Genetics Question

[marcusnasland]

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In guinea pigs short hair (L) is dominant to long hair (l), coat colour is controlled by a separate locus C such that homozygous Cy produces yellow coat, homozygous Cw produces white coat and the heterozygous CyCw produces a cream colour. A short-haired, cream-coloured guinea pig is bred to a long-haired, white guinea pig to produce several litters. if at least one of the litters includes a long-haired, cream guinea pig what overall phenotypic proportions would be expected in all litters?

 

[Maygyver]

Well, the cross would be llCwCw x LlCyCw. I am too tired to plug in and find the actual probabilities but you should be able to get it from there.

 

[FutureScaresMe]

because the problem said at least one offspring in the F1 generation has long hair, you know the parents each must carry a recessive allele for hair length, therefore you get the cross:

LlC^WC^Y x llC^WC^W

This is dihybrid test cross with one locus showing incomplete dominance, giving a characteristic genotypic and phenotypic ratio of 1:1:1:1 . Even if the color locus displayed complete dominance you would find the same ratios. If you have a problem proving it, post here.

 

[TheBoondocks]

In guinea pigs short hair (L) is dominant to long hair (l), coat colour is controlled by a separate locus C such that homozygous Cy produces yellow coat, homozygous Cw produces white coat and the heterozygous CyCw produces a cream colour. A short-haired, cream-coloured guinea pig is bred to a long-haired, white guinea pig to produce several litters. if at least one of the litters includes a long-haired, cream guinea pig what overall phenotypic proportions would be expected in all litters?

Since you have a long haired. This means that the short hair parent is heterozygous. With this in mind: short haired-cream= Ll CwCy and long haired, white = ll CwCw

The easy way is to separate the colors. 50%= long and 50%= short

cream = 50% White=50%
long and cream=.5*.5= .25
long and white= .5*.5=.25
short and white= .5*.5=.25
short and cream= .5*.5=.25
1:1:1:1

As the person above me said, when you have a dihibrid cross that involves incomplete dominance, that leads to 1:1:1:1 in this case. If it were two creams, then this wouldn't be the case. Ex

the long and white is the same, however, the colors aren't.
CwCy vs. CwCy.

.25 CWCW .5 CwCy .25 Cy Then you take these time .5 for long and short so you wouldn't get 1:1:1:1. The key is that you have a double heterozyous CwCy Ll cross with a double homozygous LL CwCw.

HTH

 

[marcusnasland]

Ok that makes sense... thanks alot! I have one more that i think i figured out the answer but i had to write it all out, just wondering if there is a simple way to think about this and i guess check if my answer is actually right.

If a tetraploid plant has 8 chromosomes (i.e. 4n=8)

how many distinct ways can these chromosomes be divided at metaphase 1 of meiosis? (assuming only bivalents are being formed)

I got 36 different ways.

 

[TheBoondocks]

Ok that makes sense... thanks alot! I have one more that i think i figured out the answer but i had to write it all out, just wondering if there is a simple way to think about this and i guess check if my answer is actually right.

If a tetraploid plant has 8 chromosomes (i.e. 4n=8)

how many distinct ways can these chromosomes be divided at metaphase 1 of meiosis? (assuming only bivalents are being formed)

I got 36 different ways.

I also got 36. It was tricky. The key here is 4n=8. This means you have 2 sets of homologous chromosomes.

A, B, C, D are homologous. The number of ways of is 4*3=12. However, half of these are repeats, so it's actually 6. Now, once you choose a pair, say AB, the other pair has to be CD. So, it's 6*1=6. Now, the same is true for the other homologous pair. so 6^2=36.

HTH