Question

Two genes are hypothesized to be linked. The map distance between the genes is thought to be 12 map units. Out of 1000 offspring produced by two heterozygous parents, predict how many would occur in each phenotypic category.

Answer 1

In a cross with heterozygous parents for two linked genes 12 map units apart, we expect 12% of 1000 offspring to be recombinant (120 offspring), and 88% to show the parental phenotype (880 offspring).

Step-by-step explanation:

Genetic Linkage and Recombinant Offspring

When assessing two linked genes with a map distance of 12 map units, we can predict the frequency of the phenotypic categories in offspring. In this scenario, 12 map units signify a 12% recombination frequency between the two genes. Therefore, in a dihybrid cross with heterozygous parents, 12% of the offspring are expected to exhibit recombinant phenotypes (nonparental), while 88% will display the parental genotypes.

Given 1000 offspring, we predict 120 will be recombinants, and 880 will have the parental phenotype. The recombinant offspring occur due to crossover events between the linked genes during meiosis. The number of recombinants helps geneticists create linkage maps to determine the relative positions of genes on chromosomes.

Answer 2

Approximately the number of individuals with each phenotype is:

• RS-->0.75 x 1000 = 750
• rs--> 0.19 x 1000 = 190
• Rs--> 0.05 x 1000 = 50
• rS--> 0.05 x 1000 = 50

Step-by-step explanation:

Available data:

• Two genes are hypothesized to be linked
• The map distance between the genes is thought to be 12 map units.
• Two heterozygous parents
• 1000 offspring produced

Let us say that the two diallelic genes are R (alleles R and r) and S (alleles S and s)

Heterozygous parents might have the following genotype RS / rs

We know that there are 12mp between genes, which means that the recombination frequency equals 0.12 or 12%.

0.12 of the gametes are recombinants and 0.88 are parentals. There are two types of parentals and two types of recombinant, so 0.06 + 0.06 are recombinants while 0.44 + 0.44 are parentals. This is:

Cross:

Parentals) RS/rs x RS/rs

Gametes) RS (parental) 0.44

rs (parental) 0.44

Rs (Recombinant) 0.06

rS (Recombinant) 0.06

Punnet square) RS rs Rs rS

RS RS/RS RS/rs RS/Rs RS/rS

rs RS/rs rs/rs Rs/rs rS/rs

Rs RS/Rs rs/Rs Rs/Rs rS/Rs

rS RS/rS rs/rS Rs/rS rS/rS

F1) Genotypes:

• 1 RS/RS --> 0.44x 0.44 = 0.1936

• 2 RS/rs --> 0.44x 0.44 = 0.1936 x2 = 0.3872
• 2 RS/Rs --> 0.44 x 0.06 = 0.264 x 2 = 0.053
• 2 RS/rS --> 0.44 x 0.06 = 0.264 x 2 = 0.053
• 1 rs/rs --> 0.44x 0.44 = 0.1936
• 2 rs/Rs --> 0.44 x 0.06 = 0.264 x 2 = 0.053
• 2 rS/rs --> 0.44 x 0.06 = 0.264 x 2 = 0.053
• 1 rS/rS --> 0.06 x 0.06 = 0.0036
• 1 Rs/Rs --> 0.06 x 0.06 = 0.0036
• 2 Rs/rS --> 0.06 x 0.06 = 0.0036 x 2 = 0.0072

Phenotypes:

• RS-->0.7506 (RS/RS + RS/rs + RS/Rs + RS/rS + Rs/rS)
• rs--> 0.1936 (rs/rs)
• Rs--> 0.0566 (rs/Rs + Rs/Rs)
• rS--> 0.0566 (rS/rS + rS/rs)

Now, to know how many individuals would occur in each phenotypic category, you just need to multiply it by 1000, that is the total number of individuals

• RS-->0.75 x 1000 = 750
• rs--> 0.19 x 1000 = 190
• Rs--> 0.05 x 1000 = 50
• rS--> 0.05 x 1000 = 50