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Chromosome Behavior and Gene Linkage Part 4 VoiceThread Transcript

Slide 1
Let's work through these problems to help prepare you for the quiz and exam questions you will see.

This question can best be answered by writing out the genotypes of the parents and then creating a Punnett square.


Slide 2
The heterozygous parent can make four different gamete types and the homozygous parent makes one gamete type. If the offspring occur in a 1:1:1:1 ratio - the genes are not linked and they assort independently.


Slide 3
If the genes are assorting independently, the genes are not linked and they may or may not be on the same chromosome. Remember for genes to be linked they have to be on the same chromosome AND closer than 50 map units. So unlinked genes can either be on different chromosomes or on the same chromosome but further than 50 map units apart.

Slide 4
To answer this question - you need to write out the genotypes of the parents and think about the way the genes are linked. To be completely linked means that they are so close together on the chromosome they crossing over never recombines them - so they stay together.


Slide 5
The original parents are CCdd and ccDD so this means parent 1 has chromosomes with alleles C and d and parent two has chromosomes with alleles c and D. Each parent can only make one gamete type.

Their offspring will all have the genotype CcDd - but you need to pay attention to the arrangement of alleles - because the genes are completely linked C stays with d and c stays with D - no crossing over occurs. When one of these offspring is crossed with a homozygous recessive individual - there are only two possible offspring that can result - Ccdd or ccDd.

Slide 6
So the answer to this question is B - two.

Slide 7
This question asks you to determine the distance between the genes for bristle structure and body color. The first thing you need to determine is which offspring are the recombinants.


Slide 8
Remember, recombinant means the offspring that were created as a result of a crossover between the two genes. By definition, the recombinants are the two smallest categories of offspring and the parentals are the two largest categories of offspring.


Slide 9
So the offspring with wildtype bristles, wildtype body and with scute bristles, ebony body are the recombinants. This means that the parental arrangement of alleles is s+ eb and s eb+. Crossing over created the s+ eb+ and s eb gametes.


Slide 10

So the answer to this question is B - 38 map units. 192 + 188 = 380/1000 x 100 = 38% recombination or 38 map units.

Slide 11
This question asks you to determine the percent of gametes the female parent made that have the genotype m eb+. To answer this question - draw out the females chromosomes and subsequent gametes.

Slide 12
You know that the female is heterozygous - but you do not know the specific arrangement of alleles on her chromosomes until you look at her offspring. She can either have both wild-type alleles on one chromosome and both mutants on the other OR she can have one wild-type and one mutant on one chromosome and the other combination of mutant and wild-type on the other chromosome.

You determine her arrangement by looking at the offspring and determining the parental arrangement. In this example, the Wild type eye color, wild type body color and the mahogany eyes, ebony body are the parental offspring. Therefore, arrangement one is her arrangement of alleles.

Her gametes that are m eb+ resulted in offspring that have the phenotype mahogany eyes and wild-type body. This category has 176 offspring out of a total of 1000 offspring.

Slide 13
Therefore the answer is A - 17.6%.


If you have questions about any of these practice problems - please let me know.

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