Question 2 (11 points) In the plant genus Triticum there are many different poly

Question

Question 2 (11 points)

In the plant genus Triticum there are many different polyploid species, as well as diploid species. Crosses were made between three different species, and hybrids were obtained. The meiotic pairing was observed in each hybrid, and this is recorded in the following table. (Hint#1: A bivalent is two homologous chromosomes paired at meiosis, and a univalent is an unpaired homolog at meiosis. HINT #2 : For this question, review your knowledge of chromosome inheritance and pairing in meiosis along with what you learned about auto- and allopolyploids in this module. For review, see pages 591-599 in the textbook.)

Species crossed to make hybrid

Pairing in hybrid F1 meiosis

Cross 1. T. durts × T. mono

7 bivalents  + 7 univalents

Cross 2. T. tivum × T. mono

7 bivalents  + 14 univalents

Cross 3. T. tivum × T. durts

14 bivalents  + 7 univalents

Using the information gathered from the crosses above, deduce the F1 chromosome number (ploidy) from each cross.

Using the information gathered from the crosses above, deduce the somatic chromosome number of each parent species used.

State which species are polyploid and whether they are autopolyploids or allopolyploids.

Describe how the observed chromosome pairing pattern arose in the the F1 of Cross 1.0.

Species crossed to make hybrid

Pairing in hybrid F1 meiosis

Cross 1. T. durts × T. mono

7 bivalents  + 7 univalents

Cross 2. T. tivum × T. mono

7 bivalents  + 14 univalents

Cross 3. T. tivum × T. durts

14 bivalents  + 7 univalents

Explanation / Answer

The total number of chromosomes in T. turdigum x T. monococcum cross (7 bivalents + 7 univalents) = 21

The Total no of chromosomes in 2. T. aestivum  x T. monococcum cross (7 bivalents + 14 univalents) = 28

The Total no of chromosomes in 3. T. aestivum  x T. turdigum cross (14 bivalents + 7 univalents) = 35

From the above information, the following can be deduced:

* Number of somatic chromosomes in T. turdigum = 14

* No of somatic chromosome in T. monococcum = 7

  * No of somatic chromosome in T. aestivum = 21

b.T. turdigum is autopolyploids and T. aestivum is allopolyploid.

c.

Cross 1. T. turdigum chromosomes forms the bivalent and T. monococcum chromosomes forms the univalent

Cross 2: Since T. aestivum is a allopolyploid, the bivalent is formed between the two species and univalent is also formd from the chromosomes of the two species

Cross 3: The bivalents are formed from the autopolyploids of T. turdigum as well as the from the chromosomes between the two species, the univalent is formed from the chromosomes of T. aestivum

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