STAV 2006 Exam Help!

[chief]

Hey Guys,
Few issues with this paper:

1. Question 7 MC
Albinism is a recessive condition found in humans. If two heterozygous normally pigmented parents had two children, the probability of one being albino and one being pigmented would be:

a) 1/2
b) 3/8
c) 1/16
d) 3/4

I got 3/16.... am i missing something?

usually STAV are pretty good with their solutions, i dont find that they make many errors.

2. Question 3 MC
Mammalian sperm develop from cells known as primary spermatocytes that contain the same number of chromosomes as normal body cells. Thse primary spermatocyes divide to form secondary spermatocytes. A final division results in the formation of sperm. The relative amounts of DNA in each type of cell is shown below:
Primary - 4
Secondary - 2
Sperm Cell - 1

It is reasonable to state that:
a) The division from secondary spermatocytes to sperm cell represents the second meiotic division.  <---- correct answer
b) wrong answer
c) the sperm cell contains a 1/4 of the total number of chromosomes for that organism  <---- my answer
d) wrong answer

where have I gone wrong here?

[minilunchbox]

Before meiosis, the cell undergoes DNA replication so there are 4 chromosomes (2 pairs) so by the end of it, each gamete/sperm cell will have 1 chromosome. But that is not a 1/4 of chromosomes for that organism, but only 1/4 of the original cell. If a sperm cell only had 1/4 number of chrosomes for an entire organism, the organism would be in some trouble.

Meiosis 1 is when the homologous chromosomes sperate and have 1/2 the amount of DNA and meiosis 2 (second meiotic division) is when the chromosomes seperate and the cell becomes haploid with 1/4 the amount of DNA.

[qshyrn]

For your first question..
1st scenario
Child 1 = 3/4 chance normal            Child 2 =1/4 albino
Therefore the probably of this case occuring is (3/4) * (1/4) = 3/16

what you didnt take into account was a second scenario

Scenario 2

Child 1  = 1/4 chance albino            Child 2= 3/4 chance normal
Therefore the probability of this case occuring is (1/4)* (3/4) = 3/16

Now we want to combine the probabilities, the as both these scenarios result in 1 albino 1 normal child.  (3/16)+(3/16) = 3/8
therefore the answer is 3/8

[chief]

that maths is way tooo complicated for biol
I feel like im back to doing methods all over again!

thanks anyhoo

[mmmmm]

For your second question, option D was "the secondary spermatocyte is diploid".
If you start of with a diploid cell, and you replicate the chromosomes, wouldn't there be double the amount of DNA in the cell? So, it divides again, shouldn't it be diploid. Hence, it divides again, making he sperm cell haploid? I can't get my head about this diploid/ haploid stuff. Can someone please explain it to me? Thank you.