50. The following table summarizes the result of a cross between two strains of Neurospora having the alleles D and d, respectively. The table shows the different patterns of octad arrangement and the number of ascus observed of each type.

Octads
D	d	D	d	D	d
D	d	D	d	D	d
D	d	d	D	d	D
D	d	d	D	d	D
d	D	D	d	d	D
d	D	D	d	d	D
d	D	d	D	D	d
d	D	d	D	D	d
115	125	14	16	17	13

Number of ascus observed

Based on the above, fill in the blanks from the options given below.
“The first two columns are from meiosis with nocrossover between locus D and __[A]__. The patternfor these two columns represent __[B]__ segregation pattern. The distance between the locus D and thecentromere is __[C]__ map units”.

A                         B                 C
(1) d allele          first division         10
(2) Centromere   first division          10
(3) d allele         second division      20
(4) Centromere  second division      10

The correct answer is option (4):

• A = centromere

• B = second division segregation

• C = 10 map units

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Understanding the Neurospora octad question

The cross is between two Neurospora strains with alleles D and d, producing ordered octads (8-spored asci) in different D/d arrangements with known frequencies. Ordered tetrads in Neurospora allow direct observation of how alleles segregate during meiosis I and II, so patterns of D and d in the octad tell whether there was a crossover between the gene D and its centromere.

In these problems, any 4D:4d straight block pattern (DDDDdddd or ddddDDDD when read along the ascus) represents first division segregation (FDS) with no crossover between the locus and the centromere. In contrast, patterns where D and d are intermingled (2:4:2 or 2:2:2:2 type) are second division segregation (SDS) asci and indicate at least one crossover between the gene and the centromere.

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Step 1: Identify what the first two columns represent (A and B)

The question statement says: “The first two columns are from meiosis with no crossover between locus D and A.” This means the partner of locus D in “no crossover” asci must be the centromere, because crossing over between a gene and its centromere is what changes a 4:4 pattern into SDS patterns. Therefore A must be “centromere,” not the “d allele” (which is simply the alternate allele at the same locus).

• Why not “d allele”? With no crossover between D and d, you still see segregation of alleles at meiosis I; the term “no crossover between locus D and d allele” is biologically incorrect, because D and d are at the same locus and cannot be separated by a chiasma.

• Correct: “no crossover between locus D and the centromere” gives classic FDS 4:4 patterns.

The question then states: “The pattern for these two columns represent B segregation pattern.” Because these columns have 4 D spores followed by 4 d spores (or the reverse), they are first division segregation asci.

• First division segregation: alleles separate at anaphase I, giving a clean 4:4 block of D and d.

• Second division segregation: alleles separate at anaphase II after a crossover, giving mixed patterns like 2:4:2 or 2:2:2:2.

Thus B must be “first division segregation.”

From options:

• (1) d allele – first division – 10

• (2) centromere – first division – 10

• (3) d allele – second division – 20

• (4) centromere – second division – 10

Only option (2) has the correct A and B so far, but C must also match the distance calculation, so we compute that next.

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Step 2: Classify first vs second division asci and calculate distance

The table’s bottom row shows the counts of each octad type: 115, 125, 14, 16, 17, 13 (total asci = 115 + 125 + 14 + 16 + 17 + 13 = 300). The first two high-frequency patterns (115 and 125 asci) are 4:4 FDS patterns. The remaining four rarer patterns (14, 16, 17, 13) are SDS patterns because D and d are interspersed.

• Total FDS asci = 115 + 125 = 240.

• Total SDS asci = 14 + 16 + 17 + 13 = 60.

The proportion of SDS asci = 60 / 300 = 0.20, i.e., 20%. For a single gene vs centromere, gene–centromere distance in map units is given by:

• Distance (cM) = ½ × (% second-division segregation) = ½ × 20 = 10 map units.

This “½” factor is used because a single crossover between gene and centromere makes half of the chromatids recombinant; thus only half the spores in each SDS ascus represent recombination.

Therefore C = 10 map units.

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Step 3: Matching with the given options

Now compare the derived answers with the options given:

• (1) A = d allele, B = first division, C = 10 → A is biologically wrong (crossover is defined between gene and centromere, not between alleles of same locus).

• (2) A = centromere, B = first division, C = 10 → A and C are correct, but B is wrong because the pattern being counted for distance is SDS, not FDS; distance is always calculated from second division segregation asci.

• (3) A = d allele, B = second division, C = 20 → A again incorrect, and 20 map units ignores the ½ factor and thus overestimates distance.

• (4) A = centromere, B = second division, C = 10 → correctly states: no crossover between locus D and centromere for FDS, uses SDS pattern for segregation type, and gives the correct 10 map unit gene–centromere distance.

Hence option (4) is the only fully correct choice.

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Key concept recap for exam use

• First division segregation (FDS): 4:4 block pattern; indicates no crossover between the gene and centromere.

• Second division segregation (SDS): mixed 2:4:2 or 2:2:2:2 patterns; indicates at least one crossover between gene and centromere.

• Gene–centromere distance (in cM or map units) = ½ × (% SDS asci).