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Flura [38]
3 years ago
15

Colorblindness is a recessive, X chromosome sex-link disorder. cross the parents to figure out which percentage of their offspri

ng will be colorblind.

Biology
1 answer:
o-na [289]3 years ago
8 0

Answer:

50%

Explanation:

XBXb is not colourblind as b is recessive.

YXB is not colourblind

YXb is colourblind

XbXb is colourblind

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What are wildfires? (With detail)
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7 0
2 years ago
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the diploid number of chromosomes in horses is 64. what is the haploid number of chromosomes in horses?
OlgaM077 [116]
The haploid number of the chromosomes should be 32. 

Diploid number means that they have the complete sets of chromosomes in their cells, these cells are usually found in somatic (body) cells and different organisms have a different number. For example, a human somatic cell has 46 chromosomes, such as a muscle cell, or a skin cell etc. 

Meanwhile, haploid number means that the number of chromosomes in the cell only have half of the chromosome number than that of the diploid cells. These haploid cells are usually found in gametes of sexually reproducing organisms, such as human, we have 23 chromosomes in our sex cells. This is important because we have to make sure the chromosome number of offsprings are not doubled, as during sexual reproduction, male and female gametes fuse together to form a zygote. 

Therefore, to calculate the haploid number of a cell, we can divide the diploid number by 2, which is 64/2, and the answer would be 32. 
7 0
3 years ago
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Consider a cross between a plant with two flower colors, yellow and red. Yellow (Y) exhibits complete dominance over red (y). It
Aneli [31]

Answer:

1) option a is correct. (260+270)/1545

<em>(Note about this option: In the statement, it is written as </em><em>(260+270) 1545. </em><em>Lacks the division symbol, </em><em>/ </em><em>)</em>

2) From these results one can conclude that the two genes are linked (option a)

Explanation:

To know if two genes are linked, we must observe the progeny distribution. If heterozygous individuals, whose genes assort independently, are test crossed, they produce a progeny with equal phenotypic frequencies 1:1:1:1. If we observe a different distribution, phenotypes appearing in different proportions, we can assume linked genes in the double heterozygote parent.  

We might verify which are the recombinant gametes produced by the di-hybrid, and we will be able to recognize them by looking at the phenotypes with lower frequencies in the progeny. Phenotypes with the highest frequencies represent the parental gametes.

To calculate the recombination frequency, we will use the following formula: P = Recombinant number / Total of individuals.  

In the present example:

Parental)        YySs                        x                 yyss

Gametes)  YS parental type                        ys, ys, ys, ys

                 ys parental type

                 Ys recombinant type

                 yS recombinante type

The total number of individuals in the offspring: 1545

Phenotypic class Number of offspring  

  • 500 Y-S- (parental)
  • 515 yyss (parental)
  • 260 Y-ss (recombinant)
  • 270 yyS- (recombinant)

According to this information, the phenotypic frequencies of the progeny differ from the phenotypic ratio 1:1:1:1. We can assume then, that t<u>hese genes are linked. </u>

The recombination frequency is:

P = Recombinant number / Total of individuals

P = 260 + 270 / 1545

P = 530/1545

P = 0.343

The genetic distance between genes is 0.343 x 100= 34.4 MU.

6 0
3 years ago
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