Question

In mice, genes for albinism (a), chronic, inflammation (i), and reduced sense of smell (r) are linked. You initially crossed a mouse that is homozygous dominant for all three traits with a mouse that is homozygous recessive for all three traits. You then crossed several of the resulting F_1 offspring to mice that are homozygous recessive for all three traits. You observe the following phenotypes in the offspring: Diagram a map between the genes and calculate the map distances. If the interference of these genes is 0.43, how many more F_2 progeny will you need to observe to see at least 2 of each double recombinant? What are your expected numbers of each single crossover after you have observed the total number of mice calculated in part (b)? If a fourth gene (z) exhibits 54% recombination with gene i, 35% recombination with gene a and 51% recombination with gene r, will this gene be on the same chromosome or a separate chromosome/If gene z is on the same chromosome as the other three genes, indicate its location by copying and modifying your genetic map from part (a).

Answer 1

Answer:

In this question, it has been mentioned that initially homozygous dominant mice for all the three linked genes (present on the same chromosome) was crossed with a homozygous reccessive mice. In the progeny F₁ generation, all the mice that will be born will be heterozygous dominant for all the three traits.

This in turn was followed by the crossing of heterozygous dominant F₁ mice with another mice which was homozygous reccessive for all the three traits. This cross gave rise to a different types of mice having different phenotypes and genotypic constitutions as well. The results thus obtained can be tabulated as follows.

Phenotype	Genotype	Numbers of progeny
Colored healthy normal smell sense	ARL/arl	22
Albino chronically inflamed reduced smell sense	arl/ arl	23
Colored chronically inflamed reduced smell sense	Arl/ arl	5
Albino healthy normal sense	aRL/ arl	6
Colored healthy reduced smell sense	ArL/ arl	0
Albino chronically inflamed normal smell sense	aRl/ arl	1
Colored chronically inflamed normal smell sense	ARl/ arl	6
Albino healthy reduced smell sense	arL/ arl	7

The double-recombinant gametes will be the least frequent type. In this case it implies to '0' & '1' number of progeny. Based on the phenotype the genotypic constitution (arrangement of genes) for both these progeny types can also be inferred and can also be used to elucidate for other types as well.

The two most-frequent types of gametes are nonrecombinant. These provide the linkage phase (cis vs. trans) of the alleles in the heterozygous parent. In this case, it represents the first two observations i.e, 22 & 23.

For further clarification another table can be set up showing the points of crossing overs giving rise to the progeny as under:-

Phenotype	Genotype	Numbers of progeny	Crossover types
Colored healthy normal smell sense	ARL/arl	22	Non-recombinant
Albino chronically inflamed reduced smell sense	arl/ arl	23	Non-recombinant
Colored chronically inflamed reduced smell sense	Arl/ arl	5	Cross over between A & R
Albino healthy normal sense	aRL/ arl	6	Cross over between A & R
Colored healthy reduced smell sense	ArL/ arl	0	Double cross over
Albino chronically inflamed normal smell sense	aRl/ arl	1	Double cross over
Colored chronically inflamed normal smell sense	ARl/ arl	6	Cross over between R & L
Albino healthy reduced smell sense	arL/arl	7	Cross over between R & L

Total number of progeny= 70

Recombination frequency between A & R= (5+6+0+1)/ 70 = 0.171 or 17.1%

Map distance= 17.1 map units

Recombination frequency between R & L = (0+1+6+7)/70 = 0.2 or 20%

Map distance = 20 map units.

Now interference= 0.43

Probability of recombination between A & R = 0.171 & between R & L = 0.2

Using product rule, the amount of detectable double crossover will be = 0.171*0.2 = 0.0342

we know, interference= 1- coefficient of interference

So, coefficient of interference =1- 0.43 = 0.57

coefficient of interference= observed double crossover/ expected crossover

so, Expected cross over = 4/0.57=7.017 approx. 7 is required for observing 4 double crossover progeny(2 for each)

So, No. of events required for 7 expected cross overs= 7/ 0.0342 = 205

In total, 205 progeny is required for observing 4 double crossover progeny.

Genetic map-

20mu 17mu

c) Probability of single cross-overs between A & R= (5+6)/70 = 0.157

Therefore, no. of single crossover events between A & R for 205 mice = 205* 0.157 = 32.185 nearly 32

Probability of single cross-overs between R & L = (6+7)/70 = 0.185

Therefore, no. of single cross over events between R & L for 205 mice = 37.925 nearly 38

d) It is mentioned that the new gene has more than 50% recombination frequency with the genes L & R,

Linkage is said to be present when the recombination frequency between two genes is less than 50% or equal to 50%. A recombination frequency of more than 50% between two genes indicates that the two genes are located on two different non- homologous chromosomes or are located on the same chromosome but far apart from one another giving rise to a occurence of independent assortment between the genes.

In case, the new gene z is located on the same chromosome as the other three,

Recombination frequency between Z & L = 54%

Map distance between Z & L = 54 mu

Recombination frequency between Z & A = 35%

Map distance = 35 mu

Recombination frequency between Z & R= 51%

Map distance = 51 mu

Based on these map distance calculations, the genetic map which satisfies these criteria is:

35mu 16mu 3mu