like this:

__|BL|BL|BL|BL

bl|.....|.....|.....|.....|

bl|.....|.....|.....|.....|

bl|.....|.....|.....|.....|

bl|.....|.....|.....|.....|

Then you just have to combine the letters from the top of the column and side of the row so you're back to having two of each letter (two genes, one for each chromosome of a diploid organism).  Since all the letters in each row and each column are alike, you get all the same result if the F1 generation - BbLl.

To make the F2 cross, you mate two of the offspring from the first cross.  So now your parents are both BbLl.

Using "FOIL" again, your results are BL, Bl, bL, and bl - for both parents.

So set up your Punnett square:

___|BL|Bl |bL|bl

BL|.....|.....|.....|.....|

Bl |.....|.....|.....|.....|

bL|.....|.....|.....|.....|

bl |.....|.....|.....|.....|

Then just combine the letters from the top of each column and the side row.  You'll get a lot of different results this time!  When you get finished, it may help you to make a note of the phenotypes of each possible offspring.  How many are black, short-hair, brown shrt-hair, black long-hair, brown long-hair.

This shows you how to do the first pairing.  Now try the second one yourself setting it up like this  example, only using the new parent genotypes.  Then you can compare the numbers of each phenotype you get in the final result.