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4 years ago

The Punnett square shows the possible genotype combinations of two parents who are heterozygous for a particular trait.

Biology

2 answers:

ANTONII [103]4 years ago

5 0

Answer:

25%

Explanation:

The child only comes out as homozygous dominant (YY) for the trait in 1/4 boxes in the Punnett Square. This translates to the child having a 25% chance of being homozygous dominant for the trait.

ANEK [815]4 years ago

5 0

Answer:

Homozygous is when the two letters are the same and dominance is represented by capital letters. There is one square that has two capital letter so 1/4 is equal to 25%. Hope this helps

Explanation:

B:)

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Answer:

Immediately following the arrival of the stimulus at the axon terminal there is a short period called the latent period.

Explanation:

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3 years ago

Why are bullets shot into a barrel of water for testing?

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Answer:

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3 years ago

In Drosophila, the genes for withered wings (whd), smooth abdomen (sm) and speck body (sp) are located on chromosome 2 and are s

lesya692 [45]

Answer:

A) 47; B) 33; C) 272; D) 122

Explanation:

The three genes are linked.

The female with withered wings and a smooth abdomen has the genotype whd sm sp+/whd sm sp+.

The male with a speck body has the genotype whd+ sm+ sp/whd+ sm+ sp.

Both individuals are homozygous for all genes, so each of them only produces one type of gamete. The resulting F1 therefore has the genotype whd sm sp+/ whd+ sm+ sp, heterozygous for all genes and with a wild-type phenotype.

The females of the F1 were mated with homozygous recessive males (test cross): whd sm sp/whd sm sp.

If we assume interference is 0, the probability of crossing over happening between the genes whd and sm is independent from the probability of crossing over happening between sm and sp.

The distance = frequency of recombination × 100, so the frequency of recombination (RF) between genes whd and sm is 0.305 and the RF between genes sm and sp is 0.155.

<u>The expected double crossover progeny among the 1000 offspring will be:</u>

RF whd-sm × RF sm-sp × 1000 =

0.305 × 0.155 × 1000 = 47 individuals will be double crossover.

Interference is 0.3

The interference is calculated as 1- coefficient of coincidence (cc).

cc = observed double crossover/expected double crossover

Therefore:

I = 1 - cc

cc = 1 - I

Observed DCO / 47 = 0.7

Observed DCO = 0.7 × 47

Observed DCO ≅ 33

The parental gametes are whd sm sp+ and whd+ sm+ sp (the genotype of the F1 female is known).

Looking at them and at the gene map we can tell that the gametes that give rise to withered wings, speck body (whd sm+ sp) and smooth abdomen (whd+ sm sp+) phenotypes are the result of recombination occurring between genes whd and sm.

To calculate the expected number of individuals with those phenotypes among the 1000 progeny we need to determine the frequency of recombination between the genes whd and sm considering there's interference.

The distance whd-sm = RF x 100

The recombination frequency is the sum of the single crossover between whd and sm and the double crossovers.

The frequency of DCO is 33/1000=0.033.

Distance whd-sm/ 100 = SCOwhd-sm + DCO

0.305 - 0.033 = SCO whd-sm

<u>Frequency of SCO whd-sm= 0.272</u>

And the expected number of individuals with those phenotypes will be 0.272 x 1000 = 272.

The gametes that originate the phenotypes withered wings, speck body, smooth abdomen (whd sm sp) and wild type (whd+ sm+ sp+) are the result of recombination between genes sm and sp.

Distance sm-sp/ 100 = SCOsm-sp + DCO

0.155 - 0.033 = SCOsm-sp

<u>Frequency of SCO sm-sp= 0.122</u>

And the expected number of individuals with those phenotypes will be 0.122 x 1000 = 122.

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4 years ago

I need this in as soon as possible, any help?

vovikov84 [41]

Answer:

Explanation:

1. The processes of diffusion in diagram A and B is that one side of the membrane is hypertonic. The other side is hypotonic. Based on this, Diagram A shows that the one space has no solute particles causing the particles to move against gradiation. The movement is shown in Diagram B. This process is used to equalize the solute particles in the solution.

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3 years ago