Genetics problem question. PLEASE help?

Question

Multiple choice questions

Hemophilia a recessive, sex-linked trait. Use n to represent the allele for hemophilia. A woman has normal blood clotting type O marries man with normal blood clotting type AB blood. The womans father was known to be hemophiliac with type B blood.

a. Womans fathers genotype as completely be determined.
a. XnY,BB b.XNY,BO c.XnXn,BO d.XnY,BO

b. term best describe woman relating to hemophilia?
a.pure breeding b.hybrid or heterozygous c.carrier d. dihybrid e. none of these

c. what are chances couple having child with hemophilia?
a.0 b.1 c.1/2 d.1/4 e. 3/4

d. chances having a daughter with hemophilia?
a.0 b.1 c.1/2 d.1/4 e.1/8

e. chances having a son with blood type A and hemophilia?
a.0 b.1 c.1/2 d.1/4 e.1/8

f. according to laws of probability, what percentage of this womans sons should inherit the sex-linked trait hemophilia
from their mother?
a.0% b.25% c.50% d.75% e.100%

kt is it b. XNY,BO or e. XnY,BO?

Answer 1

The answers are: a) b - XnY, BO, because in order for the woman to have type O blood, the father must be heterozygous. Also, since he was a hemophiliac, she must have inherited the trait from him even though it is not expressed in her (because she also inherited a normal X chromosome from her mother).

b) c - carrier, because (as above) she has one normal X chromosome and one X chromosome with the hemophilia trait. She is not pure-breeding because her X chromosomes are not identical; she is not heterozygous because she has two X chromosomes; dihybrid refers to two characteristics.

c) d - 1/4, because the cross is XXn by XY; draw the Punnett square and see

d) a - 0, because daughters get an Z chromosome from both the mother and the father, even if they get her Xn, they will also get a normal X from the father, and thus will not exhibit the recessive trait.

e) e - 1/8, because the cross is XXnOO by XYAB and only one of the possible offspring would have type A and have hemophilia (of all the offsprin, the females would be 2 XX type A and 2 XX type B, while the males would be 1 XY A, 1 XY B, 1 XnYA and 1 XnYB. Again, draw the Punnett square to be sure.

f. c - 50%, as mentioned above, 2 of the 4 possible sons would have hemophilia.

Answer 2

OK, this is going to get complicated, but I *refuse* to just give you the answers without explaining them first.

First, we'll start with blood type. The daughter was type O, her father was a type B. If her father was homozygous type B, there is no way she could be type O. This means her father was heterozygous (BO).

Next, sex-linked traits. Because the X chromosome is usually longer, traits such as this are usually found there. Because males only have one X chromosome, they are significantly more likely to have the trait. Since hemophilia, in this case, is recessive, the male only needs one "n" while the female needs two. Since the father had the disease, he has ONLY the recessive trait, and therefore only has that to give. As a result, we know the daughter (woman having the kids later) is heterozygous XNXn. We know her husband has normal clotting, so he must be XNY.

So we set up a Punnett square:

XN Xn

XN XNXN XNXn

Y XNY XnY

We'll also do one with the blood types:

Io Io

IA IAIo IAIo

IB IBIo IBIo


As for the answers:

1. The father has the disease and type BO blood (D)
2. The woman has one dominant and one recessive allele: she is a carrier (C)
3. They have a 25% chance (D)
4. None of their daugthers will have the disease, however one will be a carrier (A)
5. There is a 50% chance their son will have it, plus a 50% chance he will be type A. .50*.50 = 25% chance (D)
6. According to probability, she has a 50% chance of the son having the trait.


Hope that helps!

Answer 3

you do not say if it incredibly is intercourse-linked, so we''ll anticipate it incredibly is hassle-free Mendelian genetics, then it could be D for dominant, greater effective digits. And d for the first # of digits. Dad could desire to be Dd or DD. mom is dd. little ones could desire to be Dd : dd, 2/4 greater effective, 2/4 wide-spread, for the 1st go. little ones could be all Dd, 4/4 greater effective for the 2d go. however the daughter is wide-spread, dd. So, Dad is Dd. So it incredibly is the 1st one, Dd:Dd, the place there's a 50/50 hazard.