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Answer all the question 1000% crctly , keep in min...

創建於：2025年7月5日

使用 O4 Mini High 回答 Chat01

問題

20250705_084431.jpg20250705_084431.jpg
20250705_084457.jpg20250705_084457.jpg

Answer all the question 1000% crctly , keep in mind this test need a bacloran level to be done

Python

python
from PIL import Image
import matplotlib.pyplot as plt

img = Image.open('/mnt/data/20250705_084431.jpg')
plt.figure(figsize=(4,4))
plt.imshow(img)
plt.axis('off')

(-0.5, 719.5, 347.5, -0.5)

Python

python
plt.figure(figsize=(4,4))
plt.imshow(img.crop((400,0,800,350)))
plt.axis('off')

(-0.5, 399.5, 349.5, -0.5)

Python

python
plt.figure(figsize=(4,4))
plt.imshow(img.crop((200,150,800,600)))
plt.axis('off')

(-0.5, 599.5, 449.5, -0.5)

Python

python
# manually find approximate coordinates for map
plt.figure(figsize=(4,4))
plt.imshow(img.crop((250,80,900,400)))
plt.axis('off')

(-0.5, 649.5, 319.5, -0.5)

Python

python
# Let's manually display center image region
plt.figure(figsize=(5,2))
plt.imshow(img.crop((200,0,900,200)))
plt.axis('off')

(-0.5, 699.5, 199.5, -0.5)

Python

python
plt.figure(figsize=(5,3))
plt.imshow(img.crop((200,0,850,300)))
plt.axis('off')

(-0.5, 649.5, 299.5, -0.5)

Python

python
img2 = Image.open('/mnt/data/20250705_084457.jpg')
plt.figure(figsize=(4,3))
plt.imshow(img2)
plt.axis('off')

(-0.5, 719.5, 230.5, -0.5)

Python

python
plt.figure(figsize=(4,3))
plt.imshow(img2.crop((200,100,1000,450)))
plt.axis('off')

(-0.5, 799.5, 349.5, -0.5)

Python

python
plt.figure(figsize=(6,4))
plt.imshow(img2.crop((0,200,1100,700)))
plt.axis('off')

(-0.5, 1099.5, 499.5, -0.5)

Python

python
plt.figure(figsize=(5,5))
plt.imshow(img2.crop((200,0,1000,400)))
plt.axis('off')

(-0.5, 799.5, 399.5, -0.5)

Python

python
plt.figure(figsize=(5,3))
plt.imshow(img2.crop((0,100,800,350)))
plt.axis('off')

(-0.5, 799.5, 249.5, -0.5)

Python

python
plt.figure(figsize=(5,2))
plt.imshow(img2.crop((0,150,900,430)))
plt.axis('off')

(-0.5, 899.5, 279.5, -0.5)

思考

Assessing the data

Data 3 isn't provided, so we can't answer Q2. It's possible the user only wanted Q1 answered, though they did say to answer all questions. The images only show two pages, and they may provide Document 3 later. The second image seems to be cropped, with a table cut off, but the bottom part just includes question 2. There’s no Document 3. I’ll refocus on the content from the current images and proceed with Q1.

Analyzing possible crosses

It looks like Data 3 was mislabeled, and there's no Data 3. For Q2, the focus is on Data 1 and Data 2, but Data 2 doesn't show a cross that combines both traits of hair and shape. The genes for these traits are on different chromosomes, so they should be independent. A cross between two heterozygous plants (HhOo x HhOo) would give 3/16 of plants with both traits. For maximizing the results, a different cross might be better, like hhOO x HhOo. I'll suggest this for simplicity. For Q1, I'll show full Punnett squares.

Proposing cross to maximize yield

The user seems to be focused on maximizing trait yield through crossing, particularly focusing on homozygous recessive (hh) for hair and OO for shape genes. For maximizing the yield, crossing two pure lines, such as hhOO x hhOO, will produce F1 progeny with 100% for both traits (hairy and round). However, to ensure both genes are passed, it's crucial to select plants with the desired phenotypes and genotypes. I’ll provide the most effective strategy based on this understanding.

Maximizing trait yield through cross

To maximize the yield for hairy stems (hh) and round fruits (OO), the best approach is to cross two homozygous plants: one for hairless stems (hhOO) and one for round fruits (hhOO). This cross will give 100% progeny with desired traits. If F1 heterozygotes are used, the yield would be lower. I’ll use the relevant Punnett squares to show results and explain that genes D, H segregate independently, and D, O are linked with recombination frequency at 21%.

Thought for 2m 16s

回答

1. Interprétation chromosomique des croisements 1 et 2

Rappel des données (Doc 1)

  • D/d (longueur) et O/o (forme du fruit) sont portés sur le chromosome 1, distants de 21 cM.
  • H/h (poils sur la tige) est porté sur le chromosome 7, à 21 cM du gène couleur (non utilisé ici).

Croisement 1 : [D H] × [d h]

  • Phénotypes des parents : géant, sans poils (parent 1) × nain, avec poils (testeur).
  • Situation chromosomique : D et H sont sur deux chromosomes différents ⇒ ségrégation indépendante.
Gamètes P1 (hétéro en couplage [DH]/[dh])↓ × P2 (testeur dh)DescendancePhénotype
DH (25 %)DH / dhParentaux[D,H] (géant, sans poils)
dh (25 %)dh / dhParentaux[d,h] (nain, avec poils)
Dh (25 %)Dh / dhRecombiné[D,h] (géant, avec poils)
dH (25 %)dH / dhRecombiné[d,H] (nain, sans poils)
  • Bilan : 50 % parentaux / 50 % recombinés → loi d’assortiment indépendant (gènes non liés).

Croisement 2 : [D O] × [d o]

  • Phénotypes des parents : géant, fruits ronds (parent 1) × nain, fruits ovales (testeur).
  • Situation chromosomique : D et O sont liés sur le même chromosome (1) à 21 cM ⇒ recomb. = 21 %.
Gamètes P1 ([DO]/[do])↓ × P2 (testeur do)DescendancePhénotype
DO (39 %)DO / doParentaux[D,O] (géant, rond)
do (39 %)do / doParentaux[d,o] (nain, ovale)
Do (10.5 %)Do / doRecombiné[D,o] (géant, ovale)
dO (10.5 %)dO / doRecombiné[d,O] (nain, rond)
  • Bilan : parentaux ~ 79 % / recombinés ~ 21 % → liaison à 21 cM

2. Croisement pour maximiser tiges à poils (hh) et fruits ronds (O_)

Les deux gènes H/h (tiges poilues) et O/o (fruits rond·ovales) sont sur deux chromosomes différents ⇒ ségrégation indépendante.
Pour obtenir le plus grand pourcentage de plantes hh (poilues) et O_ (rondes), il suffit de croiser deux parents homozygotes pour chacun des allèles recherchés, par exemple :

Parent A : hh  OO×Parent B : hh  OO\boxed{ \text{Parent A : }hh\;OO \quad\times\quad \text{Parent B : }hh\;OO }
  • Génotypes connus (hh donne à coup sûr tiges poilues, OO donne à coup sûr fruits ronds).
  • Tous les descendants seront hh OO ⇒ 100 % tiges à poils et fruits ronds.

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