Question

A diploid cell contains three pairs of homologous chromosomes designated C1 and C2, M1 and M2, and S1 and S2. No crossing over occurs.
a. What combinations of chromosomes are possible in daughter cells following mitosis?
b. What combinations of chromosomes are possible in cells undergoing the first meiotic metaphase?
c. What combinations of chromosomes are possible in haploid cells following both divisions of meiosis?

Answer 1

a) In mitosis, two genetically identical daughter cells are produced. Thus, the only possible combination of chromosomes is C1/C2/M1/M2/S1/S2 since all genetic material is inherited.

b) The first meiotic division serves to split homologous pairs into two different cells, which are now haploid. There are 8 possible combinations of chromosomes (2 chromosomes for each homologous pair). For instance, a cell at this stage could inherit C1/M1/S1, or C2/M1/S1, or C1/M2/S1, etc.

c) The same 8 combinations as above. The second meiotic division splits double-stranded chromosomes of each of these haploid cells into two daughter cells each so that only single-stranded chromosomes remain. Since no crossing over occurs, the two cells descended from each haploid cell will be genetically identical to the cell it came from. For instance, if the first meiotic division produces a cell with the combination C1/M1/S1, the second meiotic division of this cell will produce two more cells with the combination C1/M1/S1. The difference is that the end of meiosis I produces cells with double-stranded chromosomes, while the end of meiosis II produces cells with single-stranded chromosomes.

Ask if you want me to clarify anything!

Answer 2

Say the recessive allele for black is b and the dominant allele for grey is B. A black female must have the genotype bb. A grey male can have the genotype BB or Bb. However, its father was black (genotype bb), meaning it must have inherited a b allele, and it is therefore heterozygous (genotype Bb).

We're dealing with a Bb x bb cross. You can make a Punnett square if you like, but you can see from one parent, you're always getting the b allele. The other parent has a 50/50 chance of passing on either B or b. Therefore, half the offspring will have the Bb genotype (and grey phenotype), and half the offspring will have the bb genotype (and black phenotype).

Answer 3

Haha I appreciate it but I'm happy to help :)

You can usually assume X-linked mutations are recessive (unless stated otherwise), as dominant ones are quickly eliminated from the population. This happens to be a recessive mutation. I'll call the mutant allele x and the wild type allele X.

a) A scalloped female has the genotype xx, and a normal male has the genotype XY. Cross them, and your F1 consists of half Xx and half xY (since the mother can only contribute the recessive allele). The Xx individuals are female carriers of the condition but phenotypically normal, while all male offspring inherit the condition. Notice the male needs only one mutant allele to develop the condition, which is why recessive X-linked diseases affect males much more than females.

The F2 generation results from a cross between Xx and xY. Now, four different genotypes are possible: XY (normal male), Xx (normal, carrier female), xx (scalloped female), and xY (scalloped male).

b) X and Y are sex chromosomes and their inheritance is (obviously) different between males and females. Autosomal chromosomes are those that are not sex-specific; in other words, all individuals, male or female, inherit all autosomal chromosomes.

Instead of X and x, I'll use W and w (for wing). Let's do the first cross again, this time assuming the mutation is on an autosomal chromosome: ww x WW. Note that here I'm assuming the male is homozygous for the W allele as I am assuming the mutation is rare. You can see, then, that all F1 offspring will be heterozygous, Ww, but NONE WILL HAVE A MUTANT PHENOTYPE as we saw above since the normal allele masks the effects of the mutant one.

Now say you want to see what happens for the F2 generation. You do an F1 cross (Ww x Ww). What do you get?

By the way, all of the crosses can be expressed in diagram form by drawing out a Punnett square. The reason I can do these so fast is because I use probabilities instead (which you'll learn about if you take a genetics course in university), but I've been doing it for so long that I can just look at a cross and rearrange the alleles in my head :P

Answer 4

F2 results from a cross between the F1 individuals: a normal female and a scalloped male (Xx crossed with xY). The scalloped male of the F1 generation should not be confused with the normal male of the F2 generation (just reread that part of the question and draw it out to help you clarify what's happening) Now answer my question at the end :P

I think I've given you enough examples to give the new question you've posted a shot. Try it out and show your work - I'm happy to check it over for you after. Remember, draw out the Punnett squares if you need help determining the genotypes of offspring, and keep your notation for the different alleles consistent!

Answer 5

Not quite (maybe you retyped a mistake?). It's a heterozygous cross meaning your offspring genotypes are WW, Ww, Ww, and ww.

You can see that one in four offspring will inherit the condition (regardless of whether they're male or female). This is quite a different result from our F2 generation in which the mutation was X-linked (where half of all offspring inherited the condition).