AAAAAAAAAAhhh! Help! Genetics problem.?
What offspring phenotypes are likely to be obtained from a black, heterozygous long tailed male and a white, heterozygous long tail female. I realise that it's sex linkage but what would the gametes be? I'm abit confused! Please help me!
Answers:
it will be black if it is female :
reason the female would have got the the dominent gene for black(through fathers X chromosome) even though it will get a white gene from the mother
If male - the offspring will get its lone chromesome from the mother , for the colour, it will be white as for this trait the mother is homogenous.
hence there is 50% percent chance of the off spring being whiteor black
separatly
there is 25 % chance the offspring will have short tail and 75% it will a have a long one( because both are heterozygous)
hence 12.5 % will be black and short tailed
12.5% will be white and short tailed
37.5% will be white and long tailed
37.5% will be black and long tailed
the gamates
male
BT , Bt
(none due to lack of X chromosome)T, (none)t
for female
bT,bt. (bT,bt)
you shouldn't play god!
ok this is what you know so far
X^B means Black coat dominant on the X gene. the Y does not have the gene but you have to include bc all males have XY.
the Tail length is not on the sex chromosome so you treat it normally.
you are mating a male YX^B Tt x female X^bX^b Tt
your male can have gametes:
Y T
X^B T
Y t
X^B t
the female has
X^b T
X^b t
put the gametes on an the sides of the punet square:
....female.. X^B T .......X^B t
male:
Y T.......YX^B TT.......YX^BTt
X^B T.....X^BX^B TT ......X^BX^B Tt
Y t.......YX^B Tt........YX^B tt
X^B t......X^BX^B Tt......X^BX^B tt
sorry i can't draw an actual punet square on this but hopefully you get the picture.
hello, now it's been a couple of years since I've had to do genetic questions, but I believe that you should get: a black female with a long tail, a black female with a long tail (again), a white male with a long tail, and a white male with a short tial. Since the color (black or white) is sex linkage and the father is black (which is the dominant allele) you can only have black colored females. This is because all the females have to have at least one dominant allele coming from the father. Since the mother is recessive homozygous (only way she can be white) the male offspring will have to be white. As for tail length: since both of the parents are heterozygous it will follow a 1:2:1 pattern of 1 homozygous dominant, 2 heterozygous, and 1 homozygous recessive.
I get the genotypes to be: BbTT, BbTt, byTt, bytt.
I hope that helps and that you don't get even more confused from me :) . good luck
First write down the genotypes:
male Tt (het long tail) B- (black)
female Tt (het long tail) bb (white)
The X will assort independently from the autosome so possible gametes are:
male TB and tB (female gametes carrying X chromosome) or T- and t- (male gametes carrying no X-linked colour gene)
female Tb and tb
Offspring: male TT b-; Tt b-; tt b-. The ratio of tails will be 1:2:1 as in normal Mendelian inheritance but all will be white as they will inherit their single X allele from their mother.
females will have exactly the same ratios of tails as males but all will be black as they will inherit the dominant black from their father and recessive white from their mother and all will have genotype Bb.
This would be an easy litter to sex!
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