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Tresset [83]
3 years ago
14

A pure-breeding, tall pea plant with white flowers is crossed with a pure-breeding, short plant with red flowers. The F1 plants

are tall, with red flowers. When allowed to fertilize themselves, these produce the following F2: 326 tall, red; 104 tall, white; 117 short, red; and 29 short, white. You are conducting a Chi-Square analysis and what is the null hypothesis?
Biology
1 answer:
DIA [1.3K]3 years ago
6 0

Answer:

Null Hypothesis -

The observed frequency is approximately equal to the expected frequency of phenotype.

Explanation:

Pure Breeding Cross - TTww x ttWW

Genotype of offspring in F2 generation - TtWw

Null Hypothesis -

The observed frequency is approximately equal to the expected frequency of phenotype.

The chi square analysis is attached

The degree of freedom for this question is 3

The p value for X^2 estimated through chi square test is 0.5

Hence the null hypothesis is accepted.

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Sexual reproduction involves a process called meiosis. What type of cells are produced during this process?
enot [183]

Answer:

haploid

Explanation:

Sexual reproduction involves the production of haploid gametes by meiosis, followed by fertilization and the formation of a diploid zygote.

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Imagine that a newly discovered, recessively inherited disease is expressed only in individuals with type O blood, although the
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Complete question:

Imagine that a newly discovered, recessively inherited disease is expressed only in individuals with type O blood, although the disease and blood group are independently inherited.

A normal man with type A blood and a normal woman with type B blood have already had one child with the disease. The woman is now pregnant for a second time.

Assuming that both parents are heterozygous for the gene that causes the disease, what is the probability that the second child will also have the disease? Express your answer as a fraction using the slash symbol and no spaces (for example, 1/2).

Answer:

The probability that the second child will also have the disease is 1/16.

Explanation:

<u>Available data:</u>

  • Two genes independently inherited: one for blood type, the other for disease
  • Man with type A blood  x  Woman with type B blood
  • Both parents are heterozygous for the gene that causes the disease; Dd

If the man has A blood, and the woman has B blood, and they already have an affected child, this means that they must be heterozygous for blood type too.

Cross:  

Parentals)           AiDd         x      BiDd

Gametes)  AD  Ad  iD  id       BD  Bd  iD  id

Punnett square)      AD       Ad       iD       id

                     BD   ABDD  ABDd  BiDD BiDd

                     Bd   ABDd   ABdd   BiDd Bidd

                     iD    AiDD    AiDd    iiDD   iiDd

                     id    AiDd     Aidd     iiDd    iidd

F1) <u>Genotype</u>:

1/16 ABDD

2/16 ABDd

1/16 ABdd

1/16 AiDD

1/16 BiDD

2/16 AiDd

2/16 BiDd

1/16 Aidd

1/16 Bidd

1/16 iiDD

2/16 iiDd

1/16 iidd

<u>Phenotype:</u>

3/16 A/B normal

4/16 A normal

4/16 B normal

3/16 0 normal

1/16 0 affected by the disease.

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