A geneticist is mapping the chromosomes of the newly captured gremlin. Stripe is heterozygous for three linked genes with allele

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A geneticist is mapping the chromosomes of the newly captured gremlin. Stripe is heterozygous for three linked genes with allele

s Ee, Hh, and Bb, that determine if gremlins are evil (E), have hair (H), and biting teeth (B). In order to determine if the 3 genes were linked, a standard testcross was done, and the 1000 offspring had the following genotypes: 48 ee Hh bb 36 ee hh Bb 400 ee Hh Bb 4 Ee Hh Bb 426 Ee hh bb 46 Ee hh Bb 38 Ee Hh bb 2 ee hh bbChoose the alleles on the homologous chromosomes of the heterozygous individuals with the correct gene order.What is the recombination frequency between genes E and B? a. 8.4% b. 9.0% c. 42.6% d. 9.4% e. 10.0%

Biology

2 answers:

igomit [66] · Answer 1 · 2022-07-22T10:39:12+03:00

Explanation:

A geneticist is mapping the chromosomes of the newly captured gremlin. Stripe is heterozygous for three linked genes with alleles Ee, Hh, and Bb, that determine if gremlins are evil (E), have hair (H), and biting teeth (B). In order to determine if the 3 genes were linked, a standard testcross was done, and the 1000 offspring ...

belka [17] · Answer 2 · 2022-07-22T12:53:48+03:00

Answer:

The correct gene order is H-e-B

Recombination frequency between genes E and B = 10%

Explanation:

In linkage parental genotype is always in greater number than recombinants, hence:

400 = eeHhBb : Parental

426 = Eehhbb : Parental

Double crossovers have the smallest number hence:

EeHhBb = 4 : Double recombinant

eehhbb = 2 : Double recombinant

Double cross over places middle allele from one sister chromatid to other one. Hence a non parental allele of the middle gene is present with parental alleles of the flanking genes.

Here recombinant type is EHB and parental type is eHB . Only gene order H-e-B would result in double cross over which would give EHB and ehb double recombinants. Hence, the correct gene order is H-e-B.

Single crossover between gene E and B would result in Heb and hEB progeny ( 48 and 46 )

Recombination frequency between E and B = [(Double recombinants + Single E and B recombinants) / Total progeny)*100]

= [ (4 + 2 + 48 + 46)/1000] * 100

= 100/1000

= 0.1 * 100 = 10%