Which kinds of cells in a parent could cause a new trait to appear in the parent’s offspring?

What are the 2 stages of solidification, and what occurs during each?

jenyasd209 [6]

Molten material is allowed to solidify into the final shape.

Casting defects occur during solidification:

Gas porosity

Shrinkage

Two stages of solidification:

Nucleation

Growth

4 0

3 years ago

What are the phases in meiosis and explain

densk [106]

Cell division happens twice during meiosis. One starting cell can produce for gametes (eggs or sperm.) In each round of division, cells go through four phases called prophase, metaphase, anaphase, and telophase.

prophase During prophase the chromosomes become visible as paired chromatids and the nuclear envelope disappears. This phase includes reduction division, which is where the number of chromosomes is decreased from 46 (diploid) to 23 (haploid.)

metaphase This is where the 23 remaining chromosomes become attached to the spindle fibers.

anaphase During this phase, the chromosomes move away from each other to one or the other pole of the spindle fiber.

telophase In which the chromatids or chromosomes move to opposite ends of the cell and two nuclei are formed.

interphase This is a resting period.

6 0

2 years ago

What hormones does the thyroid gland produce?

vivado [14]

Thyroid gland produces:

1) T3 (triiodothyronine)
2) T4 (thyroxine).
3) Thyroid stimulating hormone (TSH)

Hope this helps!

4 0

3 years ago

Look at metaphase and anaphase. Suggest the purpose of the spindle fibers during mitosis.

guajiro [1.7K]

Answer:

The spindle fibre is necessary to separate the sister chromatids into opposite sides of the cell.

Please find detailed explanation below

Explanation:

Mitosis is the cell division that produces two daughter cells that are genetically identical to the parent cell. Mitosis occurs in stages viz: prophase, metaphase, anaphase and telophase. The replicated chromosome, aligns at the centre of the cell in the METAPHASE stage and separates into opposite poles in the ANAPHASE stage.

This alignment (at the equator) and separation into opposite pole is possible with the aid of a structure called SPINDLE FIBRES. Spindle is a structure formed from microtubules, which attach to the kinetochores (located in the centromere) of each replicated chromosome to separate them from one another. Hence, the purpose of the spindle fibre during mitotic division is to separate the GENETIC MATERIAL of a parent cell into two daughter cells.

6 0

3 years ago

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

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

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 these genes are linked.

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