Question

In the tiny model plant Arabidopsis, the recessive allele hyg confers seed resistance to the drug hygromycin, and her, a recessive allele of a different gene. confers seed resistance to herbicide. A plant that was homozygous hyg/hyg her/her was cross with wild type (hyg+/hyg+ her+/her+), and the F1 was selfed. Seeds resulting from the F1 self were placed on petri dishes containing hygromycin and herbicide. a) If the two genes are unlinked, what percentage of seeds are expected to grow? b) In fact, 13 percent of the seeds grew. Does this percentage support the hypothesis of no linkage? Explain. If not, calculate the number of map units between the loci. c) Under your hypothesis, if the F1 is testcrossed, what proportion of seeds will grow on the medium containing hygromycin, and herbicide? explain every SINGLE step and show ALL WORK . show the crosses and how to calculate everything. do not skip steps.

Answer 1

a) 6.25%; b) Genes are linked. Distance = 28 map units. c) 36%

Step-by-step explanation:

The parental cross was: hyg/hyg her/her X hyg+/hyg+ her+/her+

Resulting F1 : hyg/hyg+ her/her+

a)

The F1 was then selfed. If you notice, it would be a dijhybrid cross between unlinked loci: according to Mendel's law of independent assortment, the expected phenotypes of the F2 would be:

• 9/16 hyg+/_ her+/_
• 3/16 hyg+ /_ her/her
• 3/16 hyg/hyg her+ /_
• 1/16 hyg/hyg her/her

The homozygous recessive genotypes determine seed drug and herbicide resistance. Only plants with genotype hyg/hyg her/her will be able to grow on Petri dishes containing hygromycin and herbicide, so the answer is 1/16×100=6.25%.

b)

If 13% of the seeds had the genotype hyg/hyg her/her, it deviates a lot from the expected 6.25% that we would get if genes assorted independently, so these loci are probably linked.

We have to rewrite the crosses considering that the genes are located on the same chromosome:

The parental cross was: hyg her/hyg her X hyg+ her+/hyg+ her+

Resulting F1 : hyg her/hyg+ her+

The F1 can produce the gametes:

• hyg her Parental
• hyg+ her+ Parental
• hyg her+ Recombinant
• hyg+her Recombinant

13% of the offspring had the genotype hyg her/ hyg her. That means that during the parent's meiosis, the gamete hyg her had to be produced twice and independetly. Because of rules of probability, the probability of two independent events occuring is calculated as the multiplication of the probability of occurrence of each event. Therefore:

p(hyg her) × p(hyg her) = 0.13

p(hyg her)² = 0.13

p(hyg her) =√0.13

p(hyg her) = 0.36

The probability of generating a parental hyg her gamete is 0.36. For that reason, the probability of generating the other parental gamete hyg+ her+ is also 0.36 (adding a total of 0.72 parental gametes). The recombinant gametes will be 1-0.72=0.28, and because there are 2 recombinant gametes each of them will have a frequency of 0.14.

Distance (mu) = Frequency of recombination × 100

Distance = 0.28 × 100

Distance = 28 map units.

c)

As a result of a test cross of the F1, the F2 will have the following expected phenotypic frequencies:

• 36% hyg her
• 36% hyg+ her+
• 14% hyg her+
• 14% hyg+her

So 36% of the seeds will have the genotype hyg her/ hyg her and be resistant to hygromycin and the herbicide.