Genetics practice word problems?

Question

I'm really struggling with this genetics concept, I figured if someone could walk me through the steps of these practice questions, it could help me to learn this. How do I set up the punnett squares for these and ratios?



1) In cats the genotype (BB) is black. Bb is tortoise shell, and bb is yellow. The gene is on the X chromosome. A tortoise shell female is crossed with a black male. What offspring would be expected? Would you normally expect to find any tortoise shell males?

2) Occasionally a tortoise shell male is found, and when found in sterile. Suggest an explanation for the occurrence of tortoise shell males.

3) A couple with a newborn baby are troubled that the child does not appear to resemble either of them. Suspecting that a mix up occurred at the hospital, they check the blood type of the infant. The baby has type (O) blood. Because the father is type (A), and the mother is type (B), they conclude that a mistake must have been made. Are they correct?

Answer 1

1. This is a sex-linked trait since it is on the X-chromosome.
A tortoise female crossed with black male is:
XB Xb ... XB Y
Gametes from female can go on the side of the Punnett Square XB and Xb.
Gametes from the male can go on the top of the Punnett square XB and Y.

First row of boxes on the Punnett square contains XBXB XBY (a black female and a black male).
Second row of boxes contains XbXB and XbY (a tortoise female and a yellow male).

You normally would not expect tortoise males because they would have to have two X chromosomes to be tortoise and a Y to be male: XBXbY.

2. Tortoise shell males can results from nondisjunction in either parent, but it isn't the usual situation.

3. No, those particular blood types don't prove that it isn't their baby. Dad can have genotype AO (or IAi) and Mom can have genotype BO (or IBi) and their baby can get an O allele from each parent.

Answer 2

punnet squares? your teacher lets you do those?

ratios: for these questions its relatively simple--it's simple mendelian genetics. You take the chance of getting a certain gamete from one parent * the chance of getting a certain gamete from another. I'll give a different example at bottom...

1. Bb x BY --males only get one X chromosome
offspring
1/4 BB (black females)
1/4 BY (black males)
1/4 bB (tortoise shell female)
1/4 bY (yellow males)
you wouldn't expect to find tortoise shell males because males only get 1 X chromosome and you need both X^B ad X^b to get tortoise shell)

2. XXY male --> BbY
sterile because he doesn't have an equal number of gametes.

3. not necessarily.
A can result from AA or Ao
B can result from BB or Bo
if parents are Ao and Bo
offspring =
1/4 AB 1/4 Bo 1/4 Ao 1/4 oo
(this is an example of co-dominance and is not explained below)

different example
AA x Aa
1*1/2 = half offspring will be Aa
!*1/2 = half offspring will be AA
if A is dominant to a then all offspring will exhibit the A phenotype.

Aa x Aa
1/2A*1/2A = 1/4 AA
1/2A*1/2a = 1/4 Aa
1/2a*1/2A = 1/4 Aa
1/2a*1/2a = 1/4 aa
but if A is dominant to A then only that phenotype will be shown in the Aa genotypes. therefore the ratio is 3/4 showing A phenotype an 1/4 showing aa phenotype.


AaBb x AaBb
1/2A*1/2a*1/2B*1/2b = 1/16 AaBb
etc. just like above

or if you know this displays simple dominance (3A:1a from above) just use the phenotypes to do the ratio
3/4A*3/4B = 9/16 AB
3/4A*1/4b = 3/16 Ab
1/4 a*3/4B = 3/16 aB
1/4a*1/4b = 1/16 ab
this is where you get the 9:3:3:1 ratio.

punnet squares are time consuming and i don't like them and my teachers forbade that i ever do them

but its the same concept of above you just write it out in a table like the multiplication tables that go
x1 x2 x3 x4 etc
x1 1 2 3 4
x2 2 4 6 8
x3 3 6 9 12
x4 4 8 12 16

so it'd be like
A a
A AA Aa
a Aa aa
where you get 3/4 A and 1/4 a
or
AB Ab aB ab
AB AABB AABb AaBB AaBb
Ab AABb AAbb AaBb Aabb
aB AaBB AaBb aaBB aaBb
ab AaBb Aabb aaBb aabb
and count the phenotypes
you'll get 9 AB 3 Ab 3 aB 1 ab

hope this helped

Answer 3

since the trait is X linked, only a female can express a heterozygous trait. the male can only express the independent traits,i.e. black or yellow.
1) XBXb(tortoise) x XBY

XBXB(black female)
XBXb(tortoise female)
XBY(black male)
XbY(yellow male)
all in the ratio 1:1:1:1

2)the tortoise shell males could be having the genotype XBXbY i.e.
due to non-disjunction of chromosomes during gametic meiosis (formation of ova in females) the ova fusin with the Y sperm has the genotype XBXb instesd of one X chromosome.
the presence of 2 X chromosomes however confers sterility to the 'male' cat.

3)no. a couple with A and B blood groups can have a child with o blood group. o is a recessive trait whereas A and B are codominant traits.
if the father is heterozygote for A and the mother a heterozygote for B thereis i in 4 chances that the offspring will have blood group O.
IAIo(father) x IBIo(mother)

the possible genotypes of the offsprings
IAIB(blood group AB)
IAIo(blood group A)
IBIo(blood group B)
IoIo(blood group O)

however if any of the parent is a homozygous for the particular blood group then the chances of the offspring being an O is nil.
on the whole therefore
there is 1:12 chances that the child willl be of blood group O considering all possibilities.

Answer 4

I think I know how to answer at least few questions: 2.) PP -purple, BB - brown, Pp - purple, Bb - brown, pp - white, bb - blue 3.) HH straight, PP pink, Hh straight, Pp pink, hh curly, pp yellow 7.) white For those answers I'm pretty sure. Hope I helped, since it's quite a long time since my genetics classes ...